当前位置：文江博客话题详情

如何加入（合并）数据帧（内部，外部，左，右）

发布于 2025-01-27 05:59:15 字数 1383 浏览 3 评论 0 原文

给定两个数据帧：

df1 = data.frame(CustomerId = c(1:6), Product = c(rep("Toaster", 3), rep("Radio", 3)))
df2 = data.frame(CustomerId = c(2, 4, 6), State = c(rep("Alabama", 2), rep("Ohio", 1)))

df1
#  CustomerId Product
#           1 Toaster
#           2 Toaster
#           3 Toaster
#           4   Radio
#           5   Radio
#           6   Radio

df2
#  CustomerId   State
#           2 Alabama
#           4 Alabama
#           6    Ohio

我该如何执行数据库样式，即>？也就是说，我如何得到：

noreferrer“> ninen> innin> ninen join of df1 和 df2 ：仅返回左表在右表中具有匹配键的行。
an ofter Join > df2 ：
从两个表中返回所有行，从右表中具有匹配键的左侧连接记录。
a 左outer（或简单左JOIN）和 df2
从左表返回所有行，以及从右表中匹配键的任何行。
a df1 df1 和和 df2
从右表返回所有行，并从左表中返回带有匹配键的任何行。

额外的信用：

我如何执行SQL样式选择语句？

原文

Given two data frames:

df1 = data.frame(CustomerId = c(1:6), Product = c(rep("Toaster", 3), rep("Radio", 3)))
df2 = data.frame(CustomerId = c(2, 4, 6), State = c(rep("Alabama", 2), rep("Ohio", 1)))

df1
#  CustomerId Product
#           1 Toaster
#           2 Toaster
#           3 Toaster
#           4   Radio
#           5   Radio
#           6   Radio

df2
#  CustomerId   State
#           2 Alabama
#           4 Alabama
#           6    Ohio

How can I do database style, i.e., sql style, joins? That is, how do I get:

An inner join of df1 and df2:
Return only the rows in which the left table have matching keys in the right table.
An outer join of df1 and df2:
Returns all rows from both tables, join records from the left which have matching keys in the right table.
A left outer join (or simply left join) of df1 and df2
Return all rows from the left table, and any rows with matching keys from the right table.
A right outer join of df1 and df2
Return all rows from the right table, and any rows with matching keys from the left table.

Extra credit:

How can I do a SQL style select statement?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

謸气贵蔟 2025-02-03 05:59:15

通过使用 MERGE 函数及其可选参数：

内在加入： Merge（DF1，DF2）将适用于这些示例是因为r自动通过通用变量名称加入帧，但是您很可能需要指定 MERGE（DF1，DF2，by =“ CustomerId”）以确保您仅在您想要的字段。如果匹配变量在不同的数据帧中具有不同的名称，则还可以使用 by.x 和 by.y 参数。

外部加入： 合并（x = df1，y = df2，by by =“ customerId”，all = true）

左外部： 合并（x = df1，y = df2，by =“ customerid”，all.x = true）

右外部：< /em> 合并（x = df1，y = df2，by =“ customerId”，all.y = true）

交叉加入： < /strong> 合并（x = df1，y = df2，by = null）

~~就像内在的加入一样，您可能希望将“ customerId”明确地传递给r作为匹配变量。如果输入数据。帧会意外变化，稍后更易于阅读，则更安全。~~

您可以通过提供 vector，例如， by = c（“ customerId”，“ orderId”）来合并多个列。

如果要合并的列名不相同，则可以指定，例如， by.x =“ customerid_in_df1”，by.y =“ customerid_in_df2” 其中 customeid_in_df1 is第一个数据框架中的列的名称和 customerid_in_df2 是第二个数据框中的列的名称。（如果您需要在多个列上合并，这些也可以是向量。）

回复收藏 0 原文

灰色世界里的红玫瑰 2025-02-03 05:59:15

您也可以使用Hadley Wickham的Awesome dplyr 包装。

library(dplyr)

#make sure that CustomerId cols are both the same type
#they aren’t in the provided data (one is integer and one is double)
df1$CustomerId <- as.double(df1$CustomerId)

突变连接：使用DF2

#inner
inner_join(df1, df2)

#left outer
left_join(df1, df2)

#right outer
right_join(df1, df2)

#alternate right outer
left_join(df2, df1)

#full join
full_join(df1, df2)

过滤连接中的匹配点在DF1中添加列：DF1中的行过滤器，请勿修改列

#keep only observations in df1 that match in df2.
semi_join(df1, df2)

#drop all observations in df1 that match in df2.
anti_join(df1, df2)

You can do joins as well using Hadley Wickham's awesome dplyr package.

library(dplyr)

#make sure that CustomerId cols are both the same type
#they aren’t in the provided data (one is integer and one is double)
df1$CustomerId <- as.double(df1$CustomerId)

Mutating joins: add columns to df1 using matches in df2

#inner
inner_join(df1, df2)

#left outer
left_join(df1, df2)

#right outer
right_join(df1, df2)

#alternate right outer
left_join(df2, df1)

#full join
full_join(df1, df2)

Filtering joins: filter out rows in df1, don't modify columns

#keep only observations in df1 that match in df2.
semi_join(df1, df2)

#drop all observations in df1 that match in df2.
anti_join(df1, df2)

回复收藏 0 原文

若相惜即相离 2025-02-03 05:59:15

我建议您查看 gabor grothendieck sqldf package 在SQL中表达这些操作。

library(sqldf)

## inner join
df3 <- sqldf("SELECT CustomerId, Product, State 
              FROM df1
              JOIN df2 USING(CustomerID)")

## left join (substitute 'right' for right join)
df4 <- sqldf("SELECT CustomerId, Product, State 
              FROM df1
              LEFT JOIN df2 USING(CustomerID)")

我发现SQL语法比其等效物更简单，更自然（但这可能仅反映了我的RDBMS偏置）。

参见 Gabor的SQLDF GitHub 有关加入的更多信息。

I would recommend checking out Gabor Grothendieck's sqldf package, which allows you to express these operations in SQL.

library(sqldf)

## inner join
df3 <- sqldf("SELECT CustomerId, Product, State 
              FROM df1
              JOIN df2 USING(CustomerID)")

## left join (substitute 'right' for right join)
df4 <- sqldf("SELECT CustomerId, Product, State 
              FROM df1
              LEFT JOIN df2 USING(CustomerID)")

I find the SQL syntax to be simpler and more natural than its R equivalent (but this may just reflect my RDBMS bias).

See Gabor's sqldf GitHub for more information on joins.

回复收藏 0 原文

少跟Wǒ拽 2025-02-03 05:59:15

有 data.table 内在联接的方法，这是非常时间和内存的效率（对于某些较大的数据。Frames）：

library(data.table)
  
dt1 <- data.table(df1, key = "CustomerId") 
dt2 <- data.table(df2, key = "CustomerId")

joined.dt1.dt.2 <- dt1[dt2]

MERGE 也适用于数据。表（因为它是通用的，并且调用 Merge.data.table ）

merge(dt1, dt2)

data.table在Stackoverflow上记录了：
如何进行数据。 br>
翻译SQL与外国钥匙上的sql连接到r数据。表语法
 有效合并的较大数据合并的替代方案。
中

，还有另一个选项是 join 函数在 plyr package。 [注：2022年的注：Plyr现在已退休，已被 dplyr 。 ”中的加入操作。

library(plyr)

join(df1, df2,
     type = "inner")

#   CustomerId Product   State
# 1          2 Toaster Alabama
# 2          4   Radio Alabama
# 3          6   Radio    Ohio

href =“ href = 内部，<代码>左，<代码>右，完整。

来自？JOIN ：与 Merge 不同，[ join> join ]保留X的顺序，无论使用哪种JOIN类型。

There is the data.table approach for an inner join, which is very time and memory efficient (and necessary for some larger data.frames):

library(data.table)
  
dt1 <- data.table(df1, key = "CustomerId") 
dt2 <- data.table(df2, key = "CustomerId")

joined.dt1.dt.2 <- dt1[dt2]

merge also works on data.tables (as it is generic and calls merge.data.table)

merge(dt1, dt2)

data.table documented on stackoverflow:
How to do a data.table merge operation
Translating SQL joins on foreign keys to R data.table syntax
Efficient alternatives to merge for larger data.frames R
How to do a basic left outer join with data.table in R?

Yet another option is the join function found in the plyr package. [Note from 2022: plyr is now retired and has been superseded by dplyr. Join operations in dplyr are described in this answer.]

library(plyr)

join(df1, df2,
     type = "inner")

#   CustomerId Product   State
# 1          2 Toaster Alabama
# 2          4   Radio Alabama
# 3          6   Radio    Ohio

Options for type: inner, left, right, full.

From ?join: Unlike merge, [join] preserves the order of x no matter what join type is used.

回复收藏 0 原文

清欢 2025-02-03 05:59:15

在 R Wiki 。我会在这里偷几对夫妇：

合并方法，

因为您的键被命名为同样的简短方法，即在内部加入的短途方法是merge（）：

merge(df1, df2)

完整的内部联接（所有表中的所有记录）可以是用“ ALL”关键字创建：

merge(df1, df2, all=TRUE)

DF1和DF2的左外连接：

merge(df1, df2, all.x=TRUE)

DF1和DF2的右外连接：

merge(df1, df2, all.y=TRUE)

您可以翻转'em，slap'em and ub'em down以获取其他两个外部连接:)

下标方法

使用下标方法在左侧的DF1左外连接将是：

df1[,"State"]<-df2[df1[ ,"Product"], "State"]

外部连接的其他组合可以通过修改左外的联接下标示例来创建。（是的，我知道这相当于说：“我将其作为读者的练习……”）

There are some good examples of doing this over at the R Wiki. I'll steal a couple here:

Merge Method

Since your keys are named the same the short way to do an inner join is merge():

merge(df1, df2)

a full inner join (all records from both tables) can be created with the "all" keyword:

merge(df1, df2, all=TRUE)

a left outer join of df1 and df2:

merge(df1, df2, all.x=TRUE)

a right outer join of df1 and df2:

merge(df1, df2, all.y=TRUE)

you can flip 'em, slap 'em and rub 'em down to get the other two outer joins you asked about :)

Subscript Method

A left outer join with df1 on the left using a subscript method would be:

df1[,"State"]<-df2[df1[ ,"Product"], "State"]

The other combination of outer joins can be created by mungling the left outer join subscript example. (yeah, I know that's the equivalent of saying "I'll leave it as an exercise for the reader...")

回复收藏 0 原文

故事还在继续 2025-02-03 05:59:15

更新数据。加入数据集的方法。请参阅下面的每种联接类型的示例。有两种方法，一种来自 [。data.table 传递第二个数据时。将作为第一个参数作为子集，另一种方法是使用 Merge 函数，该功能可发送到快速Data.Table方法。

df1 = data.frame(CustomerId = c(1:6), Product = c(rep("Toaster", 3), rep("Radio", 3)))
df2 = data.frame(CustomerId = c(2L, 4L, 7L), State = c(rep("Alabama", 2), rep("Ohio", 1))) # one value changed to show full outer join

library(data.table)

dt1 = as.data.table(df1)
dt2 = as.data.table(df2)
setkey(dt1, CustomerId)
setkey(dt2, CustomerId)
# right outer join keyed data.tables
dt1[dt2]

setkey(dt1, NULL)
setkey(dt2, NULL)
# right outer join unkeyed data.tables - use `on` argument
dt1[dt2, on = "CustomerId"]

# left outer join - swap dt1 with dt2
dt2[dt1, on = "CustomerId"]

# inner join - use `nomatch` argument
dt1[dt2, nomatch=NULL, on = "CustomerId"]

# anti join - use `!` operator
dt1[!dt2, on = "CustomerId"]

# inner join - using merge method
merge(dt1, dt2, by = "CustomerId")

# full outer join
merge(dt1, dt2, by = "CustomerId", all = TRUE)

# see ?merge.data.table arguments for other cases

在基准测试基础r，sqldf，dplyr和data.table.
下方。
基准测试无关/未索引数据集。
基准测试是在50m-1行数据集上执行的，在JOIN列上有50m-2的共同值，因此可以测试每个方案（内部，左，右，完整），并且连接仍然不重要。这是JOIN的类型，其压力可以加入算法。时间与 sqldf：0.4.11 ， dplyr：0.7.8 ， data.table：1.12.0 。

# inner
Unit: seconds
   expr       min        lq      mean    median        uq       max neval
   base 111.66266 111.66266 111.66266 111.66266 111.66266 111.66266     1
  sqldf 624.88388 624.88388 624.88388 624.88388 624.88388 624.88388     1
  dplyr  51.91233  51.91233  51.91233  51.91233  51.91233  51.91233     1
     DT  10.40552  10.40552  10.40552  10.40552  10.40552  10.40552     1
# left
Unit: seconds
   expr        min         lq       mean     median         uq        max 
   base 142.782030 142.782030 142.782030 142.782030 142.782030 142.782030     
  sqldf 613.917109 613.917109 613.917109 613.917109 613.917109 613.917109     
  dplyr  49.711912  49.711912  49.711912  49.711912  49.711912  49.711912     
     DT   9.674348   9.674348   9.674348   9.674348   9.674348   9.674348       
# right
Unit: seconds
   expr        min         lq       mean     median         uq        max
   base 122.366301 122.366301 122.366301 122.366301 122.366301 122.366301     
  sqldf 611.119157 611.119157 611.119157 611.119157 611.119157 611.119157     
  dplyr  50.384841  50.384841  50.384841  50.384841  50.384841  50.384841     
     DT   9.899145   9.899145   9.899145   9.899145   9.899145   9.899145     
# full
Unit: seconds
  expr       min        lq      mean    median        uq       max neval
  base 141.79464 141.79464 141.79464 141.79464 141.79464 141.79464     1
 dplyr  94.66436  94.66436  94.66436  94.66436  94.66436  94.66436     1
    DT  21.62573  21.62573  21.62573  21.62573  21.62573  21.62573     1

请注意，还有其他类型的连接可以使用 data.table ：
- 在JOIN 上更新 - 如果您想从另一个表查找到主表

- 在JOIN上汇总>
- 重叠加入JOIN - 如果您想通过ranges
合并
- 滚动JOIN JOIN - 如果您希望合并能够通过向前或向后滚动或向后滚动来与preading/laster的值匹配
- non-equi join - 如果您的联接条件是非平等

代码的复制：

library(microbenchmark)
library(sqldf)
library(dplyr)
library(data.table)
sapply(c("sqldf","dplyr","data.table"), packageVersion, simplify=FALSE)

n = 5e7
set.seed(108)
df1 = data.frame(x=sample(n,n-1L), y1=rnorm(n-1L))
df2 = data.frame(x=sample(n,n-1L), y2=rnorm(n-1L))
dt1 = as.data.table(df1)
dt2 = as.data.table(df2)

mb = list()
# inner join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x"),
               sqldf = sqldf("SELECT * FROM df1 INNER JOIN df2 ON df1.x = df2.x"),
               dplyr = inner_join(df1, df2, by = "x"),
               DT = dt1[dt2, nomatch=NULL, on = "x"]) -> mb$inner

# left outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all.x = TRUE),
               sqldf = sqldf("SELECT * FROM df1 LEFT OUTER JOIN df2 ON df1.x = df2.x"),
               dplyr = left_join(df1, df2, by = c("x"="x")),
               DT = dt2[dt1, on = "x"]) -> mb$left

# right outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all.y = TRUE),
               sqldf = sqldf("SELECT * FROM df2 LEFT OUTER JOIN df1 ON df2.x = df1.x"),
               dplyr = right_join(df1, df2, by = "x"),
               DT = dt1[dt2, on = "x"]) -> mb$right

# full outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all = TRUE),
               dplyr = full_join(df1, df2, by = "x"),
               DT = merge(dt1, dt2, by = "x", all = TRUE)) -> mb$full

lapply(mb, print) -> nul

Update on data.table methods for joining datasets. See below examples for each type of join. There are two methods, one from [.data.table when passing second data.table as the first argument to subset, another way is to use merge function which dispatches to fast data.table method.

df1 = data.frame(CustomerId = c(1:6), Product = c(rep("Toaster", 3), rep("Radio", 3)))
df2 = data.frame(CustomerId = c(2L, 4L, 7L), State = c(rep("Alabama", 2), rep("Ohio", 1))) # one value changed to show full outer join

library(data.table)

dt1 = as.data.table(df1)
dt2 = as.data.table(df2)
setkey(dt1, CustomerId)
setkey(dt2, CustomerId)
# right outer join keyed data.tables
dt1[dt2]

setkey(dt1, NULL)
setkey(dt2, NULL)
# right outer join unkeyed data.tables - use `on` argument
dt1[dt2, on = "CustomerId"]

# left outer join - swap dt1 with dt2
dt2[dt1, on = "CustomerId"]

# inner join - use `nomatch` argument
dt1[dt2, nomatch=NULL, on = "CustomerId"]

# anti join - use `!` operator
dt1[!dt2, on = "CustomerId"]

# inner join - using merge method
merge(dt1, dt2, by = "CustomerId")

# full outer join
merge(dt1, dt2, by = "CustomerId", all = TRUE)

# see ?merge.data.table arguments for other cases

Below benchmark tests base R, sqldf, dplyr and data.table.
Benchmark tests unkeyed/unindexed datasets.
Benchmark is performed on 50M-1 rows datasets, there are 50M-2 common values on join column so each scenario (inner, left, right, full) can be tested and join is still not trivial to perform. It is type of join which well stress join algorithms. Timings are as of sqldf:0.4.11, dplyr:0.7.8, data.table:1.12.0.

# inner
Unit: seconds
   expr       min        lq      mean    median        uq       max neval
   base 111.66266 111.66266 111.66266 111.66266 111.66266 111.66266     1
  sqldf 624.88388 624.88388 624.88388 624.88388 624.88388 624.88388     1
  dplyr  51.91233  51.91233  51.91233  51.91233  51.91233  51.91233     1
     DT  10.40552  10.40552  10.40552  10.40552  10.40552  10.40552     1
# left
Unit: seconds
   expr        min         lq       mean     median         uq        max 
   base 142.782030 142.782030 142.782030 142.782030 142.782030 142.782030     
  sqldf 613.917109 613.917109 613.917109 613.917109 613.917109 613.917109     
  dplyr  49.711912  49.711912  49.711912  49.711912  49.711912  49.711912     
     DT   9.674348   9.674348   9.674348   9.674348   9.674348   9.674348       
# right
Unit: seconds
   expr        min         lq       mean     median         uq        max
   base 122.366301 122.366301 122.366301 122.366301 122.366301 122.366301     
  sqldf 611.119157 611.119157 611.119157 611.119157 611.119157 611.119157     
  dplyr  50.384841  50.384841  50.384841  50.384841  50.384841  50.384841     
     DT   9.899145   9.899145   9.899145   9.899145   9.899145   9.899145     
# full
Unit: seconds
  expr       min        lq      mean    median        uq       max neval
  base 141.79464 141.79464 141.79464 141.79464 141.79464 141.79464     1
 dplyr  94.66436  94.66436  94.66436  94.66436  94.66436  94.66436     1
    DT  21.62573  21.62573  21.62573  21.62573  21.62573  21.62573     1

Be aware there are other types of joins you can perform using data.table:
- update on join - if you want to lookup values from another table to your main table
- aggregate on join - if you want to aggregate on key you are joining you do not have to materialize all join results
- overlapping join - if you want to merge by ranges
- rolling join - if you want merge to be able to match to values from preceeding/following rows by rolling them forward or backward
- non-equi join - if your join condition is non-equal

Code to reproduce:

library(microbenchmark)
library(sqldf)
library(dplyr)
library(data.table)
sapply(c("sqldf","dplyr","data.table"), packageVersion, simplify=FALSE)

n = 5e7
set.seed(108)
df1 = data.frame(x=sample(n,n-1L), y1=rnorm(n-1L))
df2 = data.frame(x=sample(n,n-1L), y2=rnorm(n-1L))
dt1 = as.data.table(df1)
dt2 = as.data.table(df2)

mb = list()
# inner join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x"),
               sqldf = sqldf("SELECT * FROM df1 INNER JOIN df2 ON df1.x = df2.x"),
               dplyr = inner_join(df1, df2, by = "x"),
               DT = dt1[dt2, nomatch=NULL, on = "x"]) -> mb$inner

# left outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all.x = TRUE),
               sqldf = sqldf("SELECT * FROM df1 LEFT OUTER JOIN df2 ON df1.x = df2.x"),
               dplyr = left_join(df1, df2, by = c("x"="x")),
               DT = dt2[dt1, on = "x"]) -> mb$left

# right outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all.y = TRUE),
               sqldf = sqldf("SELECT * FROM df2 LEFT OUTER JOIN df1 ON df2.x = df1.x"),
               dplyr = right_join(df1, df2, by = "x"),
               DT = dt1[dt2, on = "x"]) -> mb$right

# full outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all = TRUE),
               dplyr = full_join(df1, df2, by = "x"),
               DT = merge(dt1, dt2, by = "x", all = TRUE)) -> mb$full

lapply(mb, print) -> nul

回复收藏 0 原文

哀由 2025-02-03 05:59:15

2014年的新事物：

尤其是如果您对一般数据操作感兴趣（包括分类，过滤，子集，摘要等），则绝对应该看一下 dplyr ，它带有一个多样性功能的所有旨在专门针对数据框架和某些其他数据库类型的工作。甚至提供了相当复杂的SQL接口，甚至可以将SQL代码直接转换为R的函数

R。

它 = null，copy = false，...）：从
x y中有匹配值，以及x和y
left_join的所有列（x，y，by = null，copy = false，...）：返回x的所有行，以及所有行x和y semi_join的列
（x，y，by = null，copy = false，...）：从x中返回所有行，其中有匹配值
是的，仅保留x的列。
anti_join（x，y，by = null，copy = false，...）：从x返回所有行
在y中没有匹配值的情况下，仅保留x的列

，全部在这里非常详细。

选择列可以通过选择（DF，“列”）完成。如果这对您来说还不够SQL，则有 sql（）函数，您可以在其中输入SQL Code AS-IS，并且它将执行您指定的操作，就像您在一直以来（有关更多信息，请参阅 dplyr/databases小插图）。例如，如果正确应用，则 sql（“从Hflights中选择 *”）将从“ Hflights” dplyr Table（a“ tbl”）中选择所有列。

回复收藏 0 原文

那小子欠揍 2025-02-03 05:59:15

dplyr自0.4以来实施了所有这些加入，包括 exouts_join ，但值得注意的是，在0.4之前的前几个发行版中，它使用不提供 offer_join ，并且AS结果，有很多非常糟糕的黑客解决方法用户代码在相当长的一段时间以后（您仍然可以在SO中找到此类代码，Kaggle答案，GitHub，github。因此，此答案仍然有用。）

与join相关的发行亮点：

v0.5 （6/2016）

处理Posixct类型，时区，重复，不同因子水平。更好的错误和警告。
新的后缀论证控制哪个后缀重复变量名称接收（＃1296）

v0.4.0 （1/2015）

实现正确的加入和外部加入（＃96）
突变加入，从匹配另一个行中的一行中，将新变量添加到一个表中。过滤连接，该连接是根据它们是否匹配另一个表中的观察值的，从一个表中进行了过滤观察。

v0.3 < /strong>（10/2014）

现在可以在每个表中通过不同的变量来左右：df1％＆gt;％left_join（df2，c（“ var1” =“ var2”））

v0.2 （5/2014） a>

*_join（）不再重写列名（＃324）

v0.1.3 （4/2014）

具有 inner_join，left_join，semi_join，semi_join，anti_join
outer_join 尚未实现，后备为使用base :: merge（）（或plyr :: join（））
尚未实施right_join和 extran_join
hadley在这里提到其他优势
目前的一个次要功能合并的是，dplyr不是像python pandas一样具有分开的x，by.y列的能力。

每个哈德利在该问题上的评论解决方法：

right_join （x，y）与left_join（y，x）相同，就行而言，只有列将是不同的订单。轻松使用SELECT（new_column_order）
outer_join 基本上是Unim（left_join（x，y），right_join（x，y）） - IE保留两个数据帧中的所有行。

回复收藏 0 原文

趁年轻赶紧闹 2025-02-03 05:59:15

对于左将 0 ..*：0..1 基数或右连接的情况可以直接从木匠（ 0..1 表）将单方面列将其放置在joinee（ 0 ..*表）上，从而将避免创建全新的数据表。这就需要将joinee的密钥列匹配到木匠中，并索引+订购木匠的行以进行分配。

如果键是单列，那么我们可以使用一个调用 match（） 进行匹配。我将在此答案中介绍这种情况。

这是一个基于OP的示例，除了我在 df2 中添加了一个额外的行，并具有7个ID，以测试Joiner中未匹配键的情况。这是有效的 df1 左JOIN df2 ：

df1 <- data.frame(CustomerId=1:6,Product=c(rep('Toaster',3L),rep('Radio',3L)));
df2 <- data.frame(CustomerId=c(2L,4L,6L,7L),State=c(rep('Alabama',2L),'Ohio','Texas'));
df1[names(df2)[-1L]] <- df2[match(df1[,1L],df2[,1L]),-1L];
df1;
##   CustomerId Product   State
## 1          1 Toaster    <NA>
## 2          2 Toaster Alabama
## 3          3 Toaster    <NA>
## 4          4   Radio Alabama
## 5          5   Radio    <NA>
## 6          6   Radio    Ohio

在上面的我硬编码一个假设，即关键列是两个输入表的第一列。我认为，通常，这不是一个不合理的假设，因为，如果您有一个数据。使用键列的框架，如果未设置为Data.frame的第一列，那将很奇怪。一开始。您总是可以重新排序列以使其这样做。该假设的一个有利结果是，尽管我认为它只是用另一个假设替换一个假设，但密钥列的名称不必被硬编码。简洁是整数索引和速度的另一个优点。在下面的基准测试中，我将更改实施方式，以使用字符串名称索引来匹配竞争实现。

我认为，如果您有几张桌子要在一张大桌子上加入，这是一个特别合适的解决方案。反复重建每个合并的整个桌子将是不必要和效率低下的。

另一方面，如果出于任何原因需要大约在此操作中保持不变，则无法使用该解决方案，因为它会直接修改折磨。尽管在这种情况下，您可以简单地制作副本并在副本上执行就地分配。

附带说明，我简要研究了多列键的可能匹配解决方案。不幸的是，我发现的唯一匹配解决方案是：

效率低下的串联。例如匹配（互动（DF1 $ a，df1 $ b），交互（DF2 $ a，df2 $ b））或与 paste（）的相同想法。
tartesian连词效率低下，例如外部（df1 $ a，df2 $ a，`==`）＆amp;外部（df1 $ b，df2 $ b，`==`）。
基本r merge（）和基于等效的包装合并功能，该功能总是分配新表以返回合并结果，因此不适合基于原位分配的解决方案。

例如，请参见不同数据帧上的列，结果获得其他列，匹配两个带有另外两个列的列， /a>，以及这个问题的骗子，我最初提出了现场解决方案，在r 中将两个数据帧与不同数量的行相结合。

基准测试

我决定进行自己的基准测试，以查看本地分配方法与此问题中提供的其他解决方案的比较。

测试代码：

library(microbenchmark);
library(data.table);
library(sqldf);
library(plyr);
library(dplyr);

solSpecs <- list(
    merge=list(testFuncs=list(
        inner=function(df1,df2,key) merge(df1,df2,key),
        left =function(df1,df2,key) merge(df1,df2,key,all.x=T),
        right=function(df1,df2,key) merge(df1,df2,key,all.y=T),
        full =function(df1,df2,key) merge(df1,df2,key,all=T)
    )),
    data.table.unkeyed=list(argSpec='data.table.unkeyed',testFuncs=list(
        inner=function(dt1,dt2,key) dt1[dt2,on=key,nomatch=0L,allow.cartesian=T],
        left =function(dt1,dt2,key) dt2[dt1,on=key,allow.cartesian=T],
        right=function(dt1,dt2,key) dt1[dt2,on=key,allow.cartesian=T],
        full =function(dt1,dt2,key) merge(dt1,dt2,key,all=T,allow.cartesian=T) ## calls merge.data.table()
    )),
    data.table.keyed=list(argSpec='data.table.keyed',testFuncs=list(
        inner=function(dt1,dt2) dt1[dt2,nomatch=0L,allow.cartesian=T],
        left =function(dt1,dt2) dt2[dt1,allow.cartesian=T],
        right=function(dt1,dt2) dt1[dt2,allow.cartesian=T],
        full =function(dt1,dt2) merge(dt1,dt2,all=T,allow.cartesian=T) ## calls merge.data.table()
    )),
    sqldf.unindexed=list(testFuncs=list( ## note: must pass connection=NULL to avoid running against the live DB connection, which would result in collisions with the residual tables from the last query upload
        inner=function(df1,df2,key) sqldf(paste0('select * from df1 inner join df2 using(',paste(collapse=',',key),')'),connection=NULL),
        left =function(df1,df2,key) sqldf(paste0('select * from df1 left join df2 using(',paste(collapse=',',key),')'),connection=NULL),
        right=function(df1,df2,key) sqldf(paste0('select * from df2 left join df1 using(',paste(collapse=',',key),')'),connection=NULL) ## can't do right join proper, not yet supported; inverted left join is equivalent
        ##full =function(df1,df2,key) sqldf(paste0('select * from df1 full join df2 using(',paste(collapse=',',key),')'),connection=NULL) ## can't do full join proper, not yet supported; possible to hack it with a union of left joins, but too unreasonable to include in testing
    )),
    sqldf.indexed=list(testFuncs=list( ## important: requires an active DB connection with preindexed main.df1 and main.df2 ready to go; arguments are actually ignored
        inner=function(df1,df2,key) sqldf(paste0('select * from main.df1 inner join main.df2 using(',paste(collapse=',',key),')')),
        left =function(df1,df2,key) sqldf(paste0('select * from main.df1 left join main.df2 using(',paste(collapse=',',key),')')),
        right=function(df1,df2,key) sqldf(paste0('select * from main.df2 left join main.df1 using(',paste(collapse=',',key),')')) ## can't do right join proper, not yet supported; inverted left join is equivalent
        ##full =function(df1,df2,key) sqldf(paste0('select * from main.df1 full join main.df2 using(',paste(collapse=',',key),')')) ## can't do full join proper, not yet supported; possible to hack it with a union of left joins, but too unreasonable to include in testing
    )),
    plyr=list(testFuncs=list(
        inner=function(df1,df2,key) join(df1,df2,key,'inner'),
        left =function(df1,df2,key) join(df1,df2,key,'left'),
        right=function(df1,df2,key) join(df1,df2,key,'right'),
        full =function(df1,df2,key) join(df1,df2,key,'full')
    )),
    dplyr=list(testFuncs=list(
        inner=function(df1,df2,key) inner_join(df1,df2,key),
        left =function(df1,df2,key) left_join(df1,df2,key),
        right=function(df1,df2,key) right_join(df1,df2,key),
        full =function(df1,df2,key) full_join(df1,df2,key)
    )),
    in.place=list(testFuncs=list(
        left =function(df1,df2,key) { cns <- setdiff(names(df2),key); df1[cns] <- df2[match(df1[,key],df2[,key]),cns]; df1; },
        right=function(df1,df2,key) { cns <- setdiff(names(df1),key); df2[cns] <- df1[match(df2[,key],df1[,key]),cns]; df2; }
    ))
);

getSolTypes <- function() names(solSpecs);
getJoinTypes <- function() unique(unlist(lapply(solSpecs,function(x) names(x$testFuncs))));
getArgSpec <- function(argSpecs,key=NULL) if (is.null(key)) argSpecs$default else argSpecs[[key]];

initSqldf <- function() {
    sqldf(); ## creates sqlite connection on first run, cleans up and closes existing connection otherwise
    if (exists('sqldfInitFlag',envir=globalenv(),inherits=F) && sqldfInitFlag) { ## false only on first run
        sqldf(); ## creates a new connection
    } else {
        assign('sqldfInitFlag',T,envir=globalenv()); ## set to true for the one and only time
    }; ## end if
    invisible();
}; ## end initSqldf()

setUpBenchmarkCall <- function(argSpecs,joinType,solTypes=getSolTypes(),env=parent.frame()) {
    ## builds and returns a list of expressions suitable for passing to the list argument of microbenchmark(), and assigns variables to resolve symbol references in those expressions
    callExpressions <- list();
    nms <- character();
    for (solType in solTypes) {
        testFunc <- solSpecs[[solType]]$testFuncs[[joinType]];
        if (is.null(testFunc)) next; ## this join type is not defined for this solution type
        testFuncName <- paste0('tf.',solType);
        assign(testFuncName,testFunc,envir=env);
        argSpecKey <- solSpecs[[solType]]$argSpec;
        argSpec <- getArgSpec(argSpecs,argSpecKey);
        argList <- setNames(nm=names(argSpec$args),vector('list',length(argSpec$args)));
        for (i in seq_along(argSpec$args)) {
            argName <- paste0('tfa.',argSpecKey,i);
            assign(argName,argSpec$args[[i]],envir=env);
            argList[[i]] <- if (i%in%argSpec$copySpec) call('copy',as.symbol(argName)) else as.symbol(argName);
        }; ## end for
        callExpressions[[length(callExpressions)+1L]] <- do.call(call,c(list(testFuncName),argList),quote=T);
        nms[length(nms)+1L] <- solType;
    }; ## end for
    names(callExpressions) <- nms;
    callExpressions;
}; ## end setUpBenchmarkCall()

harmonize <- function(res) {
    res <- as.data.frame(res); ## coerce to data.frame
    for (ci in which(sapply(res,is.factor))) res[[ci]] <- as.character(res[[ci]]); ## coerce factor columns to character
    for (ci in which(sapply(res,is.logical))) res[[ci]] <- as.integer(res[[ci]]); ## coerce logical columns to integer (works around sqldf quirk of munging logicals to integers)
    ##for (ci in which(sapply(res,inherits,'POSIXct'))) res[[ci]] <- as.double(res[[ci]]); ## coerce POSIXct columns to double (works around sqldf quirk of losing POSIXct class) ----- POSIXct doesn't work at all in sqldf.indexed
    res <- res[order(names(res))]; ## order columns
    res <- res[do.call(order,res),]; ## order rows
    res;
}; ## end harmonize()

checkIdentical <- function(argSpecs,solTypes=getSolTypes()) {
    for (joinType in getJoinTypes()) {
        callExpressions <- setUpBenchmarkCall(argSpecs,joinType,solTypes);
        if (length(callExpressions)<2L) next;
        ex <- harmonize(eval(callExpressions[[1L]]));
        for (i in seq(2L,len=length(callExpressions)-1L)) {
            y <- harmonize(eval(callExpressions[[i]]));
            if (!isTRUE(all.equal(ex,y,check.attributes=F))) {
                ex <<- ex;
                y <<- y;
                solType <- names(callExpressions)[i];
                stop(paste0('non-identical: ',solType,' ',joinType,'.'));
            }; ## end if
        }; ## end for
    }; ## end for
    invisible();
}; ## end checkIdentical()

testJoinType <- function(argSpecs,joinType,solTypes=getSolTypes(),metric=NULL,times=100L) {
    callExpressions <- setUpBenchmarkCall(argSpecs,joinType,solTypes);
    bm <- microbenchmark(list=callExpressions,times=times);
    if (is.null(metric)) return(bm);
    bm <- summary(bm);
    res <- setNames(nm=names(callExpressions),bm[[metric]]);
    attr(res,'unit') <- attr(bm,'unit');
    res;
}; ## end testJoinType()

testAllJoinTypes <- function(argSpecs,solTypes=getSolTypes(),metric=NULL,times=100L) {
    joinTypes <- getJoinTypes();
    resList <- setNames(nm=joinTypes,lapply(joinTypes,function(joinType) testJoinType(argSpecs,joinType,solTypes,metric,times)));
    if (is.null(metric)) return(resList);
    units <- unname(unlist(lapply(resList,attr,'unit')));
    res <- do.call(data.frame,c(list(join=joinTypes),setNames(nm=solTypes,rep(list(rep(NA_real_,length(joinTypes))),length(solTypes))),list(unit=units,stringsAsFactors=F)));
    for (i in seq_along(resList)) res[i,match(names(resList[[i]]),names(res))] <- resList[[i]];
    res;
}; ## end testAllJoinTypes()

testGrid <- function(makeArgSpecsFunc,sizes,overlaps,solTypes=getSolTypes(),joinTypes=getJoinTypes(),metric='median',times=100L) {

    res <- expand.grid(size=sizes,overlap=overlaps,joinType=joinTypes,stringsAsFactors=F);
    res[solTypes] <- NA_real_;
    res$unit <- NA_character_;
    for (ri in seq_len(nrow(res))) {

        size <- res$size[ri];
        overlap <- res$overlap[ri];
        joinType <- res$joinType[ri];

        argSpecs <- makeArgSpecsFunc(size,overlap);

        checkIdentical(argSpecs,solTypes);

        cur <- testJoinType(argSpecs,joinType,solTypes,metric,times);
        res[ri,match(names(cur),names(res))] <- cur;
        res$unit[ri] <- attr(cur,'unit');

    }; ## end for

    res;

}; ## end testGrid()

这是我前面证明的OP的基准：

## OP's example, supplemented with a non-matching row in df2
argSpecs <- list(
    default=list(copySpec=1:2,args=list(
        df1 <- data.frame(CustomerId=1:6,Product=c(rep('Toaster',3L),rep('Radio',3L))),
        df2 <- data.frame(CustomerId=c(2L,4L,6L,7L),State=c(rep('Alabama',2L),'Ohio','Texas')),
        'CustomerId'
    )),
    data.table.unkeyed=list(copySpec=1:2,args=list(
        as.data.table(df1),
        as.data.table(df2),
        'CustomerId'
    )),
    data.table.keyed=list(copySpec=1:2,args=list(
        setkey(as.data.table(df1),CustomerId),
        setkey(as.data.table(df2),CustomerId)
    ))
);
## prepare sqldf
initSqldf();
sqldf('create index df1_key on df1(CustomerId);'); ## upload and create an sqlite index on df1
sqldf('create index df2_key on df2(CustomerId);'); ## upload and create an sqlite index on df2

checkIdentical(argSpecs);

testAllJoinTypes(argSpecs,metric='median');
##    join    merge data.table.unkeyed data.table.keyed sqldf.unindexed sqldf.indexed      plyr    dplyr in.place         unit
## 1 inner  644.259           861.9345          923.516        9157.752      1580.390  959.2250 270.9190       NA microseconds
## 2  left  713.539           888.0205          910.045        8820.334      1529.714  968.4195 270.9185 224.3045 microseconds
## 3 right 1221.804           909.1900          923.944        8930.668      1533.135 1063.7860 269.8495 218.1035 microseconds
## 4  full 1302.203          3107.5380         3184.729              NA            NA 1593.6475 270.7055       NA microseconds

在这里，我在随机输入数据上基准测试，尝试不同的量表和两个输入表之间的密钥重叠模式的不同模式。该基准仍仅限于单列整数键的情况。同样，为了确保就地解决方案适用于同一表的左右连接，所有随机测试数据均使用 0..1：0..1 cardinality。这是通过对第一个数据的密钥列进行采样来实现的。在生成第二个data.frame的密钥列时，框架。

makeArgSpecs.singleIntegerKey.optionalOneToOne <- function(size,overlap) {

    com <- as.integer(size*overlap);

    argSpecs <- list(
        default=list(copySpec=1:2,args=list(
            df1 <- data.frame(id=sample(size),y1=rnorm(size),y2=rnorm(size)),
            df2 <- data.frame(id=sample(c(if (com>0L) sample(df1$id,com) else integer(),seq(size+1L,len=size-com))),y3=rnorm(size),y4=rnorm(size)),
            'id'
        )),
        data.table.unkeyed=list(copySpec=1:2,args=list(
            as.data.table(df1),
            as.data.table(df2),
            'id'
        )),
        data.table.keyed=list(copySpec=1:2,args=list(
            setkey(as.data.table(df1),id),
            setkey(as.data.table(df2),id)
        ))
    );
    ## prepare sqldf
    initSqldf();
    sqldf('create index df1_key on df1(id);'); ## upload and create an sqlite index on df1
    sqldf('create index df2_key on df2(id);'); ## upload and create an sqlite index on df2

    argSpecs;

}; ## end makeArgSpecs.singleIntegerKey.optionalOneToOne()

## cross of various input sizes and key overlaps
sizes <- c(1e1L,1e3L,1e6L);
overlaps <- c(0.99,0.5,0.01);
system.time({ res <- testGrid(makeArgSpecs.singleIntegerKey.optionalOneToOne,sizes,overlaps); });
##     user   system  elapsed
## 22024.65 12308.63 34493.19

我编写了一些代码来创建上述结果的日志图。我为每个重叠百分比生成了一个单独的图。这有点混乱，但是我喜欢拥有所有解决方案类型，并在同一地块中表示加入类型。

我使用样条插值显示了每个解决方案/连接类型组合的平滑曲线，并用单个PCH符号绘制。联接类型由PCH符号捕获，使用一个点，用于左右的内部，左右角括号，以及一颗钻石。解决方案类型由传说中所示的颜色捕获。

plotRes <- function(res,titleFunc,useFloor=F) {
    solTypes <- setdiff(names(res),c('size','overlap','joinType','unit')); ## derive from res
    normMult <- c(microseconds=1e-3,milliseconds=1); ## normalize to milliseconds
    joinTypes <- getJoinTypes();
    cols <- c(merge='purple',data.table.unkeyed='blue',data.table.keyed='#00DDDD',sqldf.unindexed='brown',sqldf.indexed='orange',plyr='red',dplyr='#00BB00',in.place='magenta');
    pchs <- list(inner=20L,left='<',right='>',full=23L);
    cexs <- c(inner=0.7,left=1,right=1,full=0.7);
    NP <- 60L;
    ord <- order(decreasing=T,colMeans(res[res$size==max(res$size),solTypes],na.rm=T));
    ymajors <- data.frame(y=c(1,1e3),label=c('1ms','1s'),stringsAsFactors=F);
    for (overlap in unique(res$overlap)) {
        x1 <- res[res$overlap==overlap,];
        x1[solTypes] <- x1[solTypes]*normMult[x1$unit]; x1$unit <- NULL;
        xlim <- c(1e1,max(x1$size));
        xticks <- 10^seq(log10(xlim[1L]),log10(xlim[2L]));
        ylim <- c(1e-1,10^((if (useFloor) floor else ceiling)(log10(max(x1[solTypes],na.rm=T))))); ## use floor() to zoom in a little more, only sqldf.unindexed will break above, but xpd=NA will keep it visible
        yticks <- 10^seq(log10(ylim[1L]),log10(ylim[2L]));
        yticks.minor <- rep(yticks[-length(yticks)],each=9L)*1:9;
        plot(NA,xlim=xlim,ylim=ylim,xaxs='i',yaxs='i',axes=F,xlab='size (rows)',ylab='time (ms)',log='xy');
        abline(v=xticks,col='lightgrey');
        abline(h=yticks.minor,col='lightgrey',lty=3L);
        abline(h=yticks,col='lightgrey');
        axis(1L,xticks,parse(text=sprintf('10^%d',as.integer(log10(xticks)))));
        axis(2L,yticks,parse(text=sprintf('10^%d',as.integer(log10(yticks)))),las=1L);
        axis(4L,ymajors$y,ymajors$label,las=1L,tick=F,cex.axis=0.7,hadj=0.5);
        for (joinType in rev(joinTypes)) { ## reverse to draw full first, since it's larger and would be more obtrusive if drawn last
            x2 <- x1[x1$joinType==joinType,];
            for (solType in solTypes) {
                if (any(!is.na(x2[[solType]]))) {
                    xy <- spline(x2$size,x2[[solType]],xout=10^(seq(log10(x2$size[1L]),log10(x2$size[nrow(x2)]),len=NP)));
                    points(xy$x,xy$y,pch=pchs[[joinType]],col=cols[solType],cex=cexs[joinType],xpd=NA);
                }; ## end if
            }; ## end for
        }; ## end for
        ## custom legend
        ## due to logarithmic skew, must do all distance calcs in inches, and convert to user coords afterward
        ## the bottom-left corner of the legend will be defined in normalized figure coords, although we can convert to inches immediately
        leg.cex <- 0.7;
        leg.x.in <- grconvertX(0.275,'nfc','in');
        leg.y.in <- grconvertY(0.6,'nfc','in');
        leg.x.user <- grconvertX(leg.x.in,'in');
        leg.y.user <- grconvertY(leg.y.in,'in');
        leg.outpad.w.in <- 0.1;
        leg.outpad.h.in <- 0.1;
        leg.midpad.w.in <- 0.1;
        leg.midpad.h.in <- 0.1;
        leg.sol.w.in <- max(strwidth(solTypes,'in',leg.cex));
        leg.sol.h.in <- max(strheight(solTypes,'in',leg.cex))*1.5; ## multiplication factor for greater line height
        leg.join.w.in <- max(strheight(joinTypes,'in',leg.cex))*1.5; ## ditto
        leg.join.h.in <- max(strwidth(joinTypes,'in',leg.cex));
        leg.main.w.in <- leg.join.w.in*length(joinTypes);
        leg.main.h.in <- leg.sol.h.in*length(solTypes);
        leg.x2.user <- grconvertX(leg.x.in+leg.outpad.w.in*2+leg.main.w.in+leg.midpad.w.in+leg.sol.w.in,'in');
        leg.y2.user <- grconvertY(leg.y.in+leg.outpad.h.in*2+leg.main.h.in+leg.midpad.h.in+leg.join.h.in,'in');
        leg.cols.x.user <- grconvertX(leg.x.in+leg.outpad.w.in+leg.join.w.in*(0.5+seq(0L,length(joinTypes)-1L)),'in');
        leg.lines.y.user <- grconvertY(leg.y.in+leg.outpad.h.in+leg.main.h.in-leg.sol.h.in*(0.5+seq(0L,length(solTypes)-1L)),'in');
        leg.sol.x.user <- grconvertX(leg.x.in+leg.outpad.w.in+leg.main.w.in+leg.midpad.w.in,'in');
        leg.join.y.user <- grconvertY(leg.y.in+leg.outpad.h.in+leg.main.h.in+leg.midpad.h.in,'in');
        rect(leg.x.user,leg.y.user,leg.x2.user,leg.y2.user,col='white');
        text(leg.sol.x.user,leg.lines.y.user,solTypes[ord],cex=leg.cex,pos=4L,offset=0);
        text(leg.cols.x.user,leg.join.y.user,joinTypes,cex=leg.cex,pos=4L,offset=0,srt=90); ## srt rotation applies *after* pos/offset positioning
        for (i in seq_along(joinTypes)) {
            joinType <- joinTypes[i];
            points(rep(leg.cols.x.user[i],length(solTypes)),ifelse(colSums(!is.na(x1[x1$joinType==joinType,solTypes[ord]]))==0L,NA,leg.lines.y.user),pch=pchs[[joinType]],col=cols[solTypes[ord]]);
        }; ## end for
        title(titleFunc(overlap));
        readline(sprintf('overlap %.02f',overlap));
    }; ## end for
}; ## end plotRes()

titleFunc <- function(overlap) sprintf('R merge solutions: single-column integer key, 0..1:0..1 cardinality, %d%% overlap',as.integer(overlap*100));
plotRes(res,titleFunc,T);

这是第二个大型基准，在钥匙的数量和类型方面更为重型列以及基数。对于此基准测试，我使用三个关键列：一个字符，一个整数和一个逻辑，没有对基数的限制（即， 0 ..*：0 ..*：*）。（通常不建议通过浮点比较并发症定义具有双重或复杂值的密钥列，并且基本上没有人使用原始类型，更不用说键列了，因此我还没有在键中包含这些类型为了信息的缘故，我最初尝试通过包含Posixct键列使用四个密钥列，但是Posixct类型与 sqldf.indexed 解决方案效果不佳由于浮点比较异常，因此我将其删除。）

makeArgSpecs.assortedKey.optionalManyToMany <- function(size,overlap,uniquePct=75) {

    ## number of unique keys in df1
    u1Size <- as.integer(size*uniquePct/100);

    ## (roughly) divide u1Size into bases, so we can use expand.grid() to produce the required number of unique key values with repetitions within individual key columns
    ## use ceiling() to ensure we cover u1Size; will truncate afterward
    u1SizePerKeyColumn <- as.integer(ceiling(u1Size^(1/3)));

    ## generate the unique key values for df1
    keys1 <- expand.grid(stringsAsFactors=F,
        idCharacter=replicate(u1SizePerKeyColumn,paste(collapse='',sample(letters,sample(4:12,1L),T))),
        idInteger=sample(u1SizePerKeyColumn),
        idLogical=sample(c(F,T),u1SizePerKeyColumn,T)
        ##idPOSIXct=as.POSIXct('2016-01-01 00:00:00','UTC')+sample(u1SizePerKeyColumn)
    )[seq_len(u1Size),];

    ## rbind some repetitions of the unique keys; this will prepare one side of the many-to-many relationship
    ## also scramble the order afterward
    keys1 <- rbind(keys1,keys1[sample(nrow(keys1),size-u1Size,T),])[sample(size),];

    ## common and unilateral key counts
    com <- as.integer(size*overlap);
    uni <- size-com;

    ## generate some unilateral keys for df2 by synthesizing outside of the idInteger range of df1
    keys2 <- data.frame(stringsAsFactors=F,
        idCharacter=replicate(uni,paste(collapse='',sample(letters,sample(4:12,1L),T))),
        idInteger=u1SizePerKeyColumn+sample(uni),
        idLogical=sample(c(F,T),uni,T)
        ##idPOSIXct=as.POSIXct('2016-01-01 00:00:00','UTC')+u1SizePerKeyColumn+sample(uni)
    );

    ## rbind random keys from df1; this will complete the many-to-many relationship
    ## also scramble the order afterward
    keys2 <- rbind(keys2,keys1[sample(nrow(keys1),com,T),])[sample(size),];

    ##keyNames <- c('idCharacter','idInteger','idLogical','idPOSIXct');
    keyNames <- c('idCharacter','idInteger','idLogical');
    ## note: was going to use raw and complex type for two of the non-key columns, but data.table doesn't seem to fully support them
    argSpecs <- list(
        default=list(copySpec=1:2,args=list(
            df1 <- cbind(stringsAsFactors=F,keys1,y1=sample(c(F,T),size,T),y2=sample(size),y3=rnorm(size),y4=replicate(size,paste(collapse='',sample(letters,sample(4:12,1L),T)))),
            df2 <- cbind(stringsAsFactors=F,keys2,y5=sample(c(F,T),size,T),y6=sample(size),y7=rnorm(size),y8=replicate(size,paste(collapse='',sample(letters,sample(4:12,1L),T)))),
            keyNames
        )),
        data.table.unkeyed=list(copySpec=1:2,args=list(
            as.data.table(df1),
            as.data.table(df2),
            keyNames
        )),
        data.table.keyed=list(copySpec=1:2,args=list(
            setkeyv(as.data.table(df1),keyNames),
            setkeyv(as.data.table(df2),keyNames)
        ))
    );
    ## prepare sqldf
    initSqldf();
    sqldf(paste0('create index df1_key on df1(',paste(collapse=',',keyNames),');')); ## upload and create an sqlite index on df1
    sqldf(paste0('create index df2_key on df2(',paste(collapse=',',keyNames),');')); ## upload and create an sqlite index on df2

    argSpecs;

}; ## end makeArgSpecs.assortedKey.optionalManyToMany()

sizes <- c(1e1L,1e3L,1e5L); ## 1e5L instead of 1e6L to respect more heavy-duty inputs
overlaps <- c(0.99,0.5,0.01);
solTypes <- setdiff(getSolTypes(),'in.place');
system.time({ res <- testGrid(makeArgSpecs.assortedKey.optionalManyToMany,sizes,overlaps,solTypes); });
##     user   system  elapsed
## 38895.50   784.19 39745.53

所得图，使用上面给出的相同的绘图代码：

titleFunc <- function(overlap) sprintf('R merge solutions: character/integer/logical key, 0..*:0..* cardinality, %d%% overlap',as.integer(overlap*100));
plotRes(res,titleFunc,F);

href =“ https://i.sstatic.net/ytero.png” rel =“ noreferrer”>

https://i.sstatic.net/buxz2.png“ alt =” r-merge-benchmark-systroded-key-key-optional-optional-many-to-many-1“>

For the case of a left join with a 0..*:0..1 cardinality or a right join with a 0..1:0..* cardinality it is possible to assign in-place the unilateral columns from the joiner (the 0..1 table) directly onto the joinee (the 0..* table), and thereby avoid the creation of an entirely new table of data. This requires matching the key columns from the joinee into the joiner and indexing+ordering the joiner's rows accordingly for the assignment.

If the key is a single column, then we can use a single call to match() to do the matching. This is the case I'll cover in this answer.

Here's an example based on the OP, except I've added an extra row to df2 with an id of 7 to test the case of a non-matching key in the joiner. This is effectively df1 left join df2:

df1 <- data.frame(CustomerId=1:6,Product=c(rep('Toaster',3L),rep('Radio',3L)));
df2 <- data.frame(CustomerId=c(2L,4L,6L,7L),State=c(rep('Alabama',2L),'Ohio','Texas'));
df1[names(df2)[-1L]] <- df2[match(df1[,1L],df2[,1L]),-1L];
df1;
##   CustomerId Product   State
## 1          1 Toaster    <NA>
## 2          2 Toaster Alabama
## 3          3 Toaster    <NA>
## 4          4   Radio Alabama
## 5          5   Radio    <NA>
## 6          6   Radio    Ohio

In the above I hard-coded an assumption that the key column is the first column of both input tables. I would argue that, in general, this is not an unreasonable assumption, since, if you have a data.frame with a key column, it would be strange if it had not been set up as the first column of the data.frame from the outset. And you can always reorder the columns to make it so. An advantageous consequence of this assumption is that the name of the key column does not have to be hard-coded, although I suppose it's just replacing one assumption with another. Concision is another advantage of integer indexing, as well as speed. In the benchmarks below I'll change the implementation to use string name indexing to match the competing implementations.

I think this is a particularly appropriate solution if you have several tables that you want to left join against a single large table. Repeatedly rebuilding the entire table for each merge would be unnecessary and inefficient.

On the other hand, if you need the joinee to remain unaltered through this operation for whatever reason, then this solution cannot be used, since it modifies the joinee directly. Although in that case you could simply make a copy and perform the in-place assignment(s) on the copy.

As a side note, I briefly looked into possible matching solutions for multicolumn keys. Unfortunately, the only matching solutions I found were:

inefficient concatenations. e.g. match(interaction(df1$a,df1$b),interaction(df2$a,df2$b)), or the same idea with paste().
inefficient cartesian conjunctions, e.g. outer(df1$a,df2$a,`==`) & outer(df1$b,df2$b,`==`).
base R merge() and equivalent package-based merge functions, which always allocate a new table to return the merged result, and thus are not suitable for an in-place assignment-based solution.

For example, see Matching multiple columns on different data frames and getting other column as result, match two columns with two other columns, Matching on multiple columns, and the dupe of this question where I originally came up with the in-place solution, Combine two data frames with different number of rows in R.

Benchmarking

I decided to do my own benchmarking to see how the in-place assignment approach compares to the other solutions that have been offered in this question.

Testing code:

library(microbenchmark);
library(data.table);
library(sqldf);
library(plyr);
library(dplyr);

solSpecs <- list(
    merge=list(testFuncs=list(
        inner=function(df1,df2,key) merge(df1,df2,key),
        left =function(df1,df2,key) merge(df1,df2,key,all.x=T),
        right=function(df1,df2,key) merge(df1,df2,key,all.y=T),
        full =function(df1,df2,key) merge(df1,df2,key,all=T)
    )),
    data.table.unkeyed=list(argSpec='data.table.unkeyed',testFuncs=list(
        inner=function(dt1,dt2,key) dt1[dt2,on=key,nomatch=0L,allow.cartesian=T],
        left =function(dt1,dt2,key) dt2[dt1,on=key,allow.cartesian=T],
        right=function(dt1,dt2,key) dt1[dt2,on=key,allow.cartesian=T],
        full =function(dt1,dt2,key) merge(dt1,dt2,key,all=T,allow.cartesian=T) ## calls merge.data.table()
    )),
    data.table.keyed=list(argSpec='data.table.keyed',testFuncs=list(
        inner=function(dt1,dt2) dt1[dt2,nomatch=0L,allow.cartesian=T],
        left =function(dt1,dt2) dt2[dt1,allow.cartesian=T],
        right=function(dt1,dt2) dt1[dt2,allow.cartesian=T],
        full =function(dt1,dt2) merge(dt1,dt2,all=T,allow.cartesian=T) ## calls merge.data.table()
    )),
    sqldf.unindexed=list(testFuncs=list( ## note: must pass connection=NULL to avoid running against the live DB connection, which would result in collisions with the residual tables from the last query upload
        inner=function(df1,df2,key) sqldf(paste0('select * from df1 inner join df2 using(',paste(collapse=',',key),')'),connection=NULL),
        left =function(df1,df2,key) sqldf(paste0('select * from df1 left join df2 using(',paste(collapse=',',key),')'),connection=NULL),
        right=function(df1,df2,key) sqldf(paste0('select * from df2 left join df1 using(',paste(collapse=',',key),')'),connection=NULL) ## can't do right join proper, not yet supported; inverted left join is equivalent
        ##full =function(df1,df2,key) sqldf(paste0('select * from df1 full join df2 using(',paste(collapse=',',key),')'),connection=NULL) ## can't do full join proper, not yet supported; possible to hack it with a union of left joins, but too unreasonable to include in testing
    )),
    sqldf.indexed=list(testFuncs=list( ## important: requires an active DB connection with preindexed main.df1 and main.df2 ready to go; arguments are actually ignored
        inner=function(df1,df2,key) sqldf(paste0('select * from main.df1 inner join main.df2 using(',paste(collapse=',',key),')')),
        left =function(df1,df2,key) sqldf(paste0('select * from main.df1 left join main.df2 using(',paste(collapse=',',key),')')),
        right=function(df1,df2,key) sqldf(paste0('select * from main.df2 left join main.df1 using(',paste(collapse=',',key),')')) ## can't do right join proper, not yet supported; inverted left join is equivalent
        ##full =function(df1,df2,key) sqldf(paste0('select * from main.df1 full join main.df2 using(',paste(collapse=',',key),')')) ## can't do full join proper, not yet supported; possible to hack it with a union of left joins, but too unreasonable to include in testing
    )),
    plyr=list(testFuncs=list(
        inner=function(df1,df2,key) join(df1,df2,key,'inner'),
        left =function(df1,df2,key) join(df1,df2,key,'left'),
        right=function(df1,df2,key) join(df1,df2,key,'right'),
        full =function(df1,df2,key) join(df1,df2,key,'full')
    )),
    dplyr=list(testFuncs=list(
        inner=function(df1,df2,key) inner_join(df1,df2,key),
        left =function(df1,df2,key) left_join(df1,df2,key),
        right=function(df1,df2,key) right_join(df1,df2,key),
        full =function(df1,df2,key) full_join(df1,df2,key)
    )),
    in.place=list(testFuncs=list(
        left =function(df1,df2,key) { cns <- setdiff(names(df2),key); df1[cns] <- df2[match(df1[,key],df2[,key]),cns]; df1; },
        right=function(df1,df2,key) { cns <- setdiff(names(df1),key); df2[cns] <- df1[match(df2[,key],df1[,key]),cns]; df2; }
    ))
);

getSolTypes <- function() names(solSpecs);
getJoinTypes <- function() unique(unlist(lapply(solSpecs,function(x) names(x$testFuncs))));
getArgSpec <- function(argSpecs,key=NULL) if (is.null(key)) argSpecs$default else argSpecs[[key]];

initSqldf <- function() {
    sqldf(); ## creates sqlite connection on first run, cleans up and closes existing connection otherwise
    if (exists('sqldfInitFlag',envir=globalenv(),inherits=F) && sqldfInitFlag) { ## false only on first run
        sqldf(); ## creates a new connection
    } else {
        assign('sqldfInitFlag',T,envir=globalenv()); ## set to true for the one and only time
    }; ## end if
    invisible();
}; ## end initSqldf()

setUpBenchmarkCall <- function(argSpecs,joinType,solTypes=getSolTypes(),env=parent.frame()) {
    ## builds and returns a list of expressions suitable for passing to the list argument of microbenchmark(), and assigns variables to resolve symbol references in those expressions
    callExpressions <- list();
    nms <- character();
    for (solType in solTypes) {
        testFunc <- solSpecs[[solType]]$testFuncs[[joinType]];
        if (is.null(testFunc)) next; ## this join type is not defined for this solution type
        testFuncName <- paste0('tf.',solType);
        assign(testFuncName,testFunc,envir=env);
        argSpecKey <- solSpecs[[solType]]$argSpec;
        argSpec <- getArgSpec(argSpecs,argSpecKey);
        argList <- setNames(nm=names(argSpec$args),vector('list',length(argSpec$args)));
        for (i in seq_along(argSpec$args)) {
            argName <- paste0('tfa.',argSpecKey,i);
            assign(argName,argSpec$args[[i]],envir=env);
            argList[[i]] <- if (i%in%argSpec$copySpec) call('copy',as.symbol(argName)) else as.symbol(argName);
        }; ## end for
        callExpressions[[length(callExpressions)+1L]] <- do.call(call,c(list(testFuncName),argList),quote=T);
        nms[length(nms)+1L] <- solType;
    }; ## end for
    names(callExpressions) <- nms;
    callExpressions;
}; ## end setUpBenchmarkCall()

harmonize <- function(res) {
    res <- as.data.frame(res); ## coerce to data.frame
    for (ci in which(sapply(res,is.factor))) res[[ci]] <- as.character(res[[ci]]); ## coerce factor columns to character
    for (ci in which(sapply(res,is.logical))) res[[ci]] <- as.integer(res[[ci]]); ## coerce logical columns to integer (works around sqldf quirk of munging logicals to integers)
    ##for (ci in which(sapply(res,inherits,'POSIXct'))) res[[ci]] <- as.double(res[[ci]]); ## coerce POSIXct columns to double (works around sqldf quirk of losing POSIXct class) ----- POSIXct doesn't work at all in sqldf.indexed
    res <- res[order(names(res))]; ## order columns
    res <- res[do.call(order,res),]; ## order rows
    res;
}; ## end harmonize()

checkIdentical <- function(argSpecs,solTypes=getSolTypes()) {
    for (joinType in getJoinTypes()) {
        callExpressions <- setUpBenchmarkCall(argSpecs,joinType,solTypes);
        if (length(callExpressions)<2L) next;
        ex <- harmonize(eval(callExpressions[[1L]]));
        for (i in seq(2L,len=length(callExpressions)-1L)) {
            y <- harmonize(eval(callExpressions[[i]]));
            if (!isTRUE(all.equal(ex,y,check.attributes=F))) {
                ex <<- ex;
                y <<- y;
                solType <- names(callExpressions)[i];
                stop(paste0('non-identical: ',solType,' ',joinType,'.'));
            }; ## end if
        }; ## end for
    }; ## end for
    invisible();
}; ## end checkIdentical()

testJoinType <- function(argSpecs,joinType,solTypes=getSolTypes(),metric=NULL,times=100L) {
    callExpressions <- setUpBenchmarkCall(argSpecs,joinType,solTypes);
    bm <- microbenchmark(list=callExpressions,times=times);
    if (is.null(metric)) return(bm);
    bm <- summary(bm);
    res <- setNames(nm=names(callExpressions),bm[[metric]]);
    attr(res,'unit') <- attr(bm,'unit');
    res;
}; ## end testJoinType()

testAllJoinTypes <- function(argSpecs,solTypes=getSolTypes(),metric=NULL,times=100L) {
    joinTypes <- getJoinTypes();
    resList <- setNames(nm=joinTypes,lapply(joinTypes,function(joinType) testJoinType(argSpecs,joinType,solTypes,metric,times)));
    if (is.null(metric)) return(resList);
    units <- unname(unlist(lapply(resList,attr,'unit')));
    res <- do.call(data.frame,c(list(join=joinTypes),setNames(nm=solTypes,rep(list(rep(NA_real_,length(joinTypes))),length(solTypes))),list(unit=units,stringsAsFactors=F)));
    for (i in seq_along(resList)) res[i,match(names(resList[[i]]),names(res))] <- resList[[i]];
    res;
}; ## end testAllJoinTypes()

testGrid <- function(makeArgSpecsFunc,sizes,overlaps,solTypes=getSolTypes(),joinTypes=getJoinTypes(),metric='median',times=100L) {

    res <- expand.grid(size=sizes,overlap=overlaps,joinType=joinTypes,stringsAsFactors=F);
    res[solTypes] <- NA_real_;
    res$unit <- NA_character_;
    for (ri in seq_len(nrow(res))) {

        size <- res$size[ri];
        overlap <- res$overlap[ri];
        joinType <- res$joinType[ri];

        argSpecs <- makeArgSpecsFunc(size,overlap);

        checkIdentical(argSpecs,solTypes);

        cur <- testJoinType(argSpecs,joinType,solTypes,metric,times);
        res[ri,match(names(cur),names(res))] <- cur;
        res$unit[ri] <- attr(cur,'unit');

    }; ## end for

    res;

}; ## end testGrid()

Here's a benchmark of the example based on the OP that I demonstrated earlier:

## OP's example, supplemented with a non-matching row in df2
argSpecs <- list(
    default=list(copySpec=1:2,args=list(
        df1 <- data.frame(CustomerId=1:6,Product=c(rep('Toaster',3L),rep('Radio',3L))),
        df2 <- data.frame(CustomerId=c(2L,4L,6L,7L),State=c(rep('Alabama',2L),'Ohio','Texas')),
        'CustomerId'
    )),
    data.table.unkeyed=list(copySpec=1:2,args=list(
        as.data.table(df1),
        as.data.table(df2),
        'CustomerId'
    )),
    data.table.keyed=list(copySpec=1:2,args=list(
        setkey(as.data.table(df1),CustomerId),
        setkey(as.data.table(df2),CustomerId)
    ))
);
## prepare sqldf
initSqldf();
sqldf('create index df1_key on df1(CustomerId);'); ## upload and create an sqlite index on df1
sqldf('create index df2_key on df2(CustomerId);'); ## upload and create an sqlite index on df2

checkIdentical(argSpecs);

testAllJoinTypes(argSpecs,metric='median');
##    join    merge data.table.unkeyed data.table.keyed sqldf.unindexed sqldf.indexed      plyr    dplyr in.place         unit
## 1 inner  644.259           861.9345          923.516        9157.752      1580.390  959.2250 270.9190       NA microseconds
## 2  left  713.539           888.0205          910.045        8820.334      1529.714  968.4195 270.9185 224.3045 microseconds
## 3 right 1221.804           909.1900          923.944        8930.668      1533.135 1063.7860 269.8495 218.1035 microseconds
## 4  full 1302.203          3107.5380         3184.729              NA            NA 1593.6475 270.7055       NA microseconds

Here I benchmark on random input data, trying different scales and different patterns of key overlap between the two input tables. This benchmark is still restricted to the case of a single-column integer key. As well, to ensure that the in-place solution would work for both left and right joins of the same tables, all random test data uses 0..1:0..1 cardinality. This is implemented by sampling without replacement the key column of the first data.frame when generating the key column of the second data.frame.

makeArgSpecs.singleIntegerKey.optionalOneToOne <- function(size,overlap) {

    com <- as.integer(size*overlap);

    argSpecs <- list(
        default=list(copySpec=1:2,args=list(
            df1 <- data.frame(id=sample(size),y1=rnorm(size),y2=rnorm(size)),
            df2 <- data.frame(id=sample(c(if (com>0L) sample(df1$id,com) else integer(),seq(size+1L,len=size-com))),y3=rnorm(size),y4=rnorm(size)),
            'id'
        )),
        data.table.unkeyed=list(copySpec=1:2,args=list(
            as.data.table(df1),
            as.data.table(df2),
            'id'
        )),
        data.table.keyed=list(copySpec=1:2,args=list(
            setkey(as.data.table(df1),id),
            setkey(as.data.table(df2),id)
        ))
    );
    ## prepare sqldf
    initSqldf();
    sqldf('create index df1_key on df1(id);'); ## upload and create an sqlite index on df1
    sqldf('create index df2_key on df2(id);'); ## upload and create an sqlite index on df2

    argSpecs;

}; ## end makeArgSpecs.singleIntegerKey.optionalOneToOne()

## cross of various input sizes and key overlaps
sizes <- c(1e1L,1e3L,1e6L);
overlaps <- c(0.99,0.5,0.01);
system.time({ res <- testGrid(makeArgSpecs.singleIntegerKey.optionalOneToOne,sizes,overlaps); });
##     user   system  elapsed
## 22024.65 12308.63 34493.19

I wrote some code to create log-log plots of the above results. I generated a separate plot for each overlap percentage. It's a little bit cluttered, but I like having all the solution types and join types represented in the same plot.

I used spline interpolation to show a smooth curve for each solution/join type combination, drawn with individual pch symbols. The join type is captured by the pch symbol, using a dot for inner, left and right angle brackets for left and right, and a diamond for full. The solution type is captured by the color as shown in the legend.

plotRes <- function(res,titleFunc,useFloor=F) {
    solTypes <- setdiff(names(res),c('size','overlap','joinType','unit')); ## derive from res
    normMult <- c(microseconds=1e-3,milliseconds=1); ## normalize to milliseconds
    joinTypes <- getJoinTypes();
    cols <- c(merge='purple',data.table.unkeyed='blue',data.table.keyed='#00DDDD',sqldf.unindexed='brown',sqldf.indexed='orange',plyr='red',dplyr='#00BB00',in.place='magenta');
    pchs <- list(inner=20L,left='<',right='>',full=23L);
    cexs <- c(inner=0.7,left=1,right=1,full=0.7);
    NP <- 60L;
    ord <- order(decreasing=T,colMeans(res[res$size==max(res$size),solTypes],na.rm=T));
    ymajors <- data.frame(y=c(1,1e3),label=c('1ms','1s'),stringsAsFactors=F);
    for (overlap in unique(res$overlap)) {
        x1 <- res[res$overlap==overlap,];
        x1[solTypes] <- x1[solTypes]*normMult[x1$unit]; x1$unit <- NULL;
        xlim <- c(1e1,max(x1$size));
        xticks <- 10^seq(log10(xlim[1L]),log10(xlim[2L]));
        ylim <- c(1e-1,10^((if (useFloor) floor else ceiling)(log10(max(x1[solTypes],na.rm=T))))); ## use floor() to zoom in a little more, only sqldf.unindexed will break above, but xpd=NA will keep it visible
        yticks <- 10^seq(log10(ylim[1L]),log10(ylim[2L]));
        yticks.minor <- rep(yticks[-length(yticks)],each=9L)*1:9;
        plot(NA,xlim=xlim,ylim=ylim,xaxs='i',yaxs='i',axes=F,xlab='size (rows)',ylab='time (ms)',log='xy');
        abline(v=xticks,col='lightgrey');
        abline(h=yticks.minor,col='lightgrey',lty=3L);
        abline(h=yticks,col='lightgrey');
        axis(1L,xticks,parse(text=sprintf('10^%d',as.integer(log10(xticks)))));
        axis(2L,yticks,parse(text=sprintf('10^%d',as.integer(log10(yticks)))),las=1L);
        axis(4L,ymajors$y,ymajors$label,las=1L,tick=F,cex.axis=0.7,hadj=0.5);
        for (joinType in rev(joinTypes)) { ## reverse to draw full first, since it's larger and would be more obtrusive if drawn last
            x2 <- x1[x1$joinType==joinType,];
            for (solType in solTypes) {
                if (any(!is.na(x2[[solType]]))) {
                    xy <- spline(x2$size,x2[[solType]],xout=10^(seq(log10(x2$size[1L]),log10(x2$size[nrow(x2)]),len=NP)));
                    points(xy$x,xy$y,pch=pchs[[joinType]],col=cols[solType],cex=cexs[joinType],xpd=NA);
                }; ## end if
            }; ## end for
        }; ## end for
        ## custom legend
        ## due to logarithmic skew, must do all distance calcs in inches, and convert to user coords afterward
        ## the bottom-left corner of the legend will be defined in normalized figure coords, although we can convert to inches immediately
        leg.cex <- 0.7;
        leg.x.in <- grconvertX(0.275,'nfc','in');
        leg.y.in <- grconvertY(0.6,'nfc','in');
        leg.x.user <- grconvertX(leg.x.in,'in');
        leg.y.user <- grconvertY(leg.y.in,'in');
        leg.outpad.w.in <- 0.1;
        leg.outpad.h.in <- 0.1;
        leg.midpad.w.in <- 0.1;
        leg.midpad.h.in <- 0.1;
        leg.sol.w.in <- max(strwidth(solTypes,'in',leg.cex));
        leg.sol.h.in <- max(strheight(solTypes,'in',leg.cex))*1.5; ## multiplication factor for greater line height
        leg.join.w.in <- max(strheight(joinTypes,'in',leg.cex))*1.5; ## ditto
        leg.join.h.in <- max(strwidth(joinTypes,'in',leg.cex));
        leg.main.w.in <- leg.join.w.in*length(joinTypes);
        leg.main.h.in <- leg.sol.h.in*length(solTypes);
        leg.x2.user <- grconvertX(leg.x.in+leg.outpad.w.in*2+leg.main.w.in+leg.midpad.w.in+leg.sol.w.in,'in');
        leg.y2.user <- grconvertY(leg.y.in+leg.outpad.h.in*2+leg.main.h.in+leg.midpad.h.in+leg.join.h.in,'in');
        leg.cols.x.user <- grconvertX(leg.x.in+leg.outpad.w.in+leg.join.w.in*(0.5+seq(0L,length(joinTypes)-1L)),'in');
        leg.lines.y.user <- grconvertY(leg.y.in+leg.outpad.h.in+leg.main.h.in-leg.sol.h.in*(0.5+seq(0L,length(solTypes)-1L)),'in');
        leg.sol.x.user <- grconvertX(leg.x.in+leg.outpad.w.in+leg.main.w.in+leg.midpad.w.in,'in');
        leg.join.y.user <- grconvertY(leg.y.in+leg.outpad.h.in+leg.main.h.in+leg.midpad.h.in,'in');
        rect(leg.x.user,leg.y.user,leg.x2.user,leg.y2.user,col='white');
        text(leg.sol.x.user,leg.lines.y.user,solTypes[ord],cex=leg.cex,pos=4L,offset=0);
        text(leg.cols.x.user,leg.join.y.user,joinTypes,cex=leg.cex,pos=4L,offset=0,srt=90); ## srt rotation applies *after* pos/offset positioning
        for (i in seq_along(joinTypes)) {
            joinType <- joinTypes[i];
            points(rep(leg.cols.x.user[i],length(solTypes)),ifelse(colSums(!is.na(x1[x1$joinType==joinType,solTypes[ord]]))==0L,NA,leg.lines.y.user),pch=pchs[[joinType]],col=cols[solTypes[ord]]);
        }; ## end for
        title(titleFunc(overlap));
        readline(sprintf('overlap %.02f',overlap));
    }; ## end for
}; ## end plotRes()

titleFunc <- function(overlap) sprintf('R merge solutions: single-column integer key, 0..1:0..1 cardinality, %d%% overlap',as.integer(overlap*100));
plotRes(res,titleFunc,T);

Here's a second large-scale benchmark that's more heavy-duty, with respect to the number and types of key columns, as well as cardinality. For this benchmark I use three key columns: one character, one integer, and one logical, with no restrictions on cardinality (that is, 0..*:0..*). (In general it's not advisable to define key columns with double or complex values due to floating-point comparison complications, and basically no one ever uses the raw type, much less for key columns, so I haven't included those types in the key columns. Also, for information's sake, I initially tried to use four key columns by including a POSIXct key column, but the POSIXct type didn't play well with the sqldf.indexed solution for some reason, possibly due to floating-point comparison anomalies, so I removed it.)

makeArgSpecs.assortedKey.optionalManyToMany <- function(size,overlap,uniquePct=75) {

    ## number of unique keys in df1
    u1Size <- as.integer(size*uniquePct/100);

    ## (roughly) divide u1Size into bases, so we can use expand.grid() to produce the required number of unique key values with repetitions within individual key columns
    ## use ceiling() to ensure we cover u1Size; will truncate afterward
    u1SizePerKeyColumn <- as.integer(ceiling(u1Size^(1/3)));

    ## generate the unique key values for df1
    keys1 <- expand.grid(stringsAsFactors=F,
        idCharacter=replicate(u1SizePerKeyColumn,paste(collapse='',sample(letters,sample(4:12,1L),T))),
        idInteger=sample(u1SizePerKeyColumn),
        idLogical=sample(c(F,T),u1SizePerKeyColumn,T)
        ##idPOSIXct=as.POSIXct('2016-01-01 00:00:00','UTC')+sample(u1SizePerKeyColumn)
    )[seq_len(u1Size),];

    ## rbind some repetitions of the unique keys; this will prepare one side of the many-to-many relationship
    ## also scramble the order afterward
    keys1 <- rbind(keys1,keys1[sample(nrow(keys1),size-u1Size,T),])[sample(size),];

    ## common and unilateral key counts
    com <- as.integer(size*overlap);
    uni <- size-com;

    ## generate some unilateral keys for df2 by synthesizing outside of the idInteger range of df1
    keys2 <- data.frame(stringsAsFactors=F,
        idCharacter=replicate(uni,paste(collapse='',sample(letters,sample(4:12,1L),T))),
        idInteger=u1SizePerKeyColumn+sample(uni),
        idLogical=sample(c(F,T),uni,T)
        ##idPOSIXct=as.POSIXct('2016-01-01 00:00:00','UTC')+u1SizePerKeyColumn+sample(uni)
    );

    ## rbind random keys from df1; this will complete the many-to-many relationship
    ## also scramble the order afterward
    keys2 <- rbind(keys2,keys1[sample(nrow(keys1),com,T),])[sample(size),];

    ##keyNames <- c('idCharacter','idInteger','idLogical','idPOSIXct');
    keyNames <- c('idCharacter','idInteger','idLogical');
    ## note: was going to use raw and complex type for two of the non-key columns, but data.table doesn't seem to fully support them
    argSpecs <- list(
        default=list(copySpec=1:2,args=list(
            df1 <- cbind(stringsAsFactors=F,keys1,y1=sample(c(F,T),size,T),y2=sample(size),y3=rnorm(size),y4=replicate(size,paste(collapse='',sample(letters,sample(4:12,1L),T)))),
            df2 <- cbind(stringsAsFactors=F,keys2,y5=sample(c(F,T),size,T),y6=sample(size),y7=rnorm(size),y8=replicate(size,paste(collapse='',sample(letters,sample(4:12,1L),T)))),
            keyNames
        )),
        data.table.unkeyed=list(copySpec=1:2,args=list(
            as.data.table(df1),
            as.data.table(df2),
            keyNames
        )),
        data.table.keyed=list(copySpec=1:2,args=list(
            setkeyv(as.data.table(df1),keyNames),
            setkeyv(as.data.table(df2),keyNames)
        ))
    );
    ## prepare sqldf
    initSqldf();
    sqldf(paste0('create index df1_key on df1(',paste(collapse=',',keyNames),');')); ## upload and create an sqlite index on df1
    sqldf(paste0('create index df2_key on df2(',paste(collapse=',',keyNames),');')); ## upload and create an sqlite index on df2

    argSpecs;

}; ## end makeArgSpecs.assortedKey.optionalManyToMany()

sizes <- c(1e1L,1e3L,1e5L); ## 1e5L instead of 1e6L to respect more heavy-duty inputs
overlaps <- c(0.99,0.5,0.01);
solTypes <- setdiff(getSolTypes(),'in.place');
system.time({ res <- testGrid(makeArgSpecs.assortedKey.optionalManyToMany,sizes,overlaps,solTypes); });
##     user   system  elapsed
## 38895.50   784.19 39745.53

The resulting plots, using the same plotting code given above:

titleFunc <- function(overlap) sprintf('R merge solutions: character/integer/logical key, 0..*:0..* cardinality, %d%% overlap',as.integer(overlap*100));
plotRes(res,titleFunc,F);

回复收藏 0 原文

删除会话 2025-02-03 05:59:15

在加入两个数据帧时，每行约100万行，一个带2列，另一个具有〜20的数据，我出人意料地找到了合并（...，all.x = true，all.y = true）< /code>更快，然后 dplyr :: full_join（）。这是在dplyr v0.4

合并中需要约17秒，full_join需要〜65秒。

不过，有些食物是因为我通常默认要dplyr进行操纵任务。

回复收藏 0 原文

天气好吗我好吗 2025-02-03 05:59:15

使用 Merge 函数我们可以选择左表或右表的变量，就像我们熟悉SQL中的Select语句一样（例如：选择A。* ...或从中选择B。*）。 ....）
我们必须添加额外的代码，该代码将从新连接的表中子集。
- sql： - 在a.customerid = b.customerid
- r： - merge（df1，df2，by.x =“ customerId”，by.y =“ customerId”）[，names（df1）]
  
  < /li>

同样的方式

- 选择b。
r： - merge（df1，df2，by.x =“ customerId”，by.y = “ customerId”）[，名称（df2）]

回复收藏 0 原文

不必你懂 2025-02-03 05:59:15

更新加入。另外一个重要的SQL风格加入是“ “更新”>更新join “其中一列中的列中的列使用另一个表更新（或创建）表。

修改OP的示例表...

sales = data.frame(
  CustomerId = c(1, 1, 1, 3, 4, 6), 
  Year = 2000:2005,
  Product = c(rep("Toaster", 3), rep("Radio", 3))
)
cust = data.frame(
  CustomerId = c(1, 1, 4, 6), 
  Year = c(2001L, 2002L, 2002L, 2002L),
  State = state.name[1:4]
)

sales
# CustomerId Year Product
#          1 2000 Toaster
#          1 2001 Toaster
#          1 2002 Toaster
#          3 2003   Radio
#          4 2004   Radio
#          6 2005   Radio

cust
# CustomerId Year    State
#          1 2001  Alabama
#          1 2002   Alaska
#          4 2002  Arizona
#          6 2002 Arkansas

假设我们想将客户的状态从 CUST 添加到购买表， sales ，忽略年度列。使用基本R，我们可以识别匹配行，然后将复制值复制到：

sales$State <- cust$State[ match(sales$CustomerId, cust$CustomerId) ]

# CustomerId Year Product    State
#          1 2000 Toaster  Alabama
#          1 2001 Toaster  Alabama
#          1 2002 Toaster  Alabama
#          3 2003   Radio     <NA>
#          4 2004   Radio  Arizona
#          6 2005   Radio Arkansas

# cleanup for the next example
sales$State <- NULL

可以看到， Match 从客户表中选择第一个匹配行。

更新与多列连接。当我们仅加入单列并对第一场比赛感到满意时，上面的方法效果很好。假设我们希望在客户表中的测量年份与销售年相匹配。

正如 @bgoldst的答案所提到的那样，与互动的匹配可能是这种情况的选项。更直接地，一个人可以使用数据。表：

library(data.table)
setDT(sales); setDT(cust)

sales[, State := cust[sales, on=.(CustomerId, Year), x.State]]

#    CustomerId Year Product   State
# 1:          1 2000 Toaster    <NA>
# 2:          1 2001 Toaster Alabama
# 3:          1 2002 Toaster  Alaska
# 4:          3 2003   Radio    <NA>
# 5:          4 2004   Radio    <NA>
# 6:          6 2005   Radio    <NA>

# cleanup for next example
sales[, State := NULL]

滚动更新加入。 另外，我们可能要采用最新的状态，以下是：在以下情况下，在以下情况下，在以下情况下：

sales[, State := cust[sales, on=.(CustomerId, Year), roll=TRUE, x.State]]

#    CustomerId Year Product    State
# 1:          1 2000 Toaster     <NA>
# 2:          1 2001 Toaster  Alabama
# 3:          1 2002 Toaster   Alaska
# 4:          3 2003   Radio     <NA>
# 5:          4 2004   Radio  Arizona
# 6:          6 2005   Radio Arkansas

三个示例都专注于创建/添加新列。请参阅相关的R FAQ 以获取更新/修改现有列的示例。

Update join. One other important SQL-style join is an "update join" where columns in one table are updated (or created) using another table.

Modifying the OP's example tables...

sales = data.frame(
  CustomerId = c(1, 1, 1, 3, 4, 6), 
  Year = 2000:2005,
  Product = c(rep("Toaster", 3), rep("Radio", 3))
)
cust = data.frame(
  CustomerId = c(1, 1, 4, 6), 
  Year = c(2001L, 2002L, 2002L, 2002L),
  State = state.name[1:4]
)

sales
# CustomerId Year Product
#          1 2000 Toaster
#          1 2001 Toaster
#          1 2002 Toaster
#          3 2003   Radio
#          4 2004   Radio
#          6 2005   Radio

cust
# CustomerId Year    State
#          1 2001  Alabama
#          1 2002   Alaska
#          4 2002  Arizona
#          6 2002 Arkansas

Suppose we want to add the customer's state from cust to the purchases table, sales, ignoring the year column. With base R, we can identify matching rows and then copy values over:

sales$State <- cust$State[ match(sales$CustomerId, cust$CustomerId) ]

# CustomerId Year Product    State
#          1 2000 Toaster  Alabama
#          1 2001 Toaster  Alabama
#          1 2002 Toaster  Alabama
#          3 2003   Radio     <NA>
#          4 2004   Radio  Arizona
#          6 2005   Radio Arkansas

# cleanup for the next example
sales$State <- NULL

As can be seen here, match selects the first matching row from the customer table.

Update join with multiple columns. The approach above works well when we are joining on only a single column and are satisfied with the first match. Suppose we want the year of measurement in the customer table to match the year of sale.

As @bgoldst's answer mentions, match with interaction might be an option for this case. More straightforwardly, one could use data.table:

library(data.table)
setDT(sales); setDT(cust)

sales[, State := cust[sales, on=.(CustomerId, Year), x.State]]

#    CustomerId Year Product   State
# 1:          1 2000 Toaster    <NA>
# 2:          1 2001 Toaster Alabama
# 3:          1 2002 Toaster  Alaska
# 4:          3 2003   Radio    <NA>
# 5:          4 2004   Radio    <NA>
# 6:          6 2005   Radio    <NA>

# cleanup for next example
sales[, State := NULL]

Rolling update join. Alternately, we may want to take the last state the customer was found in:

sales[, State := cust[sales, on=.(CustomerId, Year), roll=TRUE, x.State]]

#    CustomerId Year Product    State
# 1:          1 2000 Toaster     <NA>
# 2:          1 2001 Toaster  Alabama
# 3:          1 2002 Toaster   Alaska
# 4:          3 2003   Radio     <NA>
# 5:          4 2004   Radio  Arizona
# 6:          6 2005   Radio Arkansas

The three examples above all focus on creating/adding a new column. See the related R FAQ for an example of updating/modifying an existing column.

回复收藏 0 原文

自在安然 2025-02-03 05:59:15

对于所有列上的内部联接，您也可以从 data.table package或 Intersect intersect intersect intersect 从 dplyr < dplyr << /em> - 包装作为 Merge 的替代方案，而无需通过 -columns指定。这将使两个数据范围之间的行相等：

merge(df1, df2)
#   V1 V2
# 1  B  2
# 2  C  3

dplyr::intersect(df1, df2)
#   V1 V2
# 1  B  2
# 2  C  3

data.table::fintersect(setDT(df1), setDT(df2))
#    V1 V2
# 1:  B  2
# 2:  C  3

示例数据：

df1 <- data.frame(V1 = LETTERS[1:4], V2 = 1:4)
df2 <- data.frame(V1 = LETTERS[2:3], V2 = 2:3)

For an inner join on all columns, you could also use fintersect from the data.table-package or intersect from the dplyr-package as an alternative to merge without specifying the by-columns. This will give the rows that are equal between two dataframes:

merge(df1, df2)
#   V1 V2
# 1  B  2
# 2  C  3

dplyr::intersect(df1, df2)
#   V1 V2
# 1  B  2
# 2  C  3

data.table::fintersect(setDT(df1), setDT(df2))
#    V1 V2
# 1:  B  2
# 2:  C  3

Example data:

df1 <- data.frame(V1 = LETTERS[1:4], V2 = 1:4)
df2 <- data.frame(V1 = LETTERS[2:3], V2 = 2:3)

回复收藏 0 原文

不美如何 2025-02-03 05:59:15

collapse 2.0提供了另一个加入框架，并使用 JOIN 。它比任何其他选项都要快。

library(collapse)

join(
  df1, 
  df2, 
  how = c("left", "right", "inner", "full", "semi", "anti")
)

collapse 2.0 provides another join framework with join. It is noticeably faster than any other option.

library(collapse)

join(
  df1, 
  df2, 
  how = c("left", "right", "inner", "full", "semi", "anti")
)

回复收藏 0 原文

~没有更多了~

关于作者

眼眸印温柔

暂无简介

文章

26 人气

关注发私信

友情链接

文江博客

如何加入（合并）数据帧（内部，外部，左，右）

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

评论（14）

突变连接：使用DF2

过滤连接中的匹配点在DF1中添加列：DF1中的行过滤器，请勿修改列

Mutating joins: add columns to df1 using matches in df2

Filtering joins: filter out rows in df1, don't modify columns

基准测试

Benchmarking

关于作者

相关话题

热门标签

推荐作者

yuanzihao09

1337793151

横笛休吹塞上声

你在我安

qq_QhAHT0kB

aaaa123451

友情链接

如何加入（合并）数据帧（内部，外部，左，右）

如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

评论（14）

突变连接：使用DF2

过滤连接中的匹配点在DF1中添加列：DF1中的行过滤器，请勿修改列

Mutating joins: add columns to df1 using matches in df2

Filtering joins: filter out rows in df1, don't modify columns

基准测试

Benchmarking

关于作者

相关话题

热门标签

推荐作者

yuanzihao09

1337793151

横笛休吹塞上声

你在我安

qq_QhAHT0kB

aaaa123451

友情链接

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。