将具有可变长度和字符串数量的命名字符串列表写入 R 中的文本文件

发布于 2024-11-17 23:33:21 字数 498 浏览 4 评论 0原文

我是相对较新的 R 用户，仍在学习基础知识。

我有一个命名列表 xx ，这些条目如下所示：

> xx[100:105]
$`15LOX-1`
[1] "207328_at"

$`16.1`
[1] "215946_x_at"

$`16.2`
[1] NA

$`16.3A5`
[1] "200983_x_at" "200984_s_at" "200985_s_at" "212463_at"   "228748_at"  

$`160-KD`
[1] "201224_s_at" "201225_s_at"

$`1600019D15Rik`
[1] "218465_at"   "222642_s_at" "225492_at"   "235907_at"   "238831_at"

我想将其保存到具有两列的文本文件中 - 键和值。如果多个字符串对应相同的键，它们应该位于不同的行中。不需要双引号符号。

另外，如何避免保存 NA 值？

请帮忙。

原文

I'm relatively new R user and still learning the basics.

I have a named list xx those entries looks like this:

> xx[100:105]
I'm relatively new R user and still learning the basics.
I have a named list xx those entries looks like this:
15LOX-1`
[1] "207328_at"

I'm relatively new R user and still learning the basics.
I have a named list xx those entries looks like this:
16.1`
[1] "215946_x_at"

I'm relatively new R user and still learning the basics.
I have a named list xx those entries looks like this:
16.2`
[1] NA

I'm relatively new R user and still learning the basics.
I have a named list xx those entries looks like this:
16.3A5`
[1] "200983_x_at" "200984_s_at" "200985_s_at" "212463_at"   "228748_at"  

I'm relatively new R user and still learning the basics.
I have a named list xx those entries looks like this:
160-KD`
[1] "201224_s_at" "201225_s_at"

I'm relatively new R user and still learning the basics.
I have a named list xx those entries looks like this:
1600019D15Rik`
[1] "218465_at"   "222642_s_at" "225492_at"   "235907_at"   "238831_at"

I would like to save it to a text file with two columns - Key and Value. If several strings correspond to the same key they should be in different rows. Double-quote symbols are not required.

In addition, how I can avoid NA values to be saved?

Please help.

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。

纵山崖 2024-11-24 23:33:22

使用 Andrie 的测试数据，这里有一种使用 reshape 包（或“reshape2”）来完成此操作的巧妙方法：

x <- list(
    A = "a1",
    B = "b1",
    C = NA,
    D = paste("d", 1:5, sep=""))

接下来，melt 有一个列表方法！

> melt(x)
  value L1
1    a1  A
2    b1  B
3  <NA>  C
4    d1  D
5    d2  D
6    d3  D
7    d4  D
8    d5  D

然后我们可以使用 complete.cases 或类似的东西取出 NA：

rs <- melt(x)
rs <- rs[complete.cases(x),]
colnames(rs) <- c('value','key')

Using Andrie's test data, here's a kind of slick way to do this using the reshape package (or `reshape2'):

x <- list(
    A = "a1",
    B = "b1",
    C = NA,
    D = paste("d", 1:5, sep=""))

Next, melt has a list method!

> melt(x)
  value L1
1    a1  A
2    b1  B
3  <NA>  C
4    d1  D
5    d2  D
6    d3  D
7    d4  D
8    d5  D

Then we can pull out the NAs using complete.cases or something equivalent:

rs <- melt(x)
rs <- rs[complete.cases(x),]
colnames(rs) <- c('value','key')

回复收藏 0 原文

乜一 2024-11-24 23:33:22

我会做这样的事情。

#Create a matrix
z <- cbind(key=rep(names(xx), sapply(xx, length)), value = unlist(xx))
#Remove NA
z <- z[!is.na(z[,2]),]
#Write to textfile
write.table(z, "filename.txt", row.names= F)

您可以查看 write.table 的帮助来查看其他可用选项。

询问的结果

>z
                key             value        
15LOX-1        "15LOX-1"       "207328_at"  
16.1           "16.1"          "215946_x_at"
16.3A51        "16.3A5"        "200983_x_at"
16.3A52        "16.3A5"        "200984_s_at"
16.3A53        "16.3A5"        "200985_s_at"
16.3A54        "16.3A5"        "212463_at"  
16.3A55        "16.3A5"        "228748_at"  
160-KD1        "160-KD"        "201224_s_at"
160-KD2        "160-KD"        "201225_s_at"
1600019D15Rik1 "1600019D15Rik" "218465_at"  
1600019D15Rik2 "1600019D15Rik" "222642_s_at"
1600019D15Rik3 "1600019D15Rik" "225492_at"  
1600019D15Rik4 "1600019D15Rik" "235907_at"  
1600019D15Rik5 "1600019D15Rik" "238831_at"

这是 Andrie HTH

I would do something like this.

#Create a matrix
z <- cbind(key=rep(names(xx), sapply(xx, length)), value = unlist(xx))
#Remove NA
z <- z[!is.na(z[,2]),]
#Write to textfile
write.table(z, "filename.txt", row.names= F)

You can look at the help of write.table to see the other available options.

Here's the result as asked by Andrie

>z
                key             value        
15LOX-1        "15LOX-1"       "207328_at"  
16.1           "16.1"          "215946_x_at"
16.3A51        "16.3A5"        "200983_x_at"
16.3A52        "16.3A5"        "200984_s_at"
16.3A53        "16.3A5"        "200985_s_at"
16.3A54        "16.3A5"        "212463_at"  
16.3A55        "16.3A5"        "228748_at"  
160-KD1        "160-KD"        "201224_s_at"
160-KD2        "160-KD"        "201225_s_at"
1600019D15Rik1 "1600019D15Rik" "218465_at"  
1600019D15Rik2 "1600019D15Rik" "222642_s_at"
1600019D15Rik3 "1600019D15Rik" "225492_at"  
1600019D15Rik4 "1600019D15Rik" "235907_at"  
1600019D15Rik5 "1600019D15Rik" "238831_at"

HTH

回复收藏 0 原文

佼人 2024-11-24 23:33:21

重新创建测试数据：

xx <- structure(list(
    `15LOX-1` = "207328_at", 
    `16.1` = "215946_x_at", 
    `16.2` = NA, 
    `16.3A5` = c("200983_x_at", "200984_s_at", "200985_s_at", "212463_at", "228748_at"), 
    `160-KD` = c("201224_s_at", "201225_s_at" ), 
    `1600019D15Rik` = c("218465_at", "222642_s_at", "225492_at", "235907_at", "238831_at")),
   .Names = c("15LOX-1", "16.1", "16.2", "16.3A5", "160-KD", "1600019D15Rik"))

首先，删除所有 NA 值：

xx[is.na(xx)] <- NULL

现在，创建一个临时变量，用于存储 x 中每个元素的长度：

tmp <- sapply(xx, function(xt)length(xt))

现在使用 rep 创建键（即重复 x 的名称，每次关联元素的长度），并使用 unlist 和 unname 的组合来创建值：

data.frame(
    key = rep(names(tmp), times=unname(tmp)),
    value = unlist(unname(xx))
)

这会产生：

             key       value
1        15LOX-1   207328_at
2           16.1 215946_x_at
3         16.3A5 200983_x_at
4         16.3A5 200984_s_at
5         16.3A5 200985_s_at
6         16.3A5   212463_at
7         16.3A5   228748_at
8         160-KD 201224_s_at
9         160-KD 201225_s_at
10 1600019D15Rik   218465_at
11 1600019D15Rik 222642_s_at
12 1600019D15Rik   225492_at
13 1600019D15Rik   235907_at
14 1600019D15Rik   238831_at

最后，使用 write.csv( x, file=...) 或您最喜欢的写入函数将数据保存到文件。

Recreate test data:

xx <- structure(list(
    `15LOX-1` = "207328_at", 
    `16.1` = "215946_x_at", 
    `16.2` = NA, 
    `16.3A5` = c("200983_x_at", "200984_s_at", "200985_s_at", "212463_at", "228748_at"), 
    `160-KD` = c("201224_s_at", "201225_s_at" ), 
    `1600019D15Rik` = c("218465_at", "222642_s_at", "225492_at", "235907_at", "238831_at")),
   .Names = c("15LOX-1", "16.1", "16.2", "16.3A5", "160-KD", "1600019D15Rik"))

First, remove all the NA values:

xx[is.na(xx)] <- NULL

Now, create a temporary variable that stores the length of each element in x:

tmp <- sapply(xx, function(xt)length(xt))

Now use rep to create the key (i.e. repeat the names of x, each time the length of the associated element), and use a combination of unlist and unname to create the values:

data.frame(
    key = rep(names(tmp), times=unname(tmp)),
    value = unlist(unname(xx))
)

This produces:

             key       value
1        15LOX-1   207328_at
2           16.1 215946_x_at
3         16.3A5 200983_x_at
4         16.3A5 200984_s_at
5         16.3A5 200985_s_at
6         16.3A5   212463_at
7         16.3A5   228748_at
8         160-KD 201224_s_at
9         160-KD 201225_s_at
10 1600019D15Rik   218465_at
11 1600019D15Rik 222642_s_at
12 1600019D15Rik   225492_at
13 1600019D15Rik   235907_at
14 1600019D15Rik   238831_at

Finally, use write.csv(x, file=...) or your favourite write function to save the data to file.

回复收藏 0 原文

江挽川 2024-11-24 23:33:21

reshape2 包可以使用熔化函数来做到这一点。使用 Andrie 的数据：

require(reshape2)
 > melt(x)
  value L1
1    a1  A
2    b1  B
3  <NA>  C
4    d1  D
5    d2  D
6    d3  D
7    d4  D
8    d5  D

有些事情并不完全符合您的要求。首先，列的顺序相反，这可能是问题，也可能不是问题。其次，与列关联的名称并不完全符合您的要求，这也可能不是问题。第三，NA值仍然存在，这是一个基于你的问题的问题。我将使用 Complete.cases() 来解决该问题并为其指定一些适当的名称。也许是这样的：

out <- melt(x)[, 2:1] #Reverse the key - value columns
out <- out[complete.cases(out) ,] #Subset only complete cases
names(out) <- c("Key", "Value")  #New column names

> out
  Key Value
1   A    a1
2   B    b1
4   D    d1
.....

reshape2 package can do this with the melt function. Using the data from Andrie:

require(reshape2)
 > melt(x)
  value L1
1    a1  A
2    b1  B
3  <NA>  C
4    d1  D
5    d2  D
6    d3  D
7    d4  D
8    d5  D

A few things are a not exactly as you want here. First, the columns are in reverse order which may or may not be an issue. Second, the names associated with the columns are not exactly as you wanted, again may not be an issue. Third, the NA value is still present, which is an issue based on your question. I'd use complete.cases() to address that issue and give it some appropriate names. Maybe something like this:

out <- melt(x)[, 2:1] #Reverse the key - value columns
out <- out[complete.cases(out) ,] #Subset only complete cases
names(out) <- c("Key", "Value")  #New column names

> out
  Key Value
1   A    a1
2   B    b1
4   D    d1
.....

回复收藏 0 原文

~没有更多了~