从缺少某些值的列表数据在 R 中创建数据框

发布于 2024-12-02 12:26:35 字数 332 浏览 0 评论 0原文

我有一个 CSV，其中包含一长串数据，如下所示：

Date    user_id    value
4/1       1           5
4/1       2           3
4/1       3           10
4/2       1           1
4/2       3           7

我想将其移动到一个数据框中，该数据框中只有一列用户 ID 和一列每个日期。我假设有一种方法可以处理 sapply 或 lapply 但我不确定如何处理用户 ID 并不总是在每个日期都存在。

原文

I have a CSV with a long list of data that looks like this:

Date    user_id    value
4/1       1           5
4/1       2           3
4/1       3           10
4/2       1           1
4/2       3           7

and I want to move it into a data frame that just has one column of user id's and a column for each date. I'm assuming there is a way to do with with sapply or lapply but I'm not sure how to handle that the user id's don't always exist for every date.

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韶华倾负 2024-12-09 12:26:35

也许使用 reshape 如下所示，假设您的数据存储在 dat 中：

reshape(dat,v.names = "value",idvar = "user_id",
        direction = "wide",timevar = "Date")

  user_id value.4/1 value.4/2
1       1         5         1
2       2         3        NA
3       3        10         7

或者也许更简单，使用 reshape2 中的 dcast ：

dcast(dat,user_id~Date)

  user_id 4/1 4/2
1       1   5   1
2       2   3  NA
3       3  10   7

Maybe something using reshape like the following, which assumes your data are stored in dat:

reshape(dat,v.names = "value",idvar = "user_id",
        direction = "wide",timevar = "Date")

  user_id value.4/1 value.4/2
1       1         5         1
2       2         3        NA
3       3        10         7

Or perhaps more simply, use dcast from reshape2:

dcast(dat,user_id~Date)

  user_id 4/1 4/2
1       1   5   1
2       2   3  NA
3       3  10   7

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剧终人散尽 2024-12-09 12:26:35

这也是 tidyr::spread 非常方便地完成的事情

require(tidyr)
require(dplyr)

df <- data.frame("Date" = rep(c("Nov", "Dec"), each = 3), 
             "user.id" = rep(1:3, 2),
             "value" = rnorm(6))

df.2 <- df %>% 
spread(Date, value)

df.2
  user.id        Dec       Nov
     1      -1.9094765   -1.101037
     2       0.2358694   -1.418151
     3      -0.4297790   -1.426573

This is also something that tidyr::spread does quite conveniently

require(tidyr)
require(dplyr)

df <- data.frame("Date" = rep(c("Nov", "Dec"), each = 3), 
             "user.id" = rep(1:3, 2),
             "value" = rnorm(6))

df.2 <- df %>% 
spread(Date, value)

df.2
  user.id        Dec       Nov
     1      -1.9094765   -1.101037
     2       0.2358694   -1.418151
     3      -0.4297790   -1.426573

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