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这是使用 R 透视 CSV 文件的后续问题。

在那个问题中，我想根据列（repository_name）中的值将单个列（类型）分成几列。使用了以下输入数据。

                 type          created_at repository_name
1         IssuesEvent 2012-03-11 06:48:31       bootstrap
2         IssuesEvent 2012-03-11 06:48:31       bootstrap
3   IssueCommentEvent 2012-03-11 07:03:57       bootstrap
4   IssueCommentEvent 2012-03-11 07:03:57       bootstrap
5   IssueCommentEvent 2012-03-11 07:03:57       bootstrap
6        IssuesEvent 2012-03-11 07:03:58        bootstrap
7         WatchEvent 2012-03-11 07:18:45        hogan.js
8         WatchEvent 2012-03-11 07:18:45        hogan.js
9         WatchEvent 2012-03-11 07:18:45        hogan.js
10   IssueCommentEvent 2012-03-11 07:03:57      bootstrap

完整的 CSV 文件可在https://github.com/aronlindberg/VOSS-Sequencing-Toolkit/blob/master/twitter_exploratory_analysis/all_events.csv上找到。

这是 CSV 前 30 行的 dput()：

structure(list(type = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 2L, 
2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 
1L, 4L, 4L, 4L, 2L, 2L, 2L), .Label = c("ForkEvent", "IssueCommentEvent", 
"IssuesEvent", "WatchEvent"), class = "factor"), created_at = structure(c(1L, 
1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 
6L, 7L, 7L, 7L, 8L, 8L, 8L, 9L, 9L, 9L, 10L, 10L, 10L), .Label = c("2012-03-11 06:48:31", 
"2012-03-11 06:52:50", "2012-03-11 07:03:57", "2012-03-11 07:03:58", 
"2012-03-11 07:15:44", "2012-03-11 07:18:45", "2012-03-11 07:19:01", 
"2012-03-11 07:23:56", "2012-03-11 07:32:43", "2012-03-11 07:38:52"
), class = "factor"), repository_name = structure(c(1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 
1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 1L, 1L, 1L), .Label = c("bootstrap", 
"hogan.js", "twemproxy"), class = "factor")), .Names = c("type", 
"created_at", "repository_name"), class = "data.frame", row.names = c(NA, 
-30L))

提出此代码的@flodel 很好地回答了这个问题。

data.split <- split(events.raw$type, events.raw$repository_name)
data.split

list.to.df <- function(arg.list) {
  max.len  <- max(sapply(arg.list, length))
  arg.list <- lapply(arg.list, `length<-`, max.len)
  as.data.frame(arg.list)
}

df.out <- list.to.df(data.split)
df.out

但是，现在我想对列表进行排序，以便每个 repo（repository_name）的事件（类型）每月按一列排序（从“created_at”列中提取），如下所示：

    bootstrap_2012_03   bootstrap_2012_04    hogan.js_2012_03
1    IssuesEvent          PushEvent          PushEvent
2    IssuesEvent          IssuesEvent        IssuesEvent
3    OssueCommentEvent    WatchEvent         IssuesEvent

其他一些假设是：

  • 时间戳仅用于排序，不需要跨行同步
  • 即使“IssuesEvent”重复 10 次，我也需要保留所有这些，因为我将使用 R 包 TraMineR 进行序列分析
  • 列可以不等长
  • 不同存储库的列之间没有关系（“repository_name”）
  • 同一存储库不同月份的数据完全独立

我怎样才能在 R 中做到这一点？

4

1 回答 1

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与其按repository_name列拆分，不如先创建一个结合repository_name了月份的新列：

events.raw$month      <- format(as.Date(events.raw$created_at), "%Y_%m")
events.raw$repo.month <- paste(events.raw$repository_name,
                               events.raw$month, sep = "_")

head(events)
#          type          created_at repository_name   month        repo.month
# 1 IssuesEvent 2012-03-11 06:48:31       bootstrap 2012_03 bootstrap_2012_03
# 2 IssuesEvent 2012-03-11 06:48:31       bootstrap 2012_03 bootstrap_2012_03
# 3 IssuesEvent 2012-03-11 06:48:31       bootstrap 2012_03 bootstrap_2012_03
# 4 IssuesEvent 2012-03-11 06:52:50       bootstrap 2012_03 bootstrap_2012_03
# 5 IssuesEvent 2012-03-11 06:52:50       bootstrap 2012_03 bootstrap_2012_03
# 6 IssuesEvent 2012-03-11 06:52:50       bootstrap 2012_03 bootstrap_2012_03

然后使用我上次建议的相同方法：

data.split <- split(events.raw$type, events.raw$repo.month)

list.to.df <- function(arg.list) {
  max.len  <- max(sapply(arg.list, length))
  arg.list <- lapply(arg.list, `length<-`, max.len)
  as.data.frame(arg.list)
}

df.out <- list.to.df(data.split)
head(df.out)
#    bootstrap_2012_03 hogan.js_2012_03 twemproxy_2012_03
# 1        IssuesEvent       WatchEvent        WatchEvent
# 2        IssuesEvent       WatchEvent        WatchEvent
# 3        IssuesEvent       WatchEvent        WatchEvent
# 4        IssuesEvent             <NA>              <NA>
# 5        IssuesEvent             <NA>              <NA>
# 6        IssuesEvent             <NA>              <NA>
于 2012-08-26T17:20:29.357 回答