提迪尔在多个列上使用了split_rows [英] tidyr use separate_rows over multiple columns

问题描述

我有一个data.frame，其中某些单元格包含逗号分隔值的字符串:

I have a data.frame where some cells contain strings of comma separate values:

d <- data.frame(a=c(1:3), 
       b=c("name1, name2, name3", "name4", "name5, name6"),
       c=c("name7","name8, name9", "name10" ))

我想将每个名称都拆分成自己的单元格的字符串分开.

I want to separate those strings where each name is split into its own cell. This is easy with

tidyr::separate_rows(d, b, sep=",") 

如果一次只完成一列.但是我不能同时对"b"和"c"两列执行此操作，因为这要求每个字符串中的名称数相同.不用写

if it is done for one column a time. But I can't do this for both columns "b" and "c" at the same time, since it requires that the number of names in each string is the same. Instead of writing

tidyr::separate_rows(d, b, sep=",") 
tidyr::separate_rows(d, c, sep=",") 

有没有一种方法可以做到这一点，例如与申请?像

Is there a way to do this in a one-liner, for e.g. with apply? Something like

apply(d, 2, separate_rows(...)) 

不确定如何将参数传递给separate_rows()函数.

Not sure how to pass the arguments to the separate_rows() function.

推荐答案

您可以使用管道.请注意，sep = ", "是自动检测到的.

You can use a pipe. Note that sep = ", " is automatically detected.

d %>% separate_rows(b) %>% separate_rows(c)
#   a     b      c
# 1 1 name1  name7
# 2 1 name2  name7
# 3 1 name3  name7
# 4 2 name4  name8
# 5 2 name4  name9
# 6 3 name5 name10
# 7 3 name6 name10

注意:使用tidyr版本0.6.0，其中%>%运算符包含在软件包中.

Note: Using tidyr version 0.6.0, where the %>% operator is included in the package.

更新:使用@doscendodiscimus注释，我们可以使用for()循环并在每次迭代中重新分配d.这样，我们可以拥有尽可能多的列.我们将使用列名称的字符向量，因此我们需要切换到标准评估版separate_rows_.

Update: Using @doscendodiscimus comment, we could use a for() loop and reassign d in each iteration. This way we can have as many columns as we like. We will use a character vector of column names, so we'll need to switch to the standard evaluation version, separate_rows_.

cols <- c("b", "c")
for(col in cols) {
    d <- separate_rows_(d, col)
}

提供更新的d

  a     b      c
1 1 name1  name7
2 1 name2  name7
3 1 name3  name7
4 2 name4  name8
5 2 name4  name9
6 3 name5 name10
7 3 name6 name10

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