使用dplyr总结R中的未知数列 [英] Summarizing unknown number of column in R using dplyr

问题描述

我有以下data.frame（df）

I have following data.frame (df)

ID1 ID2 Col1 Col2 Col3 Grp
A   B   1    3    6    G1
C   D   3    5    7    G1
E   F   4    5    7    G2
G   h   5    6    8    G2

我想要实现的如下：
- 由Grp组，easy
- 然后总结，以便对于每个组，我总结列并创建列具有所有ID1和ID2的字符串

What I would like to achieve is the following: - group by Grp, easy - and then summarize so that for each group I sum the columns and create the columns with strings with all ID1s and ID2s

这将是这样的：

df %>% 
   group_by(Grp) %>% 
      summarize(ID1s=toString(ID1), ID2s=toString(ID2), Col1=sum(Col1), Col2=sum(Col2), Col3=sum(Col3))

一切都很好，我知道列数Col1，Col2，Col3），但是我希望能够实现它，以便它可以用于具有已知且总是命名为相同的ID1，ID2，Grp和任何数量的具有未知名称的附加数字列的数据帧。

Everything is fine whae Iknow the number of the columns (Col1, Col2, Col3), however I would like to be able to implement it so that it would work for a data frame with known and always named the same ID1, ID2, Grp, and any number of additional numeric column with unknown names.

有没有办法在dplyr中执行。

Is there a way to do it in dplyr.

推荐答案

喜欢能够实现它，以便它可以用于已知的数据帧，并且始终命名为相同的ID1，ID2，Grp和任何数量的具有未知名称的附加数字列。

I would like to be able to implement it so that it would work for a data frame with known and always named the same ID1, ID2, Grp, and any number of additional numeric column with unknown names.

您可以先覆盖ID列，然后再分组：

You can overwrite the ID columns first and then group by them as well:

DF %>% 
  group_by(Grp) %>% mutate_each(funs(. %>% unique %>% sort %>% toString), ID1, ID2) %>% 
  group_by(ID1, ID2, add=TRUE) %>% summarise_each(funs(sum))

# Source: local data frame [2 x 6]
# Groups: Grp, ID1 [?]
# 
#     Grp   ID1   ID2  Col1  Col2  Col3
#   (chr) (chr) (chr) (int) (int) (int)
# 1    G1  A, C  B, D     4     8    13
# 2    G2  E, G  F, h     9    11    15

我想你会在折叠到一个字符串之前统一和排序，所以我添加了这些步骤。

I think you'll want to uniqify and sort before collapsing to a string, so I've added those steps.

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