如何对除一列以外的所有列进行分组？ [英] How to group by all but one columns?

问题描述

如何告诉 group_by 将数据按给定列以外的所有列分组？

How do I tell group_by to group the data by all columns except a given one?

使用 aggregate ，它将是 aggregate（x〜。，...）。

我尝试了 group_by（data，-x），但是按x的负数分组（即与按x分组相同）。 / p>

I tried group_by(data, -x), but that groups by the negative-of-x (i.e. the same as grouping by x).

推荐答案

您可以使用标准评估（ group_by _ 代替 group_by ）：

You can do this using standard evaluation (group_by_ instead of group_by):

# Fake data
set.seed(492)
dat = data.frame(value=rnorm(1000), g1=sample(LETTERS,1000,replace=TRUE),
                 g2=sample(letters,1000,replace=TRUE), g3=sample(1:10, replace=TRUE),
                 other=sample(c("red","green","black"),1000,replace=TRUE))

dat %>% group_by_(.dots=names(dat)[-grep("value", names(dat))]) %>%
  summarise(meanValue=mean(value))

       g1     g2    g3  other   meanValue
   <fctr> <fctr> <int> <fctr>       <dbl>
1       A      a     2  green  0.89281475
2       A      b     2    red -0.03558775
3       A      b     5  black -1.79184218
4       A      c    10  black  0.17518610
5       A      e     5  black  0.25830392
...

请参见此插图，以了解有关标准与非标准评估的更多信息 dplyr 。

See this vignette for more on standard vs. non-standard evaluation in dplyr.

解决@ÖmerAn的评论：看来 group_by_at 是进入 dplyr的方式 0.7.0（如果我对此有误，请纠正我）。例如：

To address @ÖmerAn's comment: It looks like group_by_at is the way to go in dplyr 0.7.0 (someone please correct me if I'm wrong about this). For example:

dat %>% 
  group_by_at(setdiff(names(dat), "value")) %>%
  summarise(meanValue=mean(value))

# Groups:   g1, g2, g3 [?]
       g1     g2    g3  other   meanValue
   <fctr> <fctr> <int> <fctr>       <dbl>
 1      A      a     2  green  0.89281475
 2      A      b     2    red -0.03558775
 3      A      b     5  black -1.79184218
 4      A      c    10  black  0.17518610
 5      A      e     5  black  0.25830392
 6      A      e     5    red -0.81879788
 7      A      e     7  green  0.30836054
 8      A      f     2  green  0.05537047
 9      A      g     1  black  1.00156405
10      A      g    10  black  1.26884303
# ... with 949 more rows

让我们确认两种方法都给出相同的输出（在 dplyr 0.7.0中）：

Let's confirm both methods give the same output (in dplyr 0.7.0):

new = dat %>% 
  group_by_at(setdiff(names(dat), "value")) %>%
  summarise(meanValue=mean(value))

old = dat %>% 
  group_by_(.dots=names(dat)[-grep("value", names(dat))]) %>%
  summarise(meanValue=mean(value))

identical(old, new)
# [1] TRUE

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