在 dplyr 中使用变量列名汇总 [英] using variable column names in dplyr summarise

问题描述

我发现这个问题已经问过了，但没有正确答案.R 在 dplyr 的汇总函数中使用变量列名

I found this question already asked but without proper answer. R using variable column names in summarise function in dplyr

我想计算两列平均值的差值，但是列名应该是由变量提供的...到目前为止我发现只有函数as.name将列名提供为文本，但这不知何故在这里不起作用......

I want to calculate the difference between two column means, but the column name should be provided by variables... So far I found only the function as.name to provide column names as text, but this somehow doesn't work here...

修复列名就可以了.

x <- c('a','b')
df <- group_by(data.frame(a=c(1,2,3,4), b=c(2,3,4,5), c=c(1,1,2,2)), c)
df %>% summarise(mean(a) - mean(b))

对于可变列，它不起作用

With variable columns, it doesn't work

df %>% summarise(mean(x[1]) - mean(x[2]))
df %>% summarise(mean(as.name(x[1])) - mean(as.name(x[2])))

由于 3 年前就有人问过这个问题，而且 dplyr 正在开发中，我想知道现在是否有答案.

Since this was asked already 3 years ago and dplyr is under good development, I am wondering if there is an answer to this now.

推荐答案

你可以使用 base::get:

df %>% summarise(mean(get(x[1])) - mean(get(x[2])))

# # A tibble: 2 x 2
#        c `mean(a) - mean(b)`
#    <dbl>               <dbl>
# 1     1                  -1
# 2     2                  -1

get 默认会在当前环境中搜索.

get will search in current environment by default.

正如错误消息所说，mean 需要一个逻辑或数字对象，as.name 返回一个名称:

As the error message says, mean expects a logical or numeric object, as.name returns a name:

class(as.name("a")) # [1] "name"

你可以评估你的名字，这也可以:

You could evaluate your name, that would work as well :

df %>% summarise(mean(eval(as.name(x[1]))) - mean(eval(as.name(x[2]))))
# # A tibble: 2 x 2
#       c `mean(eval(as.name(x[1]))) - mean(eval(as.name(x[2])))`
#   <dbl>                                                   <dbl>
# 1     1                                                      -1
# 2     2                                                      -1

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