tidyr 扩展函数如何将变量作为选择列 [英] How can tidyr spread function take variable as a select column

问题描述

tidyr 的扩展函数只接受没有引号的列名.有没有办法可以传入一个包含列名的变量例如

tidyr's spread function only takes column names without quotes. Is there a way I can pass in a variable that contains the column name for eg

# example using gather()
library("tidyr")
dummy.data <- data.frame("a" = letters[1:25], "B" = LETTERS[1:5], "x" = c(1:25))
dummy.data
var = "x"
dummy.data %>% gather(key, value, var)

出现错误

Error: All select() inputs must resolve to integer column positions.
The following do not:
*  var

这是使用匹配函数解决的，它给出了所需的列位置

Which is solved using match function which gives the required column position

dummy.data %>% gather(key, value, match(var, names(.)))

但同样的方法不适用于传播函数

But this same approach doesn't work for the spread function

dummy.data %>% spread(a, match(var, names(.)))
Error: Invalid column specification

收集和扩展函数是否采用不同的列规范.gather 需要一个列索引，而 spread 没有提到它想要什么

Do gather and spread functions take different column specification. gather takes a column index while spread doesn't mention what it wants

推荐答案

如果您想使用标准评估，您需要使用 gather_ 或 spread_

If you want to use standard evaluation you need to use gather_ or spread_

这两个给出相同的结果

dummy.data %>% gather_("key", "value", var)
dummy.data %>% gather(key, value, match(var, names(.)))

这有效:

dummy.data %>% spread_("a",var)
#   B  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y
# 1 A  1 NA NA NA NA  6 NA NA NA NA 11 NA NA NA NA 16 NA NA NA NA 21 NA NA NA NA
# 2 B NA  2 NA NA NA NA  7 NA NA NA NA 12 NA NA NA NA 17 NA NA NA NA 22 NA NA NA
# 3 C NA NA  3 NA NA NA NA  8 NA NA NA NA 13 NA NA NA NA 18 NA NA NA NA 23 NA NA
# 4 D NA NA NA  4 NA NA NA NA  9 NA NA NA NA 14 NA NA NA NA 19 NA NA NA NA 24 NA
# 5 E NA NA NA NA  5 NA NA NA NA 10 NA NA NA NA 15 NA NA NA NA 20 NA NA NA NA 25

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