如何将dplyr中的动态列名传递给自定义函数？

I have a dataset with the following structure:

我有一个具有以下结构的数据集：

Classes ‘tbl_df’ and 'data.frame':  10 obs. of  7 variables:
 $ GdeName  : chr  "Aeugst am Albis" "Aeugst am Albis" "Aeugst am Albis" "Aeugst am Albis" ...
 $ Partei   : chr  "BDP" "CSP" "CVP" "EDU" ...
 $ Stand1971: num  NA NA 4.91 NA 3.21 ...
 $ Stand1975: num  NA NA 5.389 0.438 4.536 ...
 $ Stand1979: num  NA NA 6.2774 0.0195 3.4355 ...
 $ Stand1983: num  NA NA 4.66 1.41 3.76 ...
 $ Stand1987: num  NA NA 3.48 1.65 5.75 ...

I want to provide a function which allows to compute the difference between any value, and I would like to do this using dplyrs mutate function like so: (assume the parameters from and to are passed as arguments)

我想提供一个允许计算任何值之间差异的函数，我想使用dplyrs mutate函数这样做:(假设参数from和to作为参数传递）

from <- "Stand1971"
to <- "Stand1987"

data %>%
  mutate(diff = from - to)

Of course, this doesn't work, as dplyr uses non-standard evaluation. And I know there's now an elegant solution to the problem using mutate_, and I've read this vignette, but I still can't get my head around it.

当然，这不起作用，因为dplyr使用非标准评估。而且我知道使用mutate_现在有一个优雅的问题解决方案，我已经阅读了这个小插图，但我仍然无法理解它。

What to do?

该怎么办？

Here's the first few rows of the dataset for a reproducible example

这是可重现示例的数据集的前几行

structure(list(GdeName = c("Aeugst am Albis", "Aeugst am Albis", 
"Aeugst am Albis", "Aeugst am Albis", "Aeugst am Albis", "Aeugst am Albis", 
"Aeugst am Albis", "Aeugst am Albis", "Aeugst am Albis", "Aeugst am Albis"
), Partei = c("BDP", "CSP", "CVP", "EDU", "EVP", "FDP", "FGA", 
"FPS", "GLP", "GPS"), Stand1971 = c(NA, NA, 4.907306434, NA, 
3.2109535926, 18.272143463, NA, NA, NA, NA), Stand1975 = c(NA, 
NA, 5.389079711, 0.4382328556, 4.5363022622, 18.749259742, NA, 
NA, NA, NA), Stand1979 = c(NA, NA, 6.2773722628, 0.0194647202, 
3.4355231144, 25.294403893, NA, NA, NA, 2.7055961071), Stand1983 = c(NA, 
NA, 4.6609804428, 1.412940467, 3.7563539244, 26.277246489, 0.8529335746, 
NA, NA, 2.601878177), Stand1987 = c(NA, NA, 3.4767860929, 1.6535933856, 
5.7451770193, 22.146844746, NA, 3.7453183521, NA, 13.702211858
)), .Names = c("GdeName", "Partei", "Stand1971", "Stand1975", 
"Stand1979", "Stand1983", "Stand1987"), class = c("tbl_df", "data.frame"
), row.names = c(NA, -10L))

2 个解决方案

#1

Using the latest version of dplyr (>=0.7), you can use the rlang !! (bang-bang) operator.

使用最新版本的dplyr（> = 0.7），你可以使用rlang !! （bang-bang）运营商。

library(tidyverse)
from <- "Stand1971"
to <- "Stand1987"

data %>%
  mutate(diff=(!!as.name(from))-(!!as.name(to)))

You just need to convert the strings to names with as.name and then insert them into the expression. Unfortunately I seem to have to use a few more parenthesis than I would like, but the !! operator seems to fall in a weird order-of-operations order.

您只需要将字符串转换为带有as.name的名称，然后将它们插入到表达式中。不幸的是，我似乎不得不使用比我想要的更多的括号，但!!运营商似乎陷入了一种奇怪的运营秩序。

Original answer, dplyr (0.3-<0.7):

原始答案，dplyr（0.3- <0.7）：

From that vignette (vignette("nse","dplyr")), use lazyeval's interp() function

从那个插图（vignette（“nse”，“dplyr”）），使用lazyeval的interp（）函数

library(lazyeval)

from <- "Stand1971"
to <- "Stand1987"

data %>%
  mutate_(diff=interp(~from - to, from=as.name(from), to=as.name(to)))

#2

Why not simply paste?

为什么不简单粘贴？

data %>%
  mutate_(diff = paste(from, "-", to))

#1