将dplyr中的动态列名称传递给自定义函数？ [英] Hot to pass dynamic column names in dplyr into custom function?

问题描述

我有一个具有以下结构的数据集：

 类'tbl_df'和'data.frame'：10个obs。的7个变量：
 $ GdeName：chrAeugst am AlbisAeugst am AlbisAeugst am AlbisAeugst am Albis... 
 $ Partei：chrBDPCSP CVPEDU... 
 $ 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 ... 
  / pre> 
 
 
我想提供一个允许计算任何值之间的差异的函数，我想使用 dplyr  s  mutate 函数如下：（假设参数从和 作为参数传递）
 从<  - Stand1971
到< - Stand1987
 
 data％>％
 mutate（diff = from  -  to）
  
当然，这不起作用，因为 dplyr 使用非标准评估。而且我知道现在使用 mutate _ 这个问题的优雅解决方案，我已经阅读了这个小插曲，但我仍然不能让我的头脑。 
 
 
 
该怎么办？ 
 
 
 
以下是可重现示例的数据集的前几行
 结构（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 ，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，4.6609804428,1.412940467,37563539244,26.277246489,0.8529335746，
 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））
  
 
 
解决方案
从该vignette（ vignette（nse，dplyr）），使用lazyeval的 interp（） code> function 
  library（lazyeval）
 
 from<  - Stand1971
到<  - Stand1987
 
 data％>％
 mutate_（diff = interp（〜from  -  to，from = as.name（from），to = as .name（to）））
  
 
I have a dataset with the following structure:
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)
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. 


What to do? 

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

解决方案
From that vignette (vignette("nse","dplyr")), use lazyeval's interp() function
library(lazyeval)

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

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


                        
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