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rfunctiondplyrrlang
Ana
Ana

Reputation: 1586

how to pass a column name to a function

I have written a function that does a few analysis including calculating a measure called "Net Promoter Score" using NPS package.

library(dplyr)
library(tidyr)
library(NPS)
df<-data.frame(score = sample(c(0:10),15,replace=TRUE),
           variable = sample(c('A', 'B', 'C'),15,replace=TRUE)
)
analyzer <- function(df,var, sco){
    df %>% group_by_(var) %>% transmute(n= nps(sco)) %>% unique()
}
analyzer(df,'variable','score')

This returns NA for all levels of variable.

Now dplyr functions have a way of dealing with x being handed to them as character(i.e., their _ version which I've used here), but the nps function doesn't. I also tried passing score column as nps(.[[score]]) but this returns the NPS for the whole column and doesn't break it down by group_by levels.

Upvotes: 0

Views: 136

Answers (2)

Indrajeet Patil
Indrajeet Patil

Reputation: 4879

It's because the inputs to functions are not getting evaluated properly-

(Note that the way it is implemented here, the function will work irrespective of whether you enter a bare expression x = x or a character x = "x")

library(dplyr)
library(tidyr)
library(NPS)
set.seed(123)

# data
df <- data.frame(score = sample(c(0:10), 15, replace = TRUE),
                 variable = sample(c('A', 'B', 'C'), 15, replace = TRUE))

# custom function
analyzer <- function(df, var, sco) {
  var <- rlang::ensym(var)
  sco <- rlang::ensym(sco)

  df <- df %>% 
    group_by(., !!rlang::enquo(var)) %>% 
    transmute(., n = NPS::nps(!!rlang::enquo(sco))) %>% 
    unique()

  return(df)
}

# using function
analyzer(df, 'variable', 'score')
#> # A tibble: 3 x 2
#> # Groups:   variable [3]
#>   variable      n
#>   <fct>     <dbl>
#> 1 C        -0.333
#> 2 A        -0.4  
#> 3 B        -0.25

Created on 2018-11-18 by the reprex package (v0.2.1)

Upvotes: 1

realrbird
realrbird

Reputation: 171

You can use the rlang package and then uquote your variables with !!

library(dplyr)
library(tidyr)
library(NPS)
library(rlang)

df<-data.frame(score = sample(c(0:10),15,replace=TRUE),
           variable = sample(c('A', 'B', 'C'),15,replace=TRUE)
)
analyzer <- function(df,var, sco){
    var <- rlang::enquo(var)
    sco <- rlang::enquo(sco)
    df %>% group_by(!!var) %>% transmute(n= nps(!!sco)) %>% unique()
}
analyzer(df, variable,score)

Upvotes: 6

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