CODEWITHSUNDEEP
rdplyr
alexwhan
alexwhan

Reputation: 16026

Construct variable names in select_

I am trying to write a function that will (in part) rename a variable by combining its source dataframe and existing variable name. In essence, I want:

df1 <- data.frame(a = 1, b = 2)

to become:

df1 %>%
  rename(df1_a = a)

#  df1_a b
#1     1 2

But I want to do this programatically, something along the lines of:

fun <- function(df, var) {
  outdf <- rename_(df, paste(df, var, sep = "_") = var)
  return(outdf)
}

This admittedly naive approach obviously doesn't work, but I haven't been able to figure it out. I'm sure the answer is somewhere in the nse vignette (https://cran.r-project.org/web/packages/dplyr/vignettes/nse.html), but that doesn't seem to address constructing variable names.

Upvotes: 6

Views: 115

Answers (2)

bramtayl
bramtayl

Reputation: 4024

lazyeval isn't really needed here because the environment of both inputs is known. That being said:

library(lazyeval)
library(dplyr)
library(magrittr)

fun = function(df, var) {
  df_ = lazy(df)
  var_ = lazy(var)
  fun_(df_, var_)
}

fun_ = function(df_, var_) {

  new_var_string = 
    paste(df_ %>% as.character %>% extract(1),
          var_ %>% as.character %>% extract(1), 
          sep = "_")

  dots = list(var_) %>% setNames(new_var_string)

  df_ %>%
    lazy_eval %>%
    rename_(.dots = dots)
}

fun(df1, a)

Upvotes: 0

Rich Scriven
Rich Scriven

Reputation: 99331

Not sure if this is the proper dplyr-esque way, but it'll get you going.

fun <- function(df, var) {
    x <- deparse(substitute(df))
    y <- deparse(substitute(var))
    rename_(df, .dots = with(df, setNames(as.list(y), paste(x, y, sep = "_"))))
}

fun(df1, a)
#   df1_a b
# 1     1 2

fun(df1, b)
#   a df1_b
# 1 1     2

Upvotes: 8

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