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Pass function arguments by column position to mutate_at

Tags:

r

dplyr

purrr

I'm trying to tighten up a %>% piped workflow where I need to apply the same function to several columns but with one argument changed each time. I feel like purrr's map or invoke functions should help, but I can't wrap my head around it.

My data frame has columns for life expectancy, poverty rate, and median household income. I can pass all these column names to vars in mutate_at, use round as the function to apply to each, and optionally supply a digits argument. But I can't figure out a way, if one exists, to pass different values for digits associated with each column. I'd like life expectancy rounded to 1 digit, poverty rounded to 2, and income rounded to 0.

I can call mutate on each column, but given that I might have more columns all receiving the same function with only an additional argument changed, I'd like something more concise.

library(tidyverse)

df <- tibble::tribble(
        ~name, ~life_expectancy,          ~poverty, ~household_income,
  "New Haven", 78.0580437642378, 0.264221051111753,  42588.7592521085
  )

In my imagination, I could do something like this:

df %>%
  mutate_at(vars(life_expectancy, poverty, household_income), 
            round, digits = c(1, 2, 0))

But get the error

Error in mutate_impl(.data, dots) : Column life_expectancy must be length 1 (the number of rows), not 3

Using mutate_at instead of mutate just to have the same syntax as in my ideal case:

df %>%
  mutate_at(vars(life_expectancy), round, digits = 1) %>%
  mutate_at(vars(poverty), round, digits = 2) %>%
  mutate_at(vars(household_income), round, digits = 0)
#> # A tibble: 1 x 4
#>   name      life_expectancy poverty household_income
#>   <chr>               <dbl>   <dbl>            <dbl>
#> 1 New Haven            78.1    0.26            42589

Mapping over the digits uses each of the digits options for each column, not by position, giving me 3 rows each rounded to a different number of digits.

df %>%
  mutate_at(vars(life_expectancy, poverty, household_income), 
            function(x) map(x, round, digits = c(1, 2, 0))) %>%
  unnest()
#> # A tibble: 3 x 4
#>   name      life_expectancy poverty household_income
#>   <chr>               <dbl>   <dbl>            <dbl>
#> 1 New Haven            78.1    0.3            42589.
#> 2 New Haven            78.1    0.26           42589.
#> 3 New Haven            78      0              42589

Created on 2018-11-13 by the reprex package (v0.2.1)

like image 349
camille Avatar asked Nov 13 '18 19:11

camille


3 Answers

2 solutions


mutate with !!!

invoke was a good idea but you need it less now that most tidyverse functions support the !!! operator, here's what you can do :

digits <- c(life_expectancy = 1, poverty = 2, household_income = 0)  
df %>% mutate(!!!imap(digits, ~round(..3[[.y]], .x),.))
# # A tibble: 1 x 4
#          name life_expectancy poverty household_income
#         <chr>           <dbl>   <dbl>            <dbl>
#   1 New Haven            78.1    0.26            42589

..3 is the initial data frame, passed to the function as a third argument, through the dot at the end of the call.

Written more explicitly :

df %>% mutate(!!!imap(
  digits, 
  function(digit, name, data) round(data[[name]], digit),
  data = .))

If you need to start from your old interface (though the one I propose will be more flexible), first do:

digits <- setNames(c(1, 2, 0), c("life_expectancy", "poverty", "household_income"))

mutate_at and <<-

Here we bend a bit the good practice of avoiding <<- whenever possible, but readability matters and this one is really easy to read.

digits <- c(1, 2, 0)
i <- 0
df %>%
  mutate_at(vars(life_expectancy, poverty, household_income), ~round(., digits[i<<- i+1]))
# A tibble: 1 x 4
#     name      life_expectancy poverty household_income
#     <chr>               <dbl>   <dbl>            <dbl>
#   1 New Haven            78.1    0.26            42589

(or just df %>% mutate_at(names(digits), ~round(., digits[i<<- i+1])) if you use a named vector as in my first solution)

like image 118
Moody_Mudskipper Avatar answered Nov 18 '22 18:11

Moody_Mudskipper


Here's a map2 solution along the lines of Henrik's comment. You can then wrap this inside a custom function. I provided an rough first attempt but I have done minimal tests, so it probably breaks under all sorts of situations if evaluation is strange. It also doesn't use tidyselect for .at, but neither does modify_at...

library(tidyverse)

df <- tibble::tribble(
  ~name, ~life_expectancy,          ~poverty, ~household_income,
  "New Haven", 78.0580437642378, 0.264221051111753,  42588.7592521085,
  "New York", 12.349685329, 0.324067934, 32156.230974623
)

rounded <- df %>%
  select(life_expectancy, poverty, household_income) %>%
  map2_dfc(
    .y = c(1, 2, 0),
    .f = ~ round(.x, digits = .y)
  )
df %>%
  select(-life_expectancy, -poverty, -household_income) %>%
  bind_cols(rounded)
#> # A tibble: 2 x 4
#>   name      life_expectancy poverty household_income
#>   <chr>               <dbl>   <dbl>            <dbl>
#> 1 New Haven            78.1    0.26            42589
#> 2 New York             12.3    0.32            32156


modify2_at <- function(.x, .y, .at, .f) {
  modified <- .x[.at] %>%
    map2(.y, .f)
  .x[.at] <- modified
  return(.x)
}

df %>%
  modify2_at(
    .y = c(1, 2, 0),
    .at = c("life_expectancy", "poverty", "household_income"),
    .f = ~ round(.x, digits = .y)
  )
#> # A tibble: 2 x 4
#>   name      life_expectancy poverty household_income
#>   <chr>               <dbl>   <dbl>            <dbl>
#> 1 New Haven            78.1    0.26            42589
#> 2 New York             12.3    0.32            32156

Created on 2018-11-13 by the reprex package (v0.2.1)

like image 2
Calum You Avatar answered Nov 18 '22 18:11

Calum You


Fun with tidyeval:

prepared_pairs <- 
  map2(
    set_names(syms(list("life_expectancy", "poverty", "household_income"))),
    c(1, 2, 0), 
    ~expr(round(!!.x, digits = !!.y))
  )

mutate(df, !!! prepared_pairs)

# # A tibble: 1 x 4
#   name      life_expectancy poverty household_income
#   <chr>               <dbl>   <dbl>            <dbl>
# 1 New Haven            78.1    0.26            42589
like image 2
Aurèle Avatar answered Nov 18 '22 17:11

Aurèle