Apply different functions to columns of a dataframe selecting functions by name

Question

Let's say I've got a dataframe with multiple columns, some of which I want to transform. The column names define what transformation needs to be used.

library(tidyverse)
set.seed(42)
df <- data.frame(A = 1:100, B = runif(n = 100, 0, 1), log10 = runif(n = 100, 10, 100), log2 = runif(n = 100, 10, 100), log1p = runif(n = 100, 10, 100), sqrt = runif(n = 100, 10, 100))
trans <- list()
trans$log10 <- log10
trans$log2 <- log2
trans$log1p <- log1p
trans$sqrt <- sqrt

Ideally, I would like to use an across call where the column names were matched up with the trans function names and the transformations would be performed on the fly. The desired output is the following:

df_trans <- df %>% 
  dplyr::mutate(log10 = trans$log10(log10),
                log2 = trans$log2(log2),
                log1p = trans$log1p(log1p),
                sqrt = trans$sqrt(sqrt))
df_trans

However, I don't want to manually specify each transformation separately. In the representative example I only have 4 but this number could vary and be significantly higher making manual specification cumbersome and error prone.

I have managed to match up the column names with the functions by turning the trans list into a data frame and left-joining but am then unable to call the function in the trans_function column.

trans_df <- enframe(trans, value = "trans_function")
df %>% 
  pivot_longer(cols = everything()) %>% 
  left_join(trans_df) %>% 
  dplyr::mutate(value = trans_function(value))

Error: Problem with mutate() column value.
i value = trans_function(value).
x could not find function "trans_function"

I think I either need to find a way of calling the functions from the list columns or another way of matching up the function names with the column names. All ideas are welcome.

Ronak Shah · Accepted Answer · 2021-12-19 12:54:54Z

We can use cur_column() in across to get the column name and use it to subset trans.

library(dplyr)

df %>%
  mutate(across(names(trans), ~trans[[cur_column()]](.x))) %>%
  head

#  A         B    log10     log2    log1p     sqrt
#1 1 0.9148060 1.821920 6.486402 3.998918 3.470303
#2 2 0.9370754 1.470472 5.821200 3.932046 7.496103
#3 3 0.2861395 1.469690 6.437524 2.799395 8.171007
#4 4 0.8304476 1.653261 5.639570 3.700698 6.905755
#5 5 0.6417455 1.976905 4.597484 4.500461 9.441077
#6 6 0.5190959 1.985133 5.638341 4.551289 4.440590

Comparing it with output of df_trans.

head(df_trans)

#  A         B    log10     log2    log1p     sqrt
#1 1 0.9148060 1.821920 6.486402 3.998918 3.470303
#2 2 0.9370754 1.470472 5.821200 3.932046 7.496103
#3 3 0.2861395 1.469690 6.437524 2.799395 8.171007
#4 4 0.8304476 1.653261 5.639570 3.700698 6.905755
#5 5 0.6417455 1.976905 4.597484 4.500461 9.441077
#6 6 0.5190959 1.985133 5.638341 4.551289 4.440590

storm surge · Accepted Answer · 2021-12-19 11:52:25Z

One way can be to use lapply:

library(tidyverse)
set.seed(42)

df <- data.frame(A = 1:100, B = runif(n = 100, 0, 1), log10 = runif(n = 100, 10, 100), log2 = runif(n = 100, 10, 100), log1p = runif(n = 100, 10, 100), sqrt = runif(n = 100, 10, 100))
trans <- list()
trans$log10 <- log10
trans$log2 <- log2
trans$log1p <- log1p
trans$sqrt <- sqrt


df_trans <- setNames(lapply(names(df),
            function(x) if(x %in% names(trans))
            { trans[[x]](df[,(x)])} else {df[,x]}),names(df)) %>% 
  bind_cols() %>% 
  as.data.frame() 

head(df_trans)

which gives:

  A         B    log10     log2    log1p     sqrt
  1 1 0.1365052 1.739051 6.301896 4.530600 4.318942
  2 2 0.1771364 1.549601 5.793220 4.521715 3.649834
  3 3 0.5195605 1.902438 4.819125 3.343266 6.788565
  4 4 0.8111208 1.572253 6.219991 4.075945 3.322401
  5 5 0.1153620 1.751276 6.306097 4.060292 7.817301
  6 6 0.8934218 1.724403 6.201123 3.235938 9.749128

The original dataframe being:

head(df)
  A         B    log10     log2    log1p     sqrt
  1 1 0.1365052 54.83409 78.89684 91.81428 18.65326
  2 2 0.1771364 35.44878 55.45401 90.99323 13.32129
  3 3 0.5195605 79.88006 28.22936 27.31143 46.08461
  4 4 0.8111208 37.34675 74.54249 57.90612 11.03835
  5 5 0.1153620 56.39961 79.12693 56.99123 61.11019
  6 6 0.8934218 53.01557 73.57393 24.43022 95.04549

akrun · Accepted Answer · 2021-12-19 17:40:22Z

1

In base R, we may use Map

df[names(trans)] <- Map(function(x, y) x(y), trans, df[names(trans)])

-checking

> identical(df, df_trans)
[1] TRUE

answered Dec 19, 2021 at 17:40

akrun

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Comments

PaulS · Accepted Answer · 2021-12-21 00:14:56Z

1

Another possibility is the following:

library(tidyverse)
set.seed(42)
df <- data.frame(A = 1:100, B = runif(n = 100, 0, 1), log10 = runif(n = 100, 10, 100), log2 = runif(n = 100, 10, 100), log1p = runif(n = 100, 10, 100), sqrt = runif(n = 100, 10, 100))

df %>% 
  mutate(across(-(A:B), ~ getFunction(cur_column())(.x))) %>% head

#>   A         B    log10     log2    log1p     sqrt
#> 1 1 0.9148060 1.821920 6.486402 3.998918 3.470303
#> 2 2 0.9370754 1.470472 5.821200 3.932046 7.496103
#> 3 3 0.2861395 1.469690 6.437524 2.799395 8.171007
#> 4 4 0.8304476 1.653261 5.639570 3.700698 6.905755
#> 5 5 0.6417455 1.976905 4.597484 4.500461 9.441077
#> 6 6 0.5190959 1.985133 5.638341 4.551289 4.440590

answered Dec 21, 2021 at 0:14

PaulS

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