pairwise subtraction of columns in a dataframe in R

Question

I was wondering is there a way to automate (e.g., loop) the subtraction of (X2-X1), (X3-X1), (X3-X2) in my data below and add them as three new columns to the data?

m="
id X1 X2 X3
A  1  0  4
B  2  2  2
C  3  4  1"

data <- read.table(text = m, h = T)

Maël · Answer 1 · 2022-11-10T16:58:15.280

1

With combn:

dif <- combn(data[-1], 2, \(x) x[, 2] - x[, 1])
colnames(dif) <- combn(names(data)[-1], 2, \(x) paste(x[2], x[1], sep = "-"))

cbind(data, dif)
#  id X1 X2 X3 X2-X1 X3-X1 X3-X2
#1  A  1  0  4    -1     3     4
#2  B  2  2  2     0     0     0
#3  C  3  4  1     1    -2    -3

edited Nov 10 '22 at 16:58

answered Nov 10 '22 at 16:44

Maël

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Thanks! it just seems a bit too complicated, is there a way to possibly simplify the code? – Reza Nov 10 '22 at 16:46
Well, if you already know the columns index, you can replace them in `cols`. Otherwise, if a list is okay, you can skip the `do.call` – Maël Nov 10 '22 at 16:48
It's much simpler now – Maël Nov 10 '22 at 16:55
1

This is very nice! +1 – M-- Nov 10 '22 at 16:58

score 1 · Accepted Answer · answered Nov 10 '22 at 16:52

This is very similar to this question; we basically just need to change the function that we are using in map2_dfc:

library(tidyverse)

combn(names(data)[-1], 2) %>% 
  map2_dfc(.x = .[1,], .y = .[2,], 
           .f = ~transmute(data, !!paste0(.y, "-", .x) := !!sym(.y) - !!sym(.x))) %>% 
  bind_cols(data, .)

#>   id X1 X2 X3 X2-X1 X3-X1 X3-X2
#> 1  A  1  0  4    -1     3     4
#> 2  B  2  2  2     0     0     0
#> 3  C  3  4  1     1    -2    -3

pairwise subtraction of columns in a dataframe in R

2 Answers2