Map dplyr function to each combination of variable pairs in an R dataframe

Question

I want to map a function to each combination pair of variables in a dataframe in R, returning a dataframe with the function output for each pair. I can do this manually like so:

library(tidyverse)

df <- tibble(a = c(1, 2), b = c(4, 3), c = c(5, 7))

f <- function(a, b) a - b # a simple function for sake of example

df %>% transmute(a_minus_b = f(a, b),
                 a_minus_c = f(a, c),
                 b_minus_c = f(b, c),
                 b_minus_a = f(b, a),
                 c_minus_a = f(c, a),
                 c_minus_b = f(c, b))

Doing this manually is obviously impractical for a dataframe with many variables. How can I apply my function to each combination pair of variables using iteration?

Answer 1

Another approach using dplyr and purrr may look like so:

library(tidyverse)

df <- tibble(a = c(1, 2), b = c(4, 3), c = c(5, 7))

f <- function(a, b) a - b # a simple function for sake of example

f_help <- function(x) {
  df %>% 
    transmute_at(setdiff(names(.), x), ~ f(!!sym(x), .x)) %>%
    rename_all(.funs = ~ paste0(x, "_minus_", .x))
}

map(names(df), f_help) %>% 
  bind_cols()
#> # A tibble: 2 x 6
#>   a_minus_b a_minus_c b_minus_a b_minus_c c_minus_a c_minus_b
#>       <dbl>     <dbl>     <dbl>     <dbl>     <dbl>     <dbl>
#> 1        -3        -4         3        -1         4         1
#> 2        -1        -5         1        -4         5         4

Answer 2

One dplyr and purrr solution could be:

map_dfc(.x = c(combn(rev(names(df)), 2, simplify = FALSE),
               combn(names(df), 2, simplify = FALSE)),
        ~ df %>%
         rowwise() %>%
         transmute(!!paste(.x, collapse = "_") := reduce(c_across(all_of(.x)), `-`)) %>%
         ungroup())

    c_b   c_a   b_a   a_b   a_c   b_c
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     4     3    -3    -4    -1
2     4     5     1    -1    -5    -4

Or using the specified function:

map_dfc(.x = c(combn(rev(names(df)), 2, simplify = FALSE),
               combn(names(df), 2, simplify = FALSE)),
        ~ df %>%
         rowwise() %>%
         transmute(!!paste(.x, collapse = "_") := reduce(c_across(all_of(.x)), f)) %>%
         ungroup())

Answer 3

a tidyverse using set_names

library(tidyverse)
f <- function(a, b) a - b # a simple function for sake of example
c(combn(df, 2, simplify = F),
  combn(rev(df), 2, simplify = F)) %>% 
  set_names(map_chr(., ~paste(names(.), collapse = "_minus_"))) %>% 
  map(., ~f(.x[1], .x[2]) %>% pull) %>%   
  bind_cols()
 # A tibble: 2 x 6
  a_minus_b a_minus_c b_minus_c c_minus_b c_minus_a b_minus_a
      <dbl>     <dbl>     <dbl>     <dbl>     <dbl>     <dbl>
1        -3        -4        -1         1         4         3
2        -1        -5        -4         4         5         1

Answer 4

Here is a base R solution that does what you want:

# Create combination
combos <- combn(names(df), 2, simplify = F)
combos <- c(combos, lapply(combos, rev))

# Apply function to each element of combos and add names
as.data.frame(lapply(combos, function(j) `names<-`(f(df[j[1]], df[j[2]]), paste0(j, collapse = "_minus_"))))
  a_minus_b a_minus_c b_minus_c b_minus_a c_minus_a c_minus_b
1        -3        -4        -1         3         4         1
2        -1        -5        -4         1         5         4


# Same thing but easier to read
l <- lapply(combos, function(j) {
  res <- f(df[j[1]], df[j[2]])
  names(res) <- paste0(j, collapse = "_minus_")
  res
})

as.data.frame(l)

Or, if you want the purrr equivalent:

# Tidyverse equivalent
map_dfc(combos, ~ `names<-`(f(df[.[1]], df[.[2]]), paste0(., collapse = "_minus_")))

Answer 5

via macro programming and data.table

library(data.table)
setDT(df)

df_combn <- combn(names(df),2,simplify=FALSE)
f_vector <- lapply(df_combn,function(x){paste0("f(",x[1],",",x[2],")")})
f_vector_scoped <- paste0("df[,",f_vector,"]")

out_names <- sapply(df_combn,paste0,collapse="_minus_")

for(i in 1:length(f_vector)){
  set(df,j=out_names[i],value=eval(parse(text=f_vector_scoped[i])))
}