使用r中dplyr的mutate_all（）将df中的每一列除以其他列

Question

I am trying to divide each column in a dataframe by each other column. Starting with a dataframe with columns A,B,C. I would like to end with a dataframe with columns B/A,C/A,A/B,C/B,A/C, and B/C. I have found a way to do this which requires me to write a function for each denominator, but I would prefer it if I could do so by defining a single function.

I have found a similar, but not identical question answered here. Dividing each column by the column before it in R

If possible I would like to use dplyr to solve this.

Here is the closest I have got.

## create example data frame

df <- data.frame(A=rnorm(10),
                 B=rnorm(10),
                 C=rnorm(10),)

## calculate ratios

ratio_df <- df%>%
     mutate_all(.funs = funs(A=./A,
                             B=./B,
                             C=./C))

This will return the desired results with columns A/A,B/A,C/A,A/B,B/B,C/B,A/C,B/C,C/C. I can easily filter out A/A,B/B, and C/C giving the desired result, but it is clunky when there are a large number of columns.

Is there a way to accomplish this without writing each individual function?

Answer 1

Here's an answer using purrr .

library(purrr)
library(dplyr)

df <- data.frame(A=rnorm(10),
                 B=rnorm(10),
                 C=rnorm(10))

df %>% map(~ . / df) %>% bind_cols()

Answer 2

What about:

do.call(cbind, lapply(df, function(x) x / df))

Output:

   A.A         A.B        A.C         B.A B.B        B.C        C.A        C.B C.C
1    1  0.38056317  0.4701251   2.6276847   1  1.2353406  2.1270933  0.8094934   1
2    1 -1.07851585 -1.0793270  -0.9272001   1  1.0007521 -0.9265032  0.9992484   1
3    1 -2.44512434 -4.7467554  -0.4089772   1  1.9413145 -0.2106702  0.5151149   1
4    1 -1.41765991 -2.3908820  -0.7053878   1  1.6864990 -0.4182557  0.5929443   1
5    1 -0.10640354  0.2382367  -9.3981837   1 -2.2389920  4.1975066 -0.4466296   1
6    1  2.32474492  0.3879095   0.4301547   1  0.1668611  2.5779212  5.9930092   1
7    1  0.95122898  1.6102188   1.0512716   1  1.6927773  0.6210336  0.5907452   1
8    1 -0.15996970 -0.1422422  -6.2511840   1  0.8891823 -7.0302611  1.1246287   1
9    1 -0.04431519  0.2121086 -22.5656275   1 -4.7863640  4.7145657 -0.2089269   1
10   1  0.26662240 -0.3524232   3.7506226   1 -1.3218065 -2.8374974 -0.7565404   1

Answer 3

Another answe using purrr which will also have the columns named as you specified:

library(tidyverse)

df %>% 
  select_if(is.numeric) %>% 
  map(~./df %>% select_if(is.numeric)) %>% 
  imap(~set_names(.x, paste0(names(.x), "_", .y))) %>% 
  bind_cols()