Create the mean for all possible combinations of 5 different variables

Question

I have a data set with 5 different variables (e.g., a, b, c, d, and e). I want to have simple code (preferably with the tidyverse) that allows me to take the mean for each possible combination of the variables. For example, the mean of "ab", "ac", ..., all the way to "abcde". Is there a simple way of doing this?

All I've tried is manually creating the code for each variable. However, it seems like something like a loop would be more appropriate.

For example, if my data looked like this:

a <- rnorm(10, 0, 1)
b <- rnorm(10, 0, 1)
c <- rnorm(10, 0, 1)
d <- rnorm(10, 0, 1)
e <- rnorm(10, 0, 1)
data <- cbind.data.frame(a,b,c,d,e)

I want the data to look like the output as if I had done this for each combination of a, b, c, d, e:

data$ab <- (data$a + data$b)/2
.
.
.
data$abcde <- (data$a + data$b + data$c + data$d + data$e)/5

What's your dataset look like? Can you provide some sample data? See: https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example — JasonAizkalns, Jun 02 '19 at 12:43
What do you want the *output* to look like? A column of column names with a column of the mean products? In any order? — Spacedman, Jun 02 '19 at 12:58
I have added a description for what I'd like the output to look like — user2917781, Jun 02 '19 at 13:02
You can use `combn` to get the column indexes in each product, but you'll have to loop over the combination sizes from 2 to 5. Does that help? — Spacedman, Jun 02 '19 at 13:03

score 3 · Accepted Answer · answered Jun 02 '19 at 13:04

You can generate the combinations with combn and compute means for each combination with rowMeans:

all.combs <- unlist(lapply(2:ncol(data), function(x) combn(colnames(data), x, simplify = F)), recursive = F)

m <- lapply(all.combs, function(x) rowMeans(data[, x]))
data[sapply(all.combs, paste0, collapse = '')] <- m


# example output
data[, c('ab', 'ac', 'abcde')]
#            ab          ac       abcde
# 1   0.9145668 -0.15422891  0.46534449
# 2   1.0593771  0.36834739 -0.28974715
# 3   0.8504790  0.37486041  0.58032864
# 4   0.8392725  1.67687954  0.62420232
# 5  -0.1612623 -0.31145576  0.06580884
# 6  -0.6140748 -0.05931374 -0.01082605
# 7   0.4424551  0.75504165  0.53706206
# 8  -0.1202238 -0.02772524  0.43865296
# 9  -1.3020701 -0.18290837 -0.61781512
# 10 -0.7414824 -1.56409902 -1.12516693

This definitely gets the job done. If I can figure out a solution using the tidyverse, I'll post it. — user2917781, Jun 02 '19 at 13:11
@user2917781 something like: `map(1:ncol(df), ~ combn(colnames(df), ., simplify = FALSE)) %>% flatten()` should get you going in the right direction. — JasonAizkalns, Jun 02 '19 at 13:12

Create the mean for all possible combinations of 5 different variables

1 Answers1