Combinations by group in R

Question

I have a question about combinations by group.

My mini-sample looks like this:

sample <- data.frame(
  group=c(\"a\",\"a\",\"a\",\"a\",\"b\",\"b\"         


        

Answer 1

Here's a base R option using (1) split to create a list of data.frames per unique group-entry, (2) lapply to loop over each list element and compute the combinations using combn, (3) do.call(rbind, ...) to collect the list elements back into a single data.frame.

do.call(rbind, lapply(split(sample, sample$group), {
   function(x) data.frame(group = x$group[1], t(combn(x$number, 2)))
}))

#    group X1 X2
#a.1     a  1  2
#a.2     a  1  3
#a.3     a  1  2
#a.4     a  2  3
#a.5     a  2  2
#a.6     a  3  2
#b.1     b  4  5
#b.2     b  4  3
#b.3     b  5  3

And a dplyr option:

library(dplyr)
sample %>% group_by(group) %>% do(data.frame(t(combn(.$number, 2))))
#Source: local data frame [9 x 3]
#Groups: group [2]
#
#   group    X1    X2
#  (fctr) (dbl) (dbl)
#1      a     1     2
#2      a     1     3
#3      a     1     2
#4      a     2     3
#5      a     2     2
#6      a     3     2
#7      b     4     5
#8      b     4     3
#9      b     5     3

Answer 2

We can use a group by function with data.table

library(data.table)
setDT(sample)[, {i1 <-  combn(number, 2)
                   list(i1[1,], i1[2,]) }, by =  group]
#    group V1 V2
#1:     a  1  2
#2:     a  1  3
#3:     a  1  2
#4:     a  2  3
#5:     a  2  2
#6:     a  3  2
#7:     b  4  5
#8:     b  4  3
#9:     b  5  3

Or a compact option would be

setDT(sample)[, transpose(combn(number, 2, FUN = list)), by = group]

---

Or using base R

 lst <- by(sample$number, sample$group, FUN = combn, m= 2)
 data.frame(group = rep(unique(as.character(sample$group)), 
                        sapply(lst, ncol)), t(do.call(cbind, lst)))