Merge all possible combinations of multiple data frames - r

Question

I would like to merge by columns all the possible pair combinations of these three data frames (i.e. nine combinations)
frame1 = data.frame(a=c(1,2,3), b=c(1,2,3), c=c(1,2,3))
frame2 = data.frame(a=c(2,1,3), b=c(2,1,3), c=c(2,1,3))
frame3 = data.frame(a=c(3,2,1), b=c(3,2,1), c=c(3,2,1))
which contain the same 3 rows each but not in the same order, so I would also like that the merging be by coincidence of the pair of values of the columns a and b in the two files merged. Example:
a b c
1 1 1
2 2 2
3 3 3
+
a b c
2 2 2
1 1 1
3 3 3
=
a.x b.x c.x a.y b.y c.y
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
I wanted then to obtain the difference between each pair of values of the columns c.x and c.y present in each merged file, in absolute values, and sum all these differences thus obtaining a "score" (of course this would be zero in this example), which I would like to add to an empty matrix 3x3 in the correspondant cell (i.e., the score of frame1 vs. frame 2 should be located in cell [2,1], etc.):
nframes = 3
frames = c(frame1,frame2,frame3)
matrix = matrix(, nrow = nframes, ncol = nframes)
matrix_scores = data.frame(matrix)
for (i in frames){
for (j in frames)
{
x = merge(i, j, by=c("a","b"))
score = sum(abs(x$c.x - x$c.y))
matrix_scores[j,i] <- score
}
}
However, when I run the loop I obtain the following message:
Error in fix.by(by.x, x) : 'by' must specify uniquely valid columns
Also, I understand that the line
matrix_scores[j,i] <- score
will give an error, too, but I do not know how to express that I want the score to be stored in cell [1,1], for the first iteration of the loop (frame1 vs. frame1).
The resulting matrix should be a 3x3 matrix containing all zeros:
f1 f2 f3
frame1 0 0 0
frame2 0 0 0
frame3 0 0 0

Answer 1

You can do:
# Put all frames in a list
d <- list(frame1, frame2, frame3)
# get all merge-combinations
gr <- expand.grid(1:length(d), 1:length(d))
# function to merge and get the sum diff:
foo <- function(i, x, gr){
tmp <- merge(x[[gr[i, 1]]], x[[gr[i, 2]]], by=c("a", "b"))
sum(abs(tmp$c.x - tmp$c.y))
}
# result matrix
matrix(sapply(1:nrow(gr), foo, d, gr), length(d), length(d), byrow = T)
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
# The scores are set as followed:
matrix(apply(gr, 1, paste, collapse="_"), 3, 3, byrow = T)
[,1] [,2] [,3]
[1,] "1_1" "2_1" "3_1"
[2,] "1_2" "2_2" "3_2"
[3,] "1_3" "2_3" "3_3"
# alternative using apply:
# function to merge and get the sum diff:
foo <- function(y, x){
tmp <- merge(x[[ y[1] ]], x[[ y[2] ]], by=c("a", "b"))
sum(abs(tmp$c.x - tmp$c.y))
}
# result matrix
matrix(apply(gr, 1, foo, d), length(d), length(d), byrow = T)