Similar questions have been asked about counting pairs, however none seem to be specifically useful for what I'm trying to do.
What I want is to count the number of pairs across multiple list elements and turn it into a matrix. For example, if I have a list like so:
myList <- list(
a = c(2,4,6),
b = c(1,2,3,4),
c = c(1,2,5,7),
d = c(1,2,4,5,8)
)
We can see that the pair 1:2
appears 3 times (once each in a
, b
, and c
). The pair 1:3
appears only once in b
. The pair 1:4
appears 2 times (once each in b
and d
)... etc.
I would like to count the number of times a pair appears and then turn it into a symmetrical matrix. For example, my desired output would look something like the matrix I created manually (where each element of the matrix is the total count for that pair of values):
> myMatrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 0 3 1 2 2 0 1 1
[2,] 3 0 1 3 2 1 1 1
[3,] 1 1 0 1 0 0 0 0
[4,] 2 3 1 0 0 0 0 1
[5,] 2 2 0 0 0 0 1 1
[6,] 0 1 0 0 0 0 0 0
[7,] 1 1 0 0 1 0 0 0
[8,] 1 1 0 1 1 0 0 0
Any suggestions are greatly appreciated
Inspired by @akrun's answer, I think you can use a crossproduct to get this very quickly and simply:
out <- tcrossprod(table(stack(myList)))
diag(out) <- 0
# values
#values 1 2 3 4 5 6 7 8
# 1 0 3 1 2 2 0 1 1
# 2 3 0 1 3 2 1 1 1
# 3 1 1 0 1 0 0 0 0
# 4 2 3 1 0 1 1 0 1
# 5 2 2 0 1 0 0 1 1
# 6 0 1 0 1 0 0 0 0
# 7 1 1 0 0 1 0 0 0
# 8 1 1 0 1 1 0 0 0
Original answer:
Use combn
to get the combinations, as well as reversing each combination.
Then convert to a data.frame
and table
the results.
tab <- lapply(myList, \(x) combn(x, m=2, FUN=\(cm) rbind(cm, rev(cm)), simplify=FALSE))
tab <- data.frame(do.call(rbind, unlist(tab, rec=FALSE)))
table(tab)
# X2
#X1 1 2 3 4 5 6 7 8
# 1 0 3 1 2 2 0 1 1
# 2 3 0 1 3 2 1 1 1
# 3 1 1 0 1 0 0 0 0
# 4 2 3 1 0 1 1 0 1
# 5 2 2 0 1 0 0 1 1
# 6 0 1 0 1 0 0 0 0
# 7 1 1 0 0 1 0 0 0
# 8 1 1 0 1 1 0 0 0
We could loop over the list
, get the pairwise combinations with combn
, stack
it to a two column dataset, convert the 'values' column to factor
with levels
specified as 1 to 8, get the frequency count (table
), do a cross product (crossprod
), convert the output back to logical, and then Reduce
the list
elements by adding elementwise and finally assign the diag
onal elements to 0. (If needed set the names
attributes of dimnames
to NULL
out <- Reduce(`+`, lapply(myList, function(x)
crossprod(table(transform(stack(setNames(
combn(x,
2, simplify = FALSE), combn(x, 2, paste, collapse="_"))),
values = factor(values, levels = 1:8))[2:1]))> 0))
diag(out) <- 0
names(dimnames(out)) <- NULL
-output
> out
1 2 3 4 5 6 7 8
1 0 3 1 2 2 0 1 1
2 3 0 1 3 2 1 1 1
3 1 1 0 1 0 0 0 0
4 2 3 1 0 1 1 0 1
5 2 2 0 1 0 0 1 1
6 0 1 0 1 0 0 0 0
7 1 1 0 0 1 0 0 0
8 1 1 0 1 1 0 0 0
I thought of a solution based on @TarJae answer, is not a elegant one, but it was a fun challenge!
library(tidyverse)
map_df(myList,function(x) as_tibble(t(combn(x,2)))) %>%
count(V1,V2) %>%
{. -> temp_df} %>%
bind_rows(
temp_df %>%
rename(V2 = V1, V1 = V2)
) %>%
full_join(
expand_grid(V1 = 1:8,V2 = 1:8)
) %>%
replace_na(replace = list(n = 0)) %>%
arrange(V2,V1) %>%
pivot_wider(names_from = V1,values_from = n) %>%
as.matrix()
V2 1 2 3 4 5 6 7 8
[1,] 1 0 3 1 2 2 0 1 1
[2,] 2 3 0 1 3 2 1 1 1
[3,] 3 1 1 0 1 0 0 0 0
[4,] 4 2 3 1 0 1 1 0 1
[5,] 5 2 2 0 1 0 0 1 1
[6,] 6 0 1 0 1 0 0 0 0
[7,] 7 1 1 0 0 1 0 0 0
[8,] 8 1 1 0 1 1 0 0 0
First identify the possible combination of each vector from the list to a tibble
then I bind them to one tibble
and count
the combinations.
library(tidyverse)
a <- as_tibble(t(combn(myList[[1]],2)))
b <- as_tibble(t(combn(myList[[2]],2)))
c <- as_tibble(t(combn(myList[[3]],2)))
d <- as_tibble(t(combn(myList[[4]],2)))
bind_rows(a,b,c,d) %>%
count(V1, V2)
V1 V2 n
<dbl> <dbl> <int>
1 1 2 3
2 1 3 1
3 1 4 2
4 1 5 2
5 1 7 1
6 1 8 1
7 2 3 1
8 2 4 3
9 2 5 2
10 2 6 1
11 2 7 1
12 2 8 1
13 3 4 1
14 4 5 1
15 4 6 1
16 4 8 1
17 5 7 1
18 5 8 1
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