Convert list of individuals to occurence of pairs in R - r

Question

I need specific format of data.frame for social structure analysis. How to convert data.frame containing list of individuals occuring together on multiple events:
my.df <- data.frame(individual = c("A","B","C","B","C","D"),
time = rep(c("event_01","event_02"), each = 3))
individual time
1 A event_01
2 B event_01
3 C event_01
4 B event_02
5 C event_02
6 D event_02
into a data.frame containing occurence for each pairs (including [A,A]; [B,B] etc. pairs:
ind_1 ind_2 times
A A 0
A B 1
A C 1
A D 0
B A 1
B B 0
B C 2
B D 1
C A 1
C B 2
C C 0
C D 1
D A 0
D B 1
D C 1
D D 0

Answer 1

In base R, you could do the following:
data.frame(as.table(`diag<-`(tcrossprod(table(my.df)), 0)))
# individual individual.1 Freq
# 1 A A 0
# 2 B A 1
# 3 C A 1
# 4 D A 0
# 5 A B 1
# 6 B B 0
# 7 C B 2
# 8 D B 1
# 9 A C 1
# 10 B C 2
# 11 C C 0
# 12 D C 1
# 13 A D 0
# 14 B D 1
# 15 C D 1
# 16 D D 0
tcrossprod gives you the following:
> tcrossprod(table(my.df))
individual
individual A B C D
A 1 1 1 0
B 1 2 2 1
C 1 2 2 1
D 0 1 1 1
That's essentially all the information you are looking for, but you want it in a slightly different form, without the diagonal values.
We can set the diagonals to zero with:
`diag<-`(theOutputFromAbove, 0)
Then, to get the long form, trick R into thinking that the resulting matrix is a table by using as.table, and make use of the data.frame method for tables.

Answer 2

You can do:
create the first 2 variables of the new data.frame:
df2 <- expand.grid(ind_2=levels(my.df$individual), ind_1=levels(my.df$individual))[, 2:1]
Put the value to 0 for the pairs of same individuals:
df2$times[df2[, 1]==df2[, 2]] <- 0
See the other unique combinations:
comb_diff <- combn(levels(my.df$individual), 2)
compute the times each unique combination is found together:
times_uni <- apply(comb_diff, 2, function(inds){
sum(table(my.df$time[my.df$individual %in% inds])==2)
})
Finally, fill the new data.frame:
df2$times[match(c(paste0(comb_diff[1,], comb_diff[2,]), paste0(comb_diff[2, ], comb_diff[1, ])), paste0(df2[, 1],df2[, 2]))] <- rep(times_uni, 2)
df2
# ind_1 ind_2 times
#1 A A 0
#2 A B 1
#3 A C 1
#4 A D 0
#5 B A 1
#6 B B 0
#7 B C 2
#8 B D 1
#9 C A 1
#10 C B 2
#11 C C 0
#12 C D 1
#13 D A 0
#14 D B 1
#15 D C 1
#16 D D 0

Answer 3

You can do it using data.table
dt_combs <- my.dt[,
list(ind_1 = combn(individual, 2)[1, ],
ind_2 = combn(individual, 2)[2, ]),
by = time]
dt_ncombs <- dt_combs[, .N, by = c("ind_1", "ind_2")]
dt_ncombs_inverted <- copy(dt_ncombs)
dt_ncombs_inverted[, temp := ind_1]
dt_ncombs_inverted[, ind_1 := ind_2]
dt_ncombs_inverted[, ind_2 := temp]
dt_ncombs_inverted[, temp := NULL]
dt_ncombs <- rbind(dt_ncombs, dt_ncombs_inverted)
dt_allcombs <- data.table(expand.grid(
ind_1 = my.dt[, unique(individual)],
ind_2 = my.dt[, unique(individual)]
))
dt_final <- merge(dt_allcombs,
dt_ncombs,
all.x = TRUE,
by = c("ind_1", "ind_2"))
dt_final[is.na(N), N := 0]
dt_final