Faster alternative to nested for loops in R - r

Question

Here is the scenario: I have a sample in which subjects are placed into any of three groups. Next, subjects from each group are grouped together, resulting in several "triplets" consisting of a subject from each group. I would like to count the number of times a subject from a given group (1, 2, or 3) is grouped with a subject i of a different original group.
Here is a simple code example:
data <- cbind(c(1:9), c(rep("Group 1", 3), rep("Group 2", 3), rep("Group 3", 3)))
data <- data.frame(data)
names(data) <- c("ID", "Group")
groups.of.3 <- data.frame(rbind(c(1,4,7),c(2,4,7),c(2,5,7),c(3,6,8),c(3,6,9)))
N <- nrow(data)
n1 <- nrow(data[data$Group == "Group 1", ])
n2 <- nrow(data[data$Group == "Group 2", ])
n3 <- nrow(data[data$Group == "Group 3", ])
# Check the number of times a subject from a group is grouped with a subject i
# from another group
M1 <- matrix(0, nrow = N, ncol = n1)
M2 <- matrix(0, nrow = N, ncol = n2)
M3 <- matrix(0, nrow = N, ncol = n3)
for (i in 1:N){
if (data$Group[i] != "Group 1"){
for (j in 1:n1){
M1[i,j] <- nrow(groups.of.3[groups.of.3[,1] == j &
(groups.of.3[,2] == i |
groups.of.3[,3] == i), ])
}
}
if (data$Group[i] != "Group 2"){
for (j in 1:n2){
M2[i,j] <- nrow(groups.of.3[groups.of.3[,2] == (n1 + j) &
(groups.of.3[,1] == i |
groups.of.3[,3] == i), ])
}
}
if (data$Group[i] != "Group 3"){
for (j in 1:n3){
M3[i,j] <- nrow(groups.of.3[groups.of.3[,3] == (n1 + n2 + j) &
(groups.of.3[,1] == i |
groups.of.3[,2] == i), ])
}
}
}
So I have 9 subjects, with three from each group. And then subjects from each group are subsequently grouped together (allowing for repetition of placement). This takes a lot longer with more subjects, and I am wondering if there is a faster alternative that avoids using for loops.
For instance, the matrix M1 consists of how many times subjects in Group 1 were subsequently grouped with other subjects from any other group:
M1
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
[4,] 1 1 0
[5,] 0 1 0
[6,] 0 0 2
[7,] 1 2 0
[8,] 0 0 1
[9,] 0 0 1
So the 3 columns represent the three subjects from Group 1, and the rows represent all subjects - the entries are how many times each subject from Group 1 is grouped with any of the other subjects (e.g., according to groups.of.3, subject 3 appears in a group with subject 6 twice, and subject 1 with subject 7 once).
Thanks for any help!

Answer 1

Something like this?
library(tidyr)
library(dplyr)
data <- data %>%
mutate(ID = as.numeric(levels(ID))[ID])
tmp <- groups.of.3 %>%
add_rownames() %>%
gather("X", "Person", -rowname) %>%
inner_join(data, by = c("Person" = "ID"))
tmp %>%
inner_join(tmp, by = c("rowname")) %>%
filter(Group.x != Group.y) %>%
group_by(Person.x, Group.x, Group.y) %>%
summarise(N = n()) %>%
spread(key = Group.y, value = N, fill = 0)
Person.x Group.x Group 1 Group 2 Group 3
(dbl) (fctr) (dbl) (dbl) (dbl)
1 1 Group 1 0 1 1
2 2 Group 1 0 2 2
3 3 Group 1 0 2 2
4 4 Group 2 2 0 2
5 5 Group 2 1 0 1
6 6 Group 2 2 0 2
7 7 Group 3 3 3 0
8 8 Group 3 1 1 0
9 9 Group 3 1 1 0

Answer 2

For loops aren't inherently slow:
# coerce the fields in groups.of.3 to factor
for(i in 1:3)
groups.of.3[,i] <- as.factor(groups.of.3[,i],levels =data$ID)
M <- matrix(0, N, N)
out <- NULL
for(i in 1:(3-1))
for(j in (i+1):3)
M <- M + table(groups.of.3[,i],groups.of.3[,j])
M1 <- M[,as.integer(data$Group)==1]
M2 <- M[,as.integer(data$Group)==2]
M3 <- M[,as.integer(data$Group)==3]

Answer 3

I'll answer my own question, using a very slight modification of Thierry's answer:
library(tidyr)
library(dplyr)
data <- data %>%
mutate(ID = as.numeric(levels(ID))[ID])
tmp <- groups.of.3 %>%
add_rownames() %>%
gather("X", "Person", -rowname) %>%
inner_join(data, by = c("Person" = "ID"))
tmp %>%
inner_join(tmp, by = c("rowname")) %>%
filter(Group.x != Group.y) %>%
group_by(Person.x, Group.x, Person.y) %>%
summarise(N = n()) %>%
spread(key = Person.y, value = N, fill = 0)
This gives the following output, which includes M1, M2, and M3 from the previous for loop, adjoined together.
Source: local data frame [9 x 11]
Person.x Group.x 1 2 3 4 5 6 7 8 9
(dbl) (fctr) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
1 1 Group 1 0 0 0 1 0 0 1 0 0
2 2 Group 1 0 0 0 1 1 0 2 0 0
3 3 Group 1 0 0 0 0 0 2 0 1 1
4 4 Group 2 1 1 0 0 0 0 2 0 0
5 5 Group 2 0 1 0 0 0 0 1 0 0
6 6 Group 2 0 0 2 0 0 0 0 1 1
7 7 Group 3 1 2 0 2 1 0 0 0 0
8 8 Group 3 0 0 1 0 0 1 0 0 0
9 9 Group 3 0 0 1 0 0 1 0 0 0