R - combining columns by specific conditions - r

Question

I currently has a data frame as follow:
groups <- data.frame(name=paste("person",c(1:27),sep=""),
assignment1 = c("F","A","B","H", "A", "E", "D", "G", "I", "I", "E", "A", "D", "C", "F", "C", "D", "H", "F", "H", "G", "I", "G", "C", "B", "E", "B"),
assignment2 = c("H", "F", "F", "D", "E", "G", "A", "E", "I", "C", "A", "H", "G", "B", "I", "C", "E", "I", "C", "A", "B", "B", "G", "D", "H", "F", "D"),stringsAsFactors = FALSE)
It would looks like this:
I would like to create a list for each person that only contains the people he had already worked with. For example, person1 is on group F and H for 1st and 2nd assignment respectively and
The member of groups F on 1st assignment are {"person1","person15", "person19"}.
The member of groups D on 2nd assignment are {"person1","person12", "person25"}.
I would like to create a vector for person1 like
{"person15", "person19", "person12", "person25"}.
Any one knows a convenient way to do this in R?
Any help will be appreciated. Thanks in advance.

Answer 1

You could do this:
teammates <- lapply(1:nrow(groups), function(i) {
assig1 <- subset(groups, assignment1 == groups$assignment1[i])$name
assig2 <- subset(groups, assignment2 == groups$assignment2[i])$name
unq_set <- unique(c(assig1, assig2))
return(setdiff(unq_set, groups$name[i]))
})
This takes a vector of row indices, and for each one applies a function that a) gets the names of those where assignments 1 & 2 match the given row, b) gets the unique superset of these, c) returns that, less the name of the person around whom the group is built
The output is a list like this:
[[1]]
[1] "person15" "person19" "person12" "person25"
[[2]]
[1] "person5" "person12" "person3" "person26"
[[3]]
[1] "person25" "person27" "person2" "person26"
...and so on
For more brevity, the following is equivalent (though order inside list items may be different). Same logic as #user5219763's answer for subsetting, but the setdiff part is important
teammates <- lapply(1:nrow(groups), function(i) {
setdiff(
with(groups, name[assignment1 == assignment1[i] |
assignment2 == assignment2[i] ]),
groups$name[i])
})

Answer 2

Here's a solution using dplyr and tidyr:
library(dplyr)
library(tidyr)
groups %>%
gather(var, val, -name) %>%
unite(comb, var, val) %>%
left_join(.,., by = 'comb') %>%
group_by(name.x) %>%
summarise(out = list(name.y))
The heavy lifting is done using the left_join before that we are combining columns, so that we can merge on eg assignment1_f. The output contains itself, and is not corrected for dupes - that is up to you.
However, as #akrun says, if you are doing a lot of this stuff, use igraph

Answer 3

You could use is.element()
workedWith <- function(index,data=groups){
data[is.element(data[,2],data[index,2]) | is.element(data[,3],data[index,3]),1]
}
lapply(X = seq(1:nrow(groups)),FUN = workedWith)