Finding intersection entries in a data frame - r

Question

I've run into an R-programming problem that I can't seem to wrap my head around. I have data like the following:
data = data.frame("start"=c(1,2,4,5),
"length"=c(2,2,2,3),
"decision"=c("yes","no","yes","yes"))
Which looks like:
start length decision
1 1 2 yes
2 2 2 no
3 4 2 yes
4 5 3 yes
Row one stands for a sequence of integers that start at 1 for length 2 (1,2). Row 3 is 2 integers starting at 4 (4,5). I'm looking for intersections between entries that have a 'yes' decision variable. When the decision variable is 'no', then the sequence is thrown out. Here's what I've attempted so far.
I think I need to create a sequence list first.
sequence.list = lapply(seq(dim(data)[1]),
function(d){
seq(data$start[d],(data$start[d]+data$length[d]-1),by=1)
})
This outputs:
sequence.list
[[1]]
[1] 1 2
[[2]]
[1] 2 3
[[3]]
[1] 4 5
[[4]]
[1] 5 6 7
Which is a start. Then I create a list that counts intersections between items on my list (I stole this idea from another post on here).
count.intersect = lapply(sequence.list,function(a) {
sapply(seq(length(sequence.list)),
function(b) length(intersect(sequence.list[[b]], a)))
})
This creates the list:
count.intersect
[[1]]
[1] 2 1 0 0
[[2]]
[1] 1 2 0 0
[[3]]
[1] 0 0 2 1
[[4]]
[1] 0 0 1 3
The way to read this is that entry 1 in the data frame has 2 trivial intersections with itself and 1 intersection with entry 2.
Here's where I get fuzzy on what to do. Make it a matrix?
intersect.matrix = do.call(rbind,count.intersect)
Then set the rows and columns of non-used entries to zero?
intersect.matrix[,data$decision=="no"]=0
intersect.matrix[data$decision=="no",]=0
intersect.matrix
[,1] [,2] [,3] [,4]
[1,] 2 0 0 0
[2,] 0 0 0 0
[3,] 0 0 2 1
[4,] 0 0 1 3
Now, I would like to return indices 3 and 4 somehow. I want to find the rows (or columns) containing non zeros that are also not on the diagonal.
Sorry for posting the whole procedure, I also want to know if there is a shorter way to go from the starting dataframe to finding intersections in used entries.

Answer 1

Since you are not interested in non zero values on the diagonal, first I would subtract them away:
diag.mat <- diag(intersect.matrix) * diag(ncol(intersect.matrix)
which gives:
intersect.matrix - diag.mat
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 0 0 0 1
[4,] 0 0 1 0
Then identify which of the columns still hold non zero entries using which:
which(colSums(intersect.matrix - diag.mat) != 0)
[1] 3 4

Answer 2

You asked whether there is a short way to go from your data frame data to the indices. Here it is.
(Note: This may be hard to understand if you're new to R.)
1) Create the sequence list:
sequence.list <- apply(data[1:2], 1, function(x) seq_len(x[2]) + x[1] - 1)
# [[1]]
# [1] 1 2
#
# [[2]]
# [1] 2 3
#
# [[3]]
# [1] 4 5
#
# [[4]]
# [1] 5 6 7
2) Count intersects and create the intersect matrix
intersect.matrix <- outer(s <- seq_along(sequence.list), s,
Vectorize(function(a, b)
length(Reduce(intersect, sequence.list[seq(a, b)]))))
# [,1] [,2] [,3] [,4]
# [1,] 2 1 0 0
# [2,] 1 2 0 0
# [3,] 0 0 2 1
# [4,] 0 0 1 3
3) Set cells corresponding to "no" to zero
idx <- data$decision == "no"
intersect.matrix[idx, ] <- intersect.matrix[ , idx] <- 0
# [,1] [,2] [,3] [,4]
# [1,] 2 0 0 0
# [2,] 0 0 0 0
# [3,] 0 0 2 1
# [4,] 0 0 1 3
4) Find indices of non-zero rows/columns (except diagonal)
result <- which(as.logical(colSums("diag<-"(intersect.matrix, 0))))
# [1] 3 4