'R' Matrix subset without loop - r

Question

I started using R about six months back and i have gained a little bit of experience in R. Recently, I ran into an issue regarding subsets within a matrix and would like assistance on making the solution that I have more efficient.
What I would like to do is the following. Suppose I have a matrix and two vectors as follows:
# matrix
a <- matrix(seq(1,100,by=1),10,10)
# vector (first column of matrix a)
b <- c(2,4,5,6,7,8)
# vector (column numbers of matrix a)
c <- c(5,3,1,4,6,2)
Just to reiterate,
Vector b refers to the first column of matrix a.
Vector c refers to column numbers of matrix a.
I would like to get tmp99 <- a[b,c:8]. However, when I do that I get the following warning message.
Warning message:
In c:8 : numerical expression has 6 elements: only the
first used (index has to be scalar and not vector)
So, I tried working around the problem using loops and list and I get the solution I want. I am assuming that there is a more time efficient solution than this. The solution what I have so far is the following:
a <- matrix(seq(1,100,by=1),10,10)
b <- c(2,4,5,6,7,8)
c <- c(5,3,1,4,6,2)
tmp <- list()
for (i in 1:length(b)) tmp[[i]] <- c(a[b[i],(c[i]:8)])
tmp99 <- t(sapply(tmp, '[', 1:max(sapply(tmp, length))))
tmp99[is.na(tmp99)] <- 0
What I would like to know is if there is a way to avoid using loops to achieve the above because my matrix dimension is 200000 x 200 and since I have to do this a lot (In my problem, b and c are determined as part of another part of the code and so I am not able to use absolute index numbers), I would like to cut down the time taken for the same. Any help will be greatly appreciated. Thank you.

Answer 1

You might try some kind of matrix indexing solution, like this. It's not clear if it will actually be faster or not; in small cases, I think it definitely will be, but in big cases, the overhead from creating the matrixes to index by might take longer than just running through a for loop. To get a better answer, make up a data set that is similar to yours that we could test against.
idx.in <- cbind(rep(b, 8-c+1), unlist(lapply(c, function(x) x:8)))
idx.out <- cbind(rep(seq_along(b), 8-c+1), unlist(lapply(c, function(x) 1:(8-x+1))))
tmp99 <- array(0, dim=apply(idx.out, 2, max))
tmp99[idx.out] <- a[idx.in]
Here's a version with matrix indexing but that does it separately for each row. This might be faster, depending on how many rows and columns are being replaced. What you want to avoid is running out of memory, which the for loop can help with, as it doesn't keep all the details for each step in memory at the same time.
out <- array(0, dim=c(length(b), 8-min(c)+1))
for(idx in seq_along(b)) {
out[cbind(idx, 1:(8-c[idx]+1))] <- a[cbind(b[idx], c[idx]:8)]
}
out

Answer 2

Following is one way to do it using base packages. There might be better solution using data.table but following works :)
a <- matrix(seq(1, 100, by = 1), 10, 10)
b <- c(2, 4, 5, 6, 7, 8)
c <- c(5, 3, 1, 4, 6, 2)
res <- t(sapply(X = mapply(FUN = function(b, c) expand.grid(b, seq(from = c, to = 8)), b, c, SIMPLIFY = FALSE), FUN = function(x) {
c(a[as.matrix(x)], rep(0, 8 - nrow(x)))
}))
res
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## [1,] 42 52 62 72 0 0 0 0
## [2,] 24 34 44 54 64 74 0 0
## [3,] 5 15 25 35 45 55 65 75
## [4,] 36 46 56 66 76 0 0 0
## [5,] 57 67 77 0 0 0 0 0
## [6,] 18 28 38 48 58 68 78 0
# Let's break it down in multiple steps.
coordinates <- mapply(FUN = function(b, c) expand.grid(b, seq(from = c, to = 8)), b, c, SIMPLIFY = FALSE)
# below sapply subsets c using each element in coordinates and pads result with additional 0s such that total 8 elements are returned.
res <- sapply(X = coordinates, FUN = function(x) {
c(a[as.matrix(x)], rep(0, 8 - nrow(x)))
})
res
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 42 24 5 36 57 18
## [2,] 52 34 15 46 67 28
## [3,] 62 44 25 56 77 38
## [4,] 72 54 35 66 0 48
## [5,] 0 64 45 76 0 58
## [6,] 0 74 55 0 0 68
## [7,] 0 0 65 0 0 78
## [8,] 0 0 75 0 0 0
# you probably need result as traspose
res <- t(res)
res
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## [1,] 42 52 62 72 0 0 0 0
## [2,] 24 34 44 54 64 74 0 0
## [3,] 5 15 25 35 45 55 65 75
## [4,] 36 46 56 66 76 0 0 0
## [5,] 57 67 77 0 0 0 0 0
## [6,] 18 28 38 48 58 68 78 0

Answer 3

tmp <- lapply(seq_len(length(b)),function(i) {
res <- a[b[i],c[i]:8]
res <- c(res,rep(0,c[i]-1))
res
})
tmp99 <- do.call("rbind",tmp)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# [1,] 42 52 62 72 0 0 0 0
# [2,] 24 34 44 54 64 74 0 0
# [3,] 5 15 25 35 45 55 65 75
# [4,] 36 46 56 66 76 0 0 0
# [5,] 57 67 77 0 0 0 0 0
# [6,] 18 28 38 48 58 68 78 0