Suppose I have a vector of dimension n and it is composed of 0 and 1. Then I divide this vector into m equal bins. A bin is called active if it contains at least one "1". I want to write a command that returns the place of active bins and how many "1" they contain.
For example, I have this vector: n=15, m=5
[1 0 0 | 0 1 1 | 0 0 0 | 0 1 0| 1 1 1] I want to have matrix [1 2 4 5] (the active bins) and [1 2 1 3] (how many 1 they contain).
Can I write this in R without using for loops?
I would do it like this:
a <- c(1,0,0,0,1,1,0,0,0,0,1,0,1,1,1) m <- 5 idx <- rep(1:m, each=length(a)/m) # how many ones? no <- sapply(1:5, function(x) sum(a[idx==x])) # which bins contain ones? bins <- 1:m bins[no>0] 
no <- rowSums(matrix(a,ncol=3,byrow=TRUE)); which(no>0); no[no>0]Another approach to obtain the vector with number of ones:
x <- c(1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1) n <- length(x) m <- 5 size <- n/m x.list <- split(x, cut(seq_along(x)/size, 0:m)) vapply(x.list, sum, 0) From there, do as jigr does.
