Suppose I have a vector 1 to 10 and wish to split it into subvectors of the following two conditions:
an equal length of 3.
with overlapping of 1.
I got an almost-done answer Split vector with overlapping samples in R with a function which I modified bellow:
splitWithOverlap <- function(vec, seg.length, overlap) {
starts = seq(1, length(vec), by=seg.length-overlap)
ends = starts + seg.length - 1
ends[ends > length(vec)] = length(vec)
lapply(1:length(starts), function(i) vec[starts[i]:ends[i]])
}
splitWithOverlap(1:10, 3, 2)
which produced
#[[1]]
#[1] 1 2 3
#[[2]]
#[1] 2 3 4
#[[3]]
#[1] 3 4 5
#[[4]]
#[1] 4 5 6
#[[5]]
#[1] 5 6 7
#[[6]]
#[1] 6 7 8
#[[7]]
#[1] 7 8 9
#[[8]]
#[1] 8 9 10
#[[9]]
#[1] 9 10
#[[10]]
#[1] 10
what I want is
#[[1]]
#[1] 1 2 3
#[[2]]
#[1] 2 3 4
#[[3]]
#[1] 3 4 5
#[[4]]
#[1] 4 5 6
#[[5]]
#[1] 5 6 7
#[[6]]
#[1] 6 7 8
#[[7]]
#[1] 7 8 9
#[[8]]
#[1] 8 9 10
because the two conditions are achieved and mathematically, number of blocks = vector length - subvector length + 1
(10 - 3 + 1
) = 8 and not 10
I want a modification to the function so that it will stop at subvector 8.
DATA
len = 3
ov = 1
vec = 1:10
1
lapply(1:(length(vec) - (len - ov)), function(i){
vec[i:(i + len - ov)]
})
2
ind = rep(1:len, length(vec) - (len - ov))
matrix(vec[ind + ave(ind, ind, FUN = seq_along) - 1], ncol = len, byrow = TRUE)