CODEWITHSUNDEEP
rfindintervals
Ding Li
Ding Li

Reputation: 723

finding interval indices of values in a vector, when intervals may be overlapped

I want to find the indices of values in a vector belonging to intervals which are defined by a vector of ending values and 1）"look-back" value interval and 2) previous N values.

Suppose I have 

x <- c(1,3,4,5,7,8,9,10,13,14,15,16,17,18) #the vector of interest
v_end <- c(5, 7, 15) #the end values
l<-3 #look-back value interval
N<-3 #number of value to look back

What I want is the second and third columns of the following output.

       x i n
 [1,]  1 0 1
 [2,]  3 1 1
 [3,]  4 1 1
 [4,]  5 1 1
 [5,]  7 1 1
 [6,]  8 0 0
 [7,]  9 0 0
 [8,] 10 0 1
 [9,] 13 1 1
[10,] 14 1 1
[11,] 15 1 1
[12,] 16 0 0
[13,] 17 0 0
[14,] 18 0 0

Notice that v_end and l result in three intervals [2,5],[4,7],[12,15]. [2,5] and [4,7] have overlaps, essentially, it is [2,7]. And, v_end and l result in three intervals [1,5], [3,7],[10,15]. Again there are overlapps.

The task is similar to function findInterval{base}, but can not be solved by it.

Upvotes: 1

Views: 142

Answers (1)

alexis_laz
alexis_laz

Reputation: 13122

Having ordered "v_end" and "x" (for the "N" case), the intervals for the "l" case are:

ints = cbind(start = v_end - l, end = v_end)
ints
#     start end
#[1,]     2   5
#[2,]     4   7
#[3,]    12  15

Their overlaps could be grouped with:

overlap_groups = cumsum(c(TRUE, ints[-nrow(ints), "end"] < ints[-1, "start"]))

which can be used to reduce the intervals that are overlapping:

group_end = cumsum(rle(overlap_groups)$lengths)
group_start = c(1L, group_end [-length(group_end )] + 1L)

ints2 = cbind(start = ints[group_start, "start"], end = ints[group_end, "end"])
ints2
#     start end
#[1,]     2   7
#[2,]    12  15

Then using findInterval:

istart = findInterval(x, ints2[, "start"])
iend = findInterval(x, ints2[, "end"], left.open = TRUE)

i = as.integer((istart - iend) == 1L)
i
# [1] 0 1 1 1 1 0 0 0 1 1 1 0 0 0

For the case of "N", starting with:

ints = cbind(start = x[match(v_end, x) - N], end = v_end)
ints
#     start end
#[1,]     1   5
#[2,]     3   7
#[3,]    10  15

and following the above steps, we get:

#.....
n = as.integer((istart - iend) == 1L)
n
# [1] 1 1 1 1 1 0 0 1 1 1 1 0 0 0

Generally, a convenient tool for such operations is the "IRanges" package which, here, makes the approach straightforward:

library(IRanges)

xrng = IRanges(x, x)
i = as.integer(overlapsAny(xrng, reduce(IRanges(v_end - l, v_end), min.gapwidth = 0)))
i
# [1] 0 1 1 1 1 0 0 0 1 1 1 0 0 0
n = as.integer(overlapsAny(xrng, reduce(IRanges(x[match(v_end, x) - N], v_end), min.gapwidth = 0)))
n
# [1] 1 1 1 1 1 0 0 1 1 1 1 0 0 0

Upvotes: 1

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