Check whether elements of vectors are inside intervals given by matrix - r

Question

Actually a really nice problem to which I came up with a solution (see below), which is, however, not beautiful:
Assume you have a vector x and a matrix A which contains the start of an interval in the first column and the end of the interval in the second.
How can I get the elements of A, which fall into the intervals given by A?
x <- c(4, 7, 15)
A <- cbind(c(3, 9, 14), c(5, 11, 16))
Expected output:
[1] 4 15
You could you the following information, if this would be helpful for increasing the performance:
Both, the vector and the rows of the matrix are ordered and the intervals don't overlap. All intervals have the same length. All numbers are integers, but can be huge.
Now I did not want to be lazy and came up with the following solution, which is too slow for long vectors and matrices:
x <- c(4, 7, 15) # Define input vector
A <- cbind(c(3, 9, 14), c(5, 11, 16)) # Define matrix with intervals
b <- vector()
for (i in 1:nrow(A)) {
b <- c(b, A[i, 1]:A[i, 2])
}
x[x %in% b]
I know that loops in R can be slow, but I did not know how to write the operation without one (maybe there is a way with apply).

Answer 1

We can use sapply to loop over each element of x and find if it lies in the range of any of those matrix values.
x[sapply(x, function(i) any(i > A[, 1] & i < A[,2]))]
#[1] 4 15
In case, if length(x) and nrow(A) are same then we don't even need the sapply loop and we can use this comparison directly.
x[x > A[, 1] & x < A[,2]]
#[1] 4 15

Answer 2

Here is a method that does not use an explicit loop or an apply function. outer is sometimes much faster.
x[rowSums(outer(x, A[,1], `>=`) & outer(x, A[,2], `<=`)) > 0]
[1] 4 15

Answer 3

This answer is late, but today I had the same problem to solve and my answer is maybe helpful for future readers. My solution was the following:
f3 <- function(x,A) {
Reduce(f = "|",
x = lapply(1:NROW(A),function(k) x>A[k,1] & x<A[k,2]),
init = logical(length(x)))
}
This function return a logical vector of length(x) indicating whether the corresponding value in x can be found in the intervals or not. If I want to get the elements I simply have to write
x[f3(x,A)]
I did some benchmarks and my function seems to work very well, also while testing with larger data.
Lets define the other solutions suggested here in this post:
f1 <- function(x,A) {
sapply(x, function(i) any(i > A[, 1] & i < A[,2]))
}
f2 <- function(x,A) {
rowSums(outer(x, A[,1], `>`) & outer(x, A[,2], `<`)) > 0
}
Now they are also returning a logical vector.
The benchmarks on my machine are following:
x <- c(4, 7, 15)
A <- cbind(c(3, 9, 14), c(5, 11, 16))
microbenchmark::microbenchmark(f1(x,A), f2(x,A), f3(x,A))
#Unit: microseconds
# expr min lq mean median uq max neval
#f1(x, A) 21.5 23.20 25.023 24.30 25.40 61.8 100
#f2(x, A) 18.8 21.20 23.606 22.75 23.70 75.4 100
#f3(x, A) 13.9 15.85 18.682 18.30 19.15 52.2 100
It seems like there is no big difference, but the follwoing example will make it more obvious:
x <- seq(1,100,length.out = 1e6)
A <- cbind(20:70,(20:70)+0.5)
microbenchmark::microbenchmark(f1(x,A), f2(x,A), f3(x,A), times=10)
#Unit: milliseconds
# expr min lq mean median uq max neval
#f1(x, A) 4176.172 4227.6709 4419.6010 4484.2946 4539.9668 4569.7412 10
#f2(x, A) 1418.498 1511.5647 1633.4659 1571.0249 1703.6651 1987.8895 10
#f3(x, A) 614.556 643.4138 704.3383 672.5385 770.7751 873.1291 10
That the functions all return the same result can be checked e.g. via:
all(f1(x,A)==f3(x,A))