The great findInterval()
function in R uses left-closed sub-intervals in its vec
argument, as shown in its docs:
if
i <- findInterval(x,v)
, we havev[i[j]] <= x[j] < v[i[j] + 1]
If I want right-closed sub-intervals, what are my options? The best I've come up with is this:
findInterval.rightClosed <- function(x, vec, ...) {
fi <- findInterval(x, vec, ...)
fi - (x==vec[fi])
}
Another one also works:
findInterval.rightClosed2 <- function(x, vec, ...) {
length(vec) - findInterval(-x, -rev(vec), ...)
}
Here's a little test:
x <- c(3, 6, 7, 7, 29, 37, 52)
vec <- c(2, 5, 6, 35)
findInterval(x, vec)
# [1] 1 3 3 3 3 4 4
findInterval.rightClosed(x, vec)
# [1] 1 2 3 3 3 4 4
findInterval.rightClosed2(x, vec)
# [1] 1 2 3 3 3 4 4
But I'd like to see any other solutions if there's a better one. By "better", I mean "somehow more satisfying" or "doesn't feel like a kludge" or maybe even "more efficient". =)
(Note that there's a rightmost.closed
argument to findInterval()
, but it's different - it only refers to the final sub-interval and has a different meaning.)
EDIT: Major clean-up in all aisles.
You might look at cut
. By default, cut
makes left open and right closed intervals, and that can be changed using the appropriate argument (right
). To use your example:
x <- c(3, 6, 7, 7, 29, 37, 52)
vec <- c(2, 5, 6, 35)
cutVec <- c(vec, max(x)) # for cut, range of vec should cover all of x
Now create four functions that should do the same thing: Two from the OP, one from Josh O'Brien, and then cut
. Two arguments to cut
have been changed from default settings: include.lowest = TRUE
will create an interval closed on both sides for the smallest (leftmost) interval. labels = FALSE
will cause cut
to return simply the integer values for the bins instead of creating a factor, which it otherwise does.
findInterval.rightClosed <- function(x, vec, ...) {
fi <- findInterval(x, vec, ...)
fi - (x==vec[fi])
}
findInterval.rightClosed2 <- function(x, vec, ...) {
length(vec) - findInterval(-x, -rev(vec), ...)
}
cutFun <- function(x, vec){
cut(x, vec, include.lowest = TRUE, labels = FALSE)
}
# The body of fiFun is a contribution by Josh O'Brien that got fed to the ether.
fiFun <- function(x, vec){
xxFI <- findInterval(x, vec * (1 + .Machine$double.eps))
}
Do all functions return the same result? Yup. (notice the use of cutVec
for cutFun
)
mapply(identical, list(findInterval.rightClosed(x, vec)),
list(findInterval.rightClosed2(x, vec), cutFun(x, cutVec), fiFun(x, vec)))
# [1] TRUE TRUE TRUE
Now a more demanding vector to bin:
x <- rpois(2e6, 10)
vec <- c(-Inf, quantile(x, seq(.2, 1, .2)))
Test whether identical (note use of unname
)
mapply(identical, list(unname(findInterval.rightClosed(x, vec))),
list(findInterval.rightClosed2(x, vec), cutFun(x, vec), fiFun(x, vec)))
# [1] TRUE TRUE TRUE
And benchmark:
library(microbenchmark)
microbenchmark(findInterval.rightClosed(x, vec), findInterval.rightClosed2(x, vec),
cutFun(x, vec), fiFun(x, vec), times = 50)
# Unit: milliseconds
# expr min lq median uq max
# 1 cutFun(x, vec) 35.46261 35.63435 35.81233 36.68036 53.52078
# 2 fiFun(x, vec) 51.30158 51.69391 52.24277 53.69253 67.09433
# 3 findInterval.rightClosed(x, vec) 124.57110 133.99315 142.06567 155.68592 176.43291
# 4 findInterval.rightClosed2(x, vec) 79.81685 82.01025 86.20182 95.65368 108.51624
From this run, cut
seems to be the fastest.