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copied!<p>Just a add-on note: if you know the range of the values for each vector, then it might be quicker to calculate the max and mins first, e.g.</p> <pre><code>order2 = function(x) .Internal(order(T, F, x)) findInterval2 = function(x, vec, rightmost.closed=F, all.inside=F) { nx <- length(x) index <- integer(nx) .C('find_interv_vec', xt=as.double(vec), n=length(vec), x=as.double(x), nx=nx, as.logical(rightmost.closed), as.logical(all.inside), index, DUP = FALSE, NAOK=T, PACKAGE='base') index } f <- function(a, b) { # set up vars a.length <- length(a) b.length <- length(b) b.sorted <- b[order2(b)] b.min <- b.sorted[1] b.max <- b.sorted[b.length] results <- integer(a.length) # pre-process minimums v.min <- which(a <= b.min) # pre-process maximums v.max <- which(a > b.max) results[v.max] <- b.max # compare the rest ind <- c(v.min, v.max) results[-ind] <- findInterval2(a[-ind] - 0.5, b.sorted) results } </code></pre> <p>Which gives the following timeings</p> <pre><code>> N <- 10 > n <- 1e5 > b <- runif(n, 0, 100) > a <- runif(n, 40, 60) # NB smaller range of values than b > summary( replicate(N, system.time(findInterval2(a - 0.5, b[order2(b)]))[3]) ) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0300 0.0300 0.0400 0.0390 0.0475 0.0500 > summary( replicate(N, system.time(f(a, b))[3]) ) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.010 0.030 0.030 0.027 0.030 0.040 </code></pre> <p>However, if you don't know the ranges ahead of time, or can't make an educated guess about them, then this would probably be slower.</p>

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