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copied!<p>Try this:</p> <pre><code>findInterval(a - 0.5, sort(b)) </code></pre> <hr> <p>Speed improvement from a) avoiding <code>sort</code>, and b) avoiding overhead in <code>findInterval</code> and <code>order</code> by using simpler <code>.Internal</code> wrappers:</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 } > system.time(for (i in 1:10000) findInterval(a - 0.5, sort(b))) user system elapsed 1.22 0.00 1.22 > system.time(for (i in 1:10000) sapply(a, function(x)sum(b<x))) user system elapsed 0.79 0.00 0.78 > system.time(for (i in 1:10000) rowSums(outer(a, b, ">"))) user system elapsed 0.72 0.00 0.72 > system.time(for (i in 1:10000) findInterval(a - 0.5, b[order(b)])) user system elapsed 0.42 0.00 0.42 > system.time(for (i in 1:10000) findInterval2(a - 0.5, b[order2(b)])) user system elapsed 0.16 0.00 0.15 </code></pre> <p>The complexity of defining <code>findInterval2</code> and <code>order2</code> is probably only warranted if you have heaps of iterations with fairly small N.</p> <p>Also timings for larger N:</p> <pre><code>> a = rep(a, 100) > b = rep(b, 100) > system.time(for (i in 1:100) findInterval(a - 0.5, sort(b))) user system elapsed 0.01 0.00 0.02 > system.time(for (i in 1:100) sapply(a, function(x)sum(b<x))) user system elapsed 0.67 0.00 0.68 > system.time(for (i in 1:100) rowSums(outer(a, b, ">"))) user system elapsed 3.67 0.26 3.94 > system.time(for (i in 1:100) findInterval(a - 0.5, b[order(b)])) user system elapsed 0 0 0 > system.time(for (i in 1:100) findInterval2(a - 0.5, b[order2(b)])) user system elapsed 0 0 0 </code></pre>

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