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    <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 &lt;- length(x) index &lt;- 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 } &gt; system.time(for (i in 1:10000) findInterval(a - 0.5, sort(b))) user system elapsed 1.22 0.00 1.22 &gt; system.time(for (i in 1:10000) sapply(a, function(x)sum(b&lt;x))) user system elapsed 0.79 0.00 0.78 &gt; system.time(for (i in 1:10000) rowSums(outer(a, b, "&gt;"))) user system elapsed 0.72 0.00 0.72 &gt; system.time(for (i in 1:10000) findInterval(a - 0.5, b[order(b)])) user system elapsed 0.42 0.00 0.42 &gt; 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>&gt; a = rep(a, 100) &gt; b = rep(b, 100) &gt; system.time(for (i in 1:100) findInterval(a - 0.5, sort(b))) user system elapsed 0.01 0.00 0.02 &gt; system.time(for (i in 1:100) sapply(a, function(x)sum(b&lt;x))) user system elapsed 0.67 0.00 0.68 &gt; system.time(for (i in 1:100) rowSums(outer(a, b, "&gt;"))) user system elapsed 3.67 0.26 3.94 &gt; system.time(for (i in 1:100) findInterval(a - 0.5, b[order(b)])) user system elapsed 0 0 0 &gt; 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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    1. CO@Charles: this is slower than the `sapply` solution of the OP -- I tested it by doing `system.time( replicate( 10000, ... ) )`.
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      1. USPrasad Chalasani
    2. COFor small a/b, the sort dominates the run-time. `b[order(b)]` benchmarks about 4x faster than `sort(b)`, and twice as fast as the `sapply` original on my box. For larger a/b (rep(100)), you get orders of magnitude improvement by using findInterval.
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      1. USCharles
    3. COMost of the remaining overhead is in stuff like the `is.sorted` and `is.na` checks in `findInterval`, and the various checks in order. If you define thinner wrappers around those functions (eg `order2 = function(x) .Internal(order(T, F, x))`), then it gets about another 3x faster.
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