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    <p>Well, we can't have you resorting to for-loops, now can we :)</p> <p>There is of course the question of how to represent the sparse matrix. A simple way is to have it only contain the indices of the points that are closest (and recalculate as needed). But in the solution below, I put both distance ('d1' etc) and index ('i1' etc) in a single matrix:</p> <pre><code>sparseDist &lt;- function(m, k) { m &lt;- t(m) n &lt;- ncol(m) d &lt;- vapply( seq_len(n-1L), function(i) { d&lt;-colSums((m[, seq(i+1L, n), drop=FALSE]-m[,i])^2) o&lt;-sort.list(d, na.last=NA, method='quick')[seq_len(k)] c(sqrt(d[o]), o+i) }, numeric(2*k) ) dimnames(d) &lt;- list(c(paste('d', seq_len(k), sep=''), paste('i', seq_len(k), sep='')), colnames(m)[-n]) d } </code></pre> <p>Trying it out on 9 2d-points:</p> <pre><code>&gt; m &lt;- matrix(c(0,0, 1.1,0, 2,0, 0,1.2, 1.1,1.2, 2,1.2, 0,2, 1.1,2, 2,2), 9, byrow=TRUE, dimnames=list(letters[1:9], letters[24:25])) &gt; print(dist(m), digits=2) a b c d e f g h b 1.1 c 2.0 0.9 d 1.2 1.6 2.3 e 1.6 1.2 1.5 1.1 f 2.3 1.5 1.2 2.0 0.9 g 2.0 2.3 2.8 0.8 1.4 2.2 h 2.3 2.0 2.2 1.4 0.8 1.2 1.1 i 2.8 2.2 2.0 2.2 1.2 0.8 2.0 0.9 &gt; print(sparseDist(m, 3), digits=2) a b c d e f g h d1 1.1 0.9 1.2 0.8 0.8 0.8 1.1 0.9 d2 1.2 1.2 1.5 1.1 0.9 1.2 2.0 NA d3 1.6 1.5 2.0 1.4 1.2 2.2 NA NA i1 2.0 3.0 6.0 7.0 8.0 9.0 8.0 9.0 i2 4.0 5.0 5.0 5.0 6.0 8.0 9.0 NA i3 5.0 6.0 9.0 8.0 9.0 7.0 NA NA </code></pre> <p>And trying it on a larger problem (10k points). Still, on 100k points and more dimensions it will take a long time (like 15-30 minutes).</p> <pre><code>n&lt;-1e4; m&lt;-3; m=matrix(runif(n*m), n) system.time( d &lt;- sparseDist(m, 3) ) # 9 seconds on my machine... </code></pre> <p>P.S. Just noted that you posted an answer as I was writing this: the solution here is roughly twice as fast because it doesn't calculate the same distance twice (the distance between points 1 and 13 is the same as between points 13 and 1).</p>
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    1. POComputing sparse pairwise distance matrix in R
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    1. POComputing sparse pairwise distance matrix in R
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    1. COThanks for this answer. I agree it's about twice as fast. However, for my application (kNN) I think having only the upper triangle of the distance matrix is actually slightly inconvenient. I think I may stick with a parallelized version of the code I submitted. Thanks again though!
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      1. USChristopher DuBois
 

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