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POIs there a more efficient way of nesting three for loops?
 

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  1. POIs there a more efficient way of nesting three for loops?
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    <p>I have been struggling for a while in converting the code below to use the *apply family of functions, so am now asking the StackOverflow community for a little help. Some background, this is part of a method I am developing to analyze propensity score methods for three groups. As such, I am starting with three matrices representing the distances (difference in propensity scores) between each pair of groups. That is, matrix d1 is A x B, d2 is B x C, and d3 is C x A. What I need to do is find triplets that minimize the overall distance as well as be less than some caliper. I've simplified the example as best as I could to run while getting at what I am trying to.</p> <p>Couple of notes:</p> <ul> <li><p>The distance less than the caliper check (<code>row1 &lt;- row1[row1 &lt; caliper]</code>) could be done at the end if I were to simply create a data.frame (or matrix) of all possible combinations. However, even with the small number of groups I set here would result in 3,000 rows!</p></li> <li><p>I order the vectors before moving onto the next step. Again, if I were to have a matrix of all possible combinations this could be eliminated. In my current version I have another line that will only look at the n smallest elements in order to reduce the execution time.</p></li> <li><p>This example has pretty small groups. I am working on a dataset where the groups have between 5,000 and 8,000 subjects each.</p></li> </ul> <p>Thanks in advance for any help. I am working on a paper for this and would be happy to give a acknowledgements. Also, I plan on attending the useR! conference in Spain and will buy a beer for whomever helps :-)</p> <pre><code>groups &lt;- c('Control','Treat1','Treat2') group.sizes &lt;- c(15, 10, 20) set.seed(2112) d1 &lt;- matrix(abs(rnorm(group.sizes[1] * group.sizes[2], mean=0, sd=1)), nrow=group.sizes[1], ncol=group.sizes[2], dimnames=list(1:group.sizes[1], (group.sizes[1]+1):(group.sizes[1] + group.sizes[2])) ) d2 &lt;- matrix(abs(rnorm(group.sizes[2] * group.sizes[3], mean=0, sd=1)), nrow=group.sizes[2], ncol=group.sizes[3], dimnames=list((group.sizes[1]+1):(group.sizes[1] + group.sizes[2]), (group.sizes[2] + group.sizes[1] + 1):(sum(group.sizes)) ) ) d3 &lt;- matrix(abs(rnorm(group.sizes[3] * group.sizes[1], mean=0, sd=1)), nrow=group.sizes[3], ncol=group.sizes[1], dimnames=list((group.sizes[2] + group.sizes[1] + 1):(sum(group.sizes)), 1:group.sizes[1]) ) caliper &lt;- 1 results &lt;- data.frame(v1=character(), v2=character(), v3=character(), d1=numeric(), d2=numeric(), d3=numeric()) for(i1 in dimnames(d1)[[1]]) { row1 &lt;- d1[i1,] row1 &lt;- row1[row1 &lt; caliper] row1 &lt;- row1[order(row1)] for(i2 in names(row1)) { row2 &lt;- d2[i2,] row2 &lt;- row2[row2 &lt; caliper] row2 &lt;- row2[order(row2)] for(i3 in names(row2)) { val &lt;- d3[i3,i1] if(val &lt; caliper) { results &lt;- rbind(results, data.frame(v1=i1, v2=i2, v3=i3, d1=row1[i2], d2=row2[i3], d3=val)) } } } } head(results) </code></pre>
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    1. USjbryer
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    1. COBuy a beer ? Is this not considered as a form of bribery on StackOverflow ? :)
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    2. COApparently the `df.sizes` vector is missing.
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      1. USjuba
    3. COSorry about that juba, fixed.
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