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copied!<p>Following up @CarlWitthoft's answer, you probably want:</p> <pre><code>N <- 5 rlex <-NULL for (i in 1:N) { x <-sample(0:1, 100000, 1/2) rlex <-append(rlex, rle(x)$lengths) } </code></pre> <p>since I think you don't care about the <code>$values</code> component (i.e. whether each run is a run of zeros or ones).</p> <p><strong>Result</strong>: one long vector of run lengths.</p> <p>But this would probably be a lot more efficient:</p> <pre><code>maxlen <- 30 rlemat <- matrix(nrow=N,ncol=maxlen) for (i in 1:N) { x <-sample(0:1, 100000, 1/2) rlemat[i,] <- table(factor(rle(x)$lengths,levels=1:maxlen)) } </code></pre> <p><strong>Result</strong>: an <code>N</code> by <code>maxlen</code> table of run lengths from each iteration.</p> <p>If you only want to save the total number of runs of each length you could try:</p> <pre><code>rlecumsum <- rep(0,maxlen) for (i in 1:N) { x <-sample(0:1, 100000, 1/2) rlecumsum <- rlecumsum + table(factor(rle(x)$lengths,levels=1:maxlen)) } </code></pre> <p><strong>Result</strong>: an vector of length <code>maxlen</code> of the total numbers of run lengths across all iterations.</p> <p>And here's my final answer:</p> <pre><code>rlecumtab <- matrix(0,ncol=2,nrow=maxlen) for (i in 1:N) { x <- sample(0:1, 100000, 1/2) r1 <- rle(x) rtab <- table(factor(r1$lengths,levels=1:maxlen),r1$values) rlecumtab <- rlecumtab + rtab } </code></pre> <p><strong>Result</strong>: a <code>maxlen</code> by 2 table of the total numbers of run lengths across all iterations, divided by type (0-run vs 1-run).</p>

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