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POVectorize matrix operation in R
 

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    <p>I have a R x C matrix filled to the k-th row and empty below this row. What i need to do is to fill the remaining rows. In order to do this, i have a function that takes 2 entire rows as arguments, process these rows and output 2 fresh rows (these outputs will fill the empty rows of the matrix, in batches of 2). I have a fixed matrix containing all 'pairs' of rows to be processed, but my for loop is not helping performance:</p> <pre><code># the processRows function: processRows = function(r1, r2) { # just change a little bit the two rows and return it in a compact way nr1 = r1 * 0.1 nr2 = -r2 * 0.1 matrix (c(nr1, nr2), ncol = 2) } # M is the matrix # nrow(M) and k are even, so nLeft is even M = matrix(1:48, ncol = 3) # half to fill (can be more or less, but k is always even) k = nrow(M)/2 # simulate empty rows to be filled M[-(1:k), ] = 0 cat('before fill') print(M) # number of empty rows to fill nLeft = nrow(M) - k nextRow = k + 1 # each row in idxList represents a 'pair' of rows to be processed # any pairwise combination of non-empty rows could happen # make it reproducible set.seed(1) idxList = matrix (sample(1:k, k), ncol = 2, byrow = TRUE) for ( i in 1 : (nLeft / 2)) { row1 = M[idxList[i, 1],] row2 = M[idxList[i, 2],] # the two columns in 'results' will become 2 rows in M results = processRows(row1, row2) # fill the matrix M[nextRow, ] = results[, 1] nextRow = nextRow + 1 M[nextRow, ] = results[, 2] nextRow = nextRow + 1 } cat('after fill') print(M) </code></pre>
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    1. COYou'll get much better answers if you make your question [reproducible](http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example). If `processRows()` is really involved or not entirely relevant to the question, simplify that bit and give us something we can work with.
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    2. COalso look at [this recent question](http://stackoverflow.com/questions/11377677/r-vectorised-conditional-replace/11379749) for some free speed suggestions via `complier` package or something more involved via `Rcpp` package
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    3. COIt is relatively trivial to make two long vectors of all pairs of rows. The real question is what is `processRows()` doing? The assignment operations you are doing in the loop are relatively trivial (you'll likely get virtually no gain from using the `compiler` package in an instance like this). **IF** `processRows()` can take all the data in one pass, you may get a significant performance improvement, but that is likely the bottleneck (and what you have told us _nothing_ about). You may get a slight gain by `M[nextRow:(nextRow+1), ] <- t(results)` to replace what you do in two assignments
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