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POHow can I make this R matrix filling function faster?
 

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  1. POHow can I make this R matrix filling function faster?
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    <p>A while back I wrote a function to fill time series matrices that had NA values up according to the specifications needed and it's had occational uses on a few matrices that are about 50000 rows, 350 columns. The matrix can contain either numeric or character values. The main problem is fixing the matrix is slow and I thought I'd gauge some experts on how to do this faster.</p> <p>I guess going to rcpp or paralleling it might help but I think it's might be my design rather than R itself that's inefficient. I generally vecotrize everything in R but since the missing values follow no pattern I've found no other way than to work with the matrix on a per row basis. </p> <p>The function needs to be called so it can carry forwards missing values and also be called to quickly just fill the latest values with the last known one.</p> <p>Here is an example matrix:</p> <pre><code>testMatrix &lt;- structure(c(NA, NA, NA, 29.98, 66.89, NA, -12.78, -11.65, NA, 4.03, NA, NA, NA, 29.98, 66.89, NA, -12.78, -11.65, NA, NA, NA, NA, NA, 29.98, 66.89, NA, -12.78, NA, NA, 4.76, NA, NA, NA, NA, 66.89, NA, -12.78, NA, NA, 4.76, NA, NA, NA, 29.98, 66.89, NA, -12.78, NA, NA, 4.76, NA, NA, NA, 29.98, 66.89, NA, -12.78, NA, NA, 4.39, NA, NA, NA, 29.98, 66.89, NA, -10.72, -11.65, NA, 4.39, NA, NA, NA, 29.98, 50.65, NA, -10.72, -11.65, NA, 4.39, NA, NA, 4.72, NA, 50.65, NA, -10.72, -38.61, 45.3, NA), .Dim = c(10L, 9L), .Dimnames = list(c("ID_a", "ID_b", "ID_c", "ID_d", "ID_e", "ID_f", "ID_g", "ID_h", "ID_i", "ID_j"), c("2010-09-30", "2010-10-31", "2010-11-30", "2010-12-31", "2011-01-31", "2011-02-28", "2011-03-31", "2011-04-30", "2011-05-31"))) print(testMatrix) 2010-09-30 2010-10-31 2010-11-30 2010-12-31 2011-01-31 2011-02-28 2011-03-31 2011-04-30 2011-05-31 ID_a NA NA NA NA NA NA NA NA NA ID_b NA NA NA NA NA NA NA NA NA ID_c NA NA NA NA NA NA NA NA 4.72 ID_d 29.98 29.98 29.98 NA 29.98 29.98 29.98 29.98 NA ID_e 66.89 66.89 66.89 66.89 66.89 66.89 66.89 50.65 50.65 ID_f NA NA NA NA NA NA NA NA NA ID_g -12.78 -12.78 -12.78 -12.78 -12.78 -12.78 -10.72 -10.72 -10.72 ID_h -11.65 -11.65 NA NA NA NA -11.65 -11.65 -38.61 ID_i NA NA NA NA NA NA NA NA 45.30 ID_j 4.03 NA 4.76 4.76 4.76 4.39 4.39 4.39 NA </code></pre> <p>This is the function I currently use:</p> <pre><code># ---------------------------------------------------------------------------- # GetMatrixWithBlanksFilled # ---------------------------------------------------------------------------- # # Arguments: # inputMatrix --- A matrix with gaps in the time series rows # fillGapMax --- The max number of columns to carry a number # forward if there are no more values in the # time series row. # # Returns: # A matrix with gaps filled. GetMatrixWithBlanksFilled &lt;- function(inputMatrix, fillGapMax = 6, forwardLooking = TRUE) { if("DEBUG_ON" %in% ls(globalenv())){browser()} cntRow &lt;- nrow(inputMatrix) cntCol &lt;- ncol(inputMatrix) # if (forwardLooking) { for (i in 1:cntRow) { # Store the location of the first non NA element in the row firstValueCol &lt;- (1:cntCol)[!is.na(inputMatrix[i,])][1] if (!(is.na(firstValueCol))) { if (!(firstValueCol == cntCol)) { nextValueCol &lt;- firstValueCol # If there is a a value number in the row and it's not at the end of the time # series, start iterating through the row while there are more NA values and # more data values and not at the end of the row continue. while ((sum(as.numeric(is.na(inputMatrix[i,nextValueCol:cntCol]))))&gt;0 &amp;&amp; (sum(as.numeric(!is.na(inputMatrix[i,nextValueCol:cntCol]))))&gt;0 &amp;&amp; !(nextValueCol == cntCol)) { # Find the next NA element nextNaCol &lt;- (nextValueCol:cntCol)[is.na(inputMatrix[i,nextValueCol:cntCol])][1] # Find the next value element nextValueCol &lt;- (nextNaCol:cntCol)[!is.na(inputMatrix[i,nextNaCol:cntCol])][1] # If there is another value element then fill up all NA elements in between with the last known value if (!is.na(nextValueCol)) { inputMatrix[i,nextNaCol:(nextValueCol-1)] &lt;- inputMatrix[i,(nextNaCol-1)] } else { # If there is no other value element then fill up all NA elements up to the max number supplied # with the last known value unless it's close to the end of the row then just fill up to the end. inputMatrix[i,nextNaCol:min(nextNaCol+fillGapMax,cntCol)] &lt;- inputMatrix[i,(nextNaCol-1)] nextValueCol &lt;- cntCol } } } } } } else { for (i in 1:cntRow) { if (is.na(inputMatrix[i,ncol(inputMatrix)])) { tempRow &lt;- inputMatrix[i,max(1,length(inputMatrix[i,])-fillGapMax):length(inputMatrix[i,])] if (length(tempRow[!is.na(tempRow)])&gt;0) { lastNonNaLocation &lt;- (length(tempRow):1)[!is.na(tempRow)][length(tempRow[!is.na(tempRow)])] inputMatrix[i,(ncol(inputMatrix)-lastNonNaLocation+2):ncol(inputMatrix)] &lt;- tempRow[!is.na(tempRow)][length(tempRow[!is.na(tempRow)])] } } } } return(inputMatrix) } </code></pre> <p>I'm then calling it with something like:</p> <pre><code>&gt; fixedMatrix1 &lt;- GetMatrixWithBlanksFilled(testMatrix,fillGapMax=12,forwardLooking=TRUE) &gt; print(fixedMatrix1) 2010-09-30 2010-10-31 2010-11-30 2010-12-31 2011-01-31 2011-02-28 2011-03-31 2011-04-30 2011-05-31 ID_a NA NA NA NA NA NA NA NA NA ID_b NA NA NA NA NA NA NA NA NA ID_c NA NA NA NA NA NA NA NA 4.72 ID_d 29.98 29.98 29.98 29.98 29.98 29.98 29.98 29.98 29.98 ID_e 66.89 66.89 66.89 66.89 66.89 66.89 66.89 50.65 50.65 ID_f NA NA NA NA NA NA NA NA NA ID_g -12.78 -12.78 -12.78 -12.78 -12.78 -12.78 -10.72 -10.72 -10.72 ID_h -11.65 -11.65 -11.65 -11.65 -11.65 -11.65 -11.65 -11.65 -38.61 ID_i NA NA NA NA NA NA NA NA 45.30 ID_j 4.03 4.03 4.76 4.76 4.76 4.39 4.39 4.39 4.39 </code></pre> <p>or</p> <pre><code>&gt; fixedMatrix2 &lt;- GetMatrixWithBlanksFilled(testMatrix,fillGapMax=1,forwardLooking=FALSE) &gt; print(fixedMatrix2) 2010-09-30 2010-10-31 2010-11-30 2010-12-31 2011-01-31 2011-02-28 2011-03-31 2011-04-30 2011-05-31 ID_a NA NA NA NA NA NA NA NA NA ID_b NA NA NA NA NA NA NA NA NA ID_c NA NA NA NA NA NA NA NA 4.72 ID_d 29.98 29.98 29.98 NA 29.98 29.98 29.98 29.98 29.98 ID_e 66.89 66.89 66.89 66.89 66.89 66.89 66.89 50.65 50.65 ID_f NA NA NA NA NA NA NA NA NA ID_g -12.78 -12.78 -12.78 -12.78 -12.78 -12.78 -10.72 -10.72 -10.72 ID_h -11.65 -11.65 NA NA NA NA -11.65 -11.65 -38.61 ID_i NA NA NA NA NA NA NA NA 45.30 ID_j 4.03 NA 4.76 4.76 4.76 4.39 4.39 4.39 4.39 </code></pre> <p>This example runs quickly but is there any way to make it faster for large matrices?</p> <pre><code>&gt; n &lt;- 38 &gt; m &lt;- 5000 &gt; bigM &lt;- matrix(rep(testMatrix,n*m),m*nrow(testMatrix),n*ncol(testMatrix),FALSE) &gt; system.time(output &lt;- GetMatrixWithBlanksFilled(bigM,fillGapMax=12,forwardLooking=TRUE)) user system elapsed 86.47 0.06 87.24 </code></pre> <p>This dummy one has a lot of NA only rows and completely filled ones but the normal ones can take about 15-20 min.</p> <p><strong>UPDATE</strong></p> <p>Regarding Charles' comment about na.locf not completely mirroring the logic of the above: Below is a simplified version of how the final function is excluding checks for inputs etc:</p> <pre><code>FillGaps &lt;- function( dataMatrix, fillGapMax ) { require("zoo") numRow &lt;- nrow(dataMatrix) numCol &lt;- ncol(dataMatrix) iteration &lt;- (numCol-fillGapMax) if(length(iteration)&gt;0) { for (i in iteration:1) { tempMatrix &lt;- dataMatrix[,i:(i+fillGapMax),drop=FALSE] tempMatrix &lt;- t(zoo::na.locf(t(tempMatrix), na.rm=FALSE, maxgap=fillGapMax)) dataMatrix[,i:(i+fillGapMax)] &lt;- tempMatrix } } return(dataMatrix) } </code></pre>
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    1. COThis seems like a case when you need native code accelerator -- if this method has a name, try finding it in CRAN; if it is not there or not, you should implement it in C or Fortran and run from R (although it is easy -- see R-ext). Or use Rcpp for even easier conversion.
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