How can I make indexing faster, that is, searching for a groups of numbers within a matrix?

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Joe 2015-6-5

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评论： Joe 2015-6-15

采纳的回答： Walter Roberson

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Hello!

I was wondering if there was anything faster than this:

X = ( Y1 < a & Y1 >= b & Y2 >= c & Y2 < d );

I run it in a loop a lot and it is by far taking up the most time. X is a group of numbers that meets 4 conditions. I am searching two large matrices, Y1 and Y2 for numbers that meets that condition. Anyone have an easier way?

I am on Matlab 2015a, 64bit, 120GB Ram, and blazing fast processors.

Thanks!

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James Tursa 2015-6-8

编辑：James Tursa 2015-6-9

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Do you want X returned from the mex routine, or are you just interested in the Numbers(X) result? I.e., the mex routine can be set up to either return X, or just use X internally but only return the Numbers(X) result. Currently what I have is this, which only returns the Numbers(X) result, but it could easily be modified to return X as well if you needed it for some reason. To create the mex routine, copy this code into a file called subset4a.c (or any_other_name.c) and then issue the following command:

mex subset4a.c

// Returns subset of an array based on relational results
// Syntax:
//     Z = subset4a(Numbers,Y1,Y2,a,b,c,d);
//
// Returns equivalent of:
//     Z = Numbers(  Y1 < a  &  Y1 >= b  &  Y2 >= c  &  Y2 < d  )
//
// Programmer:  James Tursa
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
    double a, b, c, d;
    double *X, *Z, *Y1, *Y2;
    size_t *x, *z;
    size_t i, n, k, j;
      if( nrhs != 7 ) {
          mexErrMsgTxt("Need exactly 7 full real double inputs");
      }
      if( nlhs > 1 ) {
          mexErrMsgTxt("Too many outputs");
      }
      for( i=0; i<7; i++ ) {
          if( !mxIsDouble(prhs[i]) || mxIsComplex(prhs[i]) || mxIsSparse(prhs[i]) ) {
              mexErrMsgTxt("Need exactly 7 full real double inputs");
          }
      }
      n = mxGetNumberOfElements(prhs[0]);
      if( mxGetNumberOfElements(prhs[1]) != n || mxGetNumberOfElements(prhs[2]) != n ) {
          mexErrMsgTxt("1st three arguments must have same number of elements");
      }
      for( i=3; i<7; i++ ) {
          if( mxGetNumberOfElements(prhs[i]) != 1 ) {
              mexErrMsgTxt("Last four arguments must be scalars");
          }
      }
      X = mxGetPr(prhs[0]);
      Y1 = mxGetPr(prhs[1]);
      Y2 = mxGetPr(prhs[2]);
      a = mxGetScalar(prhs[3]);
      b = mxGetScalar(prhs[4]);
      c = mxGetScalar(prhs[5]);
      d = mxGetScalar(prhs[6]);
      z = x = mxMalloc(n*sizeof(*x)); // Work array to save indexes
      for( i=0; i<n; i++ ) {
          if( *Y1 < a  &&  *Y1 >= b  &&  *Y2 >= c  &&  *Y2 < d ) {
              *z++ = i; // Remember the index of the element to save
          }
          Y1++;
          Y2++;
      }
      k = z - x; // Number of indexes we saved
      plhs[0] = mxCreateDoubleMatrix( k, 1, mxREAL ); // EDIT FIX
      if( k ) {
          Z = mxGetPr(plhs[0]);
          z = x;
          for( i=0; i<k; i++ ) {
              *Z++ = X[*z++]; // Copy the elements to the result
          }
      }
      mxFree(x); // Free the work array
  }

James Tursa 2015-6-9

Use the edited version.

Joe 2015-6-15

James,

I implemented this code and it reduced the time it takes to run that code ~50 - 80%!! Great! Thank you very much. I am now looking at mex'ing other lines of code to get (hopefully) similar speed ups. Thanks again!

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采纳的回答

Walter Roberson 2015-6-5

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In the special case that a relatively small fraction satisfies one of the conditions, and letting letting C1, C2, C3, C4 be the conditions ranked from least to most probable, such as C1 being "Y1 >= B", then you might try:

X( (((sparse(C1) & C2) & C3) & C4 )

for example

X( ((sparse(Y1 >= B) & (Y2 < d)) & (Y2 >= c)) & (Y1 < a) )

I don't promise it will be faster, but in theory it could be.

Note: this has more overhead than the way you used, so your occupancy needs to b less than... ummm, perhaps 1/4 maybe... before this gets speedup.

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Joe 2015-6-5

Walter -

I implemented this and got 20% speed up! Great suggestion!

Thanks, Joe

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