Matrix Indexing

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Joerg
Joerg 2011-1-31
I have two index files:
On times:
t_on = [2 10 20 ...];
Off times:
t_off = [5 15 23 ...];
I want to get a vector with ones from 2..5, 10...15, 20...23
[0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 ...]
I am looking for a way to do so without using loops.
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Joerg
Joerg 2011-1-31
Because the vectors are large (1e6) and I assue working with loops is timeconsuming.
Doug Hull
Doug Hull 2011-1-31
I would not assume. Run the profiler.

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Matt Fig
Matt Fig 2011-1-31
Another alternative:
A = zeros(1,t_off(end)+1);
A(t_on) = 1;
A(t_off+1) = -1;
A = cumsum(A(1:t_off(end)));
  2 个评论
Sean de Wolski
Sean de Wolski 2011-1-31
This is a better answer; it'll be much faster.
Matt Fig
Matt Fig 2011-1-31
I want to point out that the above method depends on there being a zero between each run of ones. If this is not the case, use this instead:
L = length(t_on);
A = t_on(2:L)-t_off(1:L-1) > 1;
t_on = t_on([true A]);
t_off = t_off([A true]);
A = zeros(1,t_off(end));
A(t_on) = 1;
A(t_off(1:end-1)+1) = -1;
A = cumsum(A);
This doesn't add much time. Note that a well designed FOR loop is not much slower on my machine. For example, this:
R = zeros(1,t_off(end));
for ii = 1:length(t_off)
R(t_on(ii):t_off(ii)) = 1;
end
when called with this data:
t_on = 2:8:10000000;
t_off = t_on+ceil(rand(1,length(t_on))*8)+1;
runs in 1 second, whereas the vectorized solution above runs in .5 seconds. The CELLFUN method runs in 13 seconds.

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更多回答（3 个）

Matt Fig
Matt Fig 2011-1-31
Sean de Wolski
Sean de Wolski 2011-1-31
This will work, may not be faster than a loop.
t_on = [2 10 20];
t_off = [5 15 23];
pts = cellfun(@(x)(x(1):x(2)),num2cell([t_on.',t_off.'],2),'uni',false);
t = false(1,t_off(end));
t([pts{:}]) = true;
James Tursa
James Tursa 2011-1-31
As an exercise in speed I wrote a mex routine to do this. It is faster then the previously posted m-code, but not by a huge amount. Here it is:
/* Program: onesx
* Programmer: James Tursa
*
* Syntax: C = onesx( A , B [,largest_index] )
* A = real double vector of indexes to turn on 1's
* B = real double vector of indexes to turn off 1's (end inclusive)
* largest_index = optional largest index for result
*
* Example: C = onesx( [3 7 11] , [4 9 14] )
* C = 0 0 1 1 0 0 1 1 1 0 1 1 1 1
*
*
* Example: C = onesx( [3 7 11] , [4 9 14] , 16 )
* C = 0 0 1 1 0 0 1 1 1 0 1 1 1 1 0 0
*
*/
#include "mex.h"
#ifndef MWSIZE_MAX
#define mwIndex int
#define mwSignedIndex int
#define mwSize int
#endif
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
mwSize i, k, k1, k2, n1, n2, largest_index;
double *pr1, *pr2, *ones;
double x, z;
if( nrhs < 2 || nrhs > 3 ) {
mexErrMsgTxt("Expected two double vector inputs (and optional third max index).");
}
if( nlhs > 1 ) {
mexErrMsgTxt("Too many outputs.");
}
n1 = mxGetNumberOfElements(prhs[0]);
n2 = mxGetNumberOfElements(prhs[1]);
if( n1 == 0 || !mxIsDouble(prhs[0]) || mxGetNumberOfDimensions(prhs[0]) != 2 ||
mxIsComplex(prhs[0]) || (mxGetM(prhs[0]) != 1 && mxGetN(prhs[0]) != 1) ||
n2 == 0 || !mxIsDouble(prhs[1]) || mxGetNumberOfDimensions(prhs[1]) != 2 ||
mxIsComplex(prhs[1]) || (mxGetM(prhs[1]) != 1 && mxGetN(prhs[1]) != 1) ) {
mexErrMsgTxt("Inputs must be non-empty real double vectors.");
}
if( n1 != n2 ) {
mexErrMsgTxt("Inputs must have the same number of elements.");
}
if( nrhs == 3 ) {
largest_index = z = mxGetScalar(prhs[2]);
if( largest_index != z || largest_index < 0 ) {
mexErrMsgTxt("3rd argument largest index must be positive integral value and not too large.");
}
}
pr1 = mxGetPr(prhs[0]);
pr2 = mxGetPr(prhs[1]);
x = 0.0;
for( i=0; i<n1; i++ ) {
if( ((mwSize) pr1[i]) != pr1[i] ) {
mexPrintf("1st(%d) = %f\n",i+1,pr1[i]);
mexErrMsgTxt("Invalid input element. Must be positive integral value and not too large.");
}
if( ((mwSize) pr2[i]) != pr2[i] ) {
mexPrintf("2nd(%d) = %f\n",i+1,pr2[i]);
mexErrMsgTxt("Invalid input element. Must be positive integral value and not too large.");
}
if( pr1[i] > pr2[i] || pr1[i] <= 0 || pr2[i] <= 0 ) {
mexPrintf("1st(%d) = %d , 2nd(%d) = %d\n",i+1,(mwSize)pr1[i],i+1,(mwSize)pr2[i]);
mexErrMsgTxt("Invalid input element pair. Must have 0 < 1st <= 2nd.");
}
if( pr2[i] > x ) x = pr2[i];
if( nrhs == 3 && x > z ) {
mexPrintf("2nd(%d) = %d > 3rd = %d\n",i+1,(mwSize)x,largest_index);
mexErrMsgTxt("Input element is larger than 3rd argument largest index.");
}
}
if( nrhs == 3 ) x = z;
plhs[0] = mxCreateDoubleMatrix( 1, x, mxREAL);
ones = mxGetPr(plhs[0]);
for( i=0; i<n1; i++ ) {
k1 = ((mwSize) pr1[i]) - 1;
k2 = ((mwSize) pr2[i]) - 1;
for( k=k1; k<=k2; k++ ) {
ones[k] = 1.0;
}
}
}
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Ned Gulley
Ned Gulley 2011-1-31
Hi James:
If you add a space to the beginning of each line, it will be formatted as monospace text. Another way to do this is to select all of the code and then select the "code" button in the toolbar above the text input. I added the necessary spaces to your answer to make it look better.
James Tursa
James Tursa 2011-2-1
Aha! Thanks ... (I'm still getting used to this Answers section)

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