Yes, this will be faster as C-Mex.
im1 = im > 50 & im < 100
Here at first a 640x480 LOGICAL matrix is calculated in "im > 50". Then a 2nd matrix is created for "im < 100". But is a pixels is <= 50, the test for > 100 is a waste if time. Most of all if the different intervals are not overlapping a lot of time can be saved by avoiding unnecessary tests.
Before I post a C-code, it would be helpful to know, if the intervals are continous, in opposite to your example. E.g. 50 < im < 100, 100 <= im < 150, 150 <= im < 200? Or is im==100 exluded on purpose?
[TEST-IMPLEMENTATION UNTIL QUESTIONS ARE CLEARED]: I assume, that the intervals are dense here.
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[]) {
uint8_T *im, limit1, limit2, limit3, limit4, p;
mxLogical *im1, *im2, *im3;
mwSize m, n, len;
mwIndex i;
// Check inputs:
if (nrhs != 5 || nlhs != 3) {
mexErrMsgIdAndTxt("JSimon:ImInterval:BadNArgs",
"Need 5 inputs and 3 outputs.");
}
if (!mxIsUint8(prhs[0])) {
mexErrMsgIdAndTxt("JSimon:ImInterval:BadTypeInput1",
"1st input must be an UINT8 matrix.");
}
// Get size of input and pointer to data:
m = mxGetM(plhs[0]);
n = mxGetN(plhs[0]);
im = (uint8_T *) mxGetData(prhs[0]);
len = m * n;
limit1 = (uint8_T) mxGetScalar(prhs[1]);
limit2 = (uint8_T) mxGetScalar(prhs[2]);
limit3 = (uint8_T) mxGetScalar(prhs[3]);
limit4 = (uint8_T) mxGetScalar(prhs[4]);
// Create outputs:
plhs[0] = mxCreateNumericMatrix(m, n, mxLOGICAL_CLASS, mxREAL);
im1 = (mxLogical *) mxGetData(plhs[0]);
plhs[1] = mxCreateNumericMatrix(m, n, mxLOGICAL_CLASS, mxREAL);
im2 = (mxLogical *) mxGetData(plhs[1]);
plhs[2] = mxCreateNumericMatrix(m, n, mxLOGICAL_CLASS, mxREAL);
im3 = (mxLogical *) mxGetData(plhs[2]);
// The calculations:
for (i = 0; i < len; i++) {
p = im[i];
if (p > limit1) {
if (p < limit2) {
im1[i] = 1;
} else if (p < limit3) {
im2[i] = 1;
} else if (p < limit4) {
im3[i] = 1;
}
}
}
return;
}
