Count conservative occurrences of zeros and if zero appeared for defined number of times it should be changed.

Hello,

There is an array A size (NxM), each column represents when a signal was created. I need to close that signal if certain time it was not created again and assign -1 value. For example like this:

1 0 1 0 1
0 1 0 1 0
0 0 1 0 0
1 0 0 1 1
0 0 1 1 0
0 1 0 0 0

I need to count consecutive occurrences of zero column-vise, and if zero occurred for defined number of times for example two, it should be changed to -1 and output should look like this:

1  0 1 0  1
0  1 0 1  0
-1  0 1 0 -1
1 -1 0 1  1
0  0 1 1  0
-1  1 0 0 -1

Count conservative occurrences of zeros and if zero appeared for defined number of times it should be changed.

[Jan]

I assume you mean "consecutive" instead of "conservative". What is the wanted output for:

x1 = [0; 0; 1]
x2 = [1; 0; 0; 0; 0]

Answer 1

[Jan]

If you want to replace the 2nd element after a 1 by -1, if it was a 0 before:

A = double(rand(10, 5) < 0.3);   % Some test data

n       = 2;
pattern = [1, zeros(1, n)];
for k = 1:size(A, 2)
  index = strfind(A(:, k).', pattern);
  A(index + n, k) = -1;
end

[Mantas Vaitonis]

Hello @Jan. Thank You for the answer. What variable "Signals" suppose to be? Would there be a way to avoid for loop, because this part will be bottle neck for my algorithm, due to big data. I am able to count the numbers of appearance, but only for vectors.

x = [1 0 0 1 1 1 0 1 0 0 0 1 1 0 1 1 1 1];
[v,h]=size(x);
ind=find(diff(x,1,2)) ;
di = [ind h] - [0 ind];
b = arrayfun(@(x) x-1:-1:0, di , 'un',0);
d=cellfun(@(x) fliplr(x),b,'un',0);
y = cat(2,d{:});

[Mantas Vaitonis]

Now after you edited your algorithm it does work. Is there a way to get same results without for loops? You did use did use strfind for patterns, which does work only for rows. Maybe there would be way to count 0 consecutive occurrence?

[Jan]

@Mantas Vaitonis: "Signal" was a typo only, I've adjusted the name of the variable to your "A" already.

strfind operates on vectors only, but here the loop is very efficient. I do not see a reason to search for a vectorized version, if it is slower. For:

A = double(rand(1e6, 5) < 0.3);

my code needs 0.046 seconds only.

If speed really matters, a C-mex function would be better. If you answer my question about the wanted answer for:

x1 = [0; 0; 1]
x2 = [1; 0; 0; 0; 0]

i will post a C-mex method. But without clarifying this detail, such a code would be based on guessing, what you exactly need, and therefore potentially useless.

My method: [1;0;0;0;0]   ==>   [1;0;-1;0;0]
Paolo's:   [1;0;0;0;0]   ==>   [1;0;-1;0;-1]

So please be so kind and define exactly, what you need.

By the way, I do not understand your code in the comment. It calculates:

x = [1 0 0 1 1 1 0 1 0 0 0 1 1 0 1 1 1 1]
y = [0 0 1 0 1 2 0 0 0 1 2 0 1 0 0 1 2 3]

How does this match the question? I'd expect an output like:

y = [1 0 -1 1 1 1 0 1 0 -1 0 1 1 0 1 1 1 1]

[Jan]

Well, I tried it with guessing, that you want [1;0;-1;0;0] as output for [1;0;0;0;0]:

// File:    InsertM1.c
// Compile: mex -O InsertM1.c
#include "mex.h"  
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[]) {
  double *A, *B, a;
  size_t mA, nA, iC, iR, n, count;

    // Get inputs:
    A  = mxGetPr(prhs[0]);
    mA = mxGetM(prhs[0]);
    nA = mxGetN(prhs[0]);
    n  = (size_t) mxGetScalar(prhs[1]);

    // Create outputs:
    // plhs[0] = mxCreateDoubleMatrix(mA, nA, mxREAL);
    plhs[0] = mxCreateUninitNumericMatrix(mA, nA, mxDOUBLE_CLASS, mxREAL);
    B       = mxGetPr(plhs[0]);

    for (iC = 0; iC < nA; iC++) {     // Loop over columns
       count = n + 1;                 // Reset counter, ignore initial zeros
       for (iR = 0; iR < mA; iR++) {  // Loop over rows
          a = *A++;
          if (a == 1.0) {             // 1 is found, reset the counter
             count = 0;
          } else if (++count == n) {  // If counter==n insert -1
             a = -1.0;
          }
          *B++ = a;
       }
    }
  }

For

A = double(rand(1e6, 5) < 0.3);

this takes 0.037 seconds, while the M version with strfind needs 0.086 seconds.

The code is faster if you operate on int8 data:

// replace double *A, *B, a;  by
int8_T *A, *B, a;

// Replace: ... mxGetPr(...  by
(int8_T *) mxGetData(...

// Replace a == 1.0 by a == 1 and a == -1.0 by a == -1

Then the code runs in the half time. Even if you convert the data at each call, the code is 10% faster than with doubles:

A = double(InsertM1(int8(A), 2))

[Mantas Vaitonis]

@Jan Thank You for your time and effort. Your guess was right I want [1;0;0;0;0] to get output like this [1;0;-1;0;0]. Yes now Your solution is even faster. If I may ask one more question, due to the fact that I am using GPU and not that advanced on MEX. How to change the code to work with gpuArray?

[Jan]

@Mantas Vaitonis: strfind is not implemented for GPUs. If you have a C compiler installed, I'd suggest to try the MEX function. Which OS do you use? Can you work with UINT8 instead of DOUBLEs?

[Mantas Vaitonis]

@Jan, I am using WIN 10 64BIT, yes it would be possible to use UNIT8 instead of double, becasue this varibale valeus are 1, 0 or -1.