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

Question

Mantas Vaitonis 2018-7-8

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/409297-count-conservative-occurrences-of-zeros-and-if-zero-appeared-for-defined-number-of-times-it-should-b

评论： Mantas Vaitonis 2018-7-9

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:

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.

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Jan 2018-7-8

编辑：Jan 2018-7-8

在 MATLAB Online 中打开

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

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

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Jan 2018-7-8

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/409297-count-conservative-occurrences-of-zeros-and-if-zero-appeared-for-defined-number-of-times-it-should-b#answer_327940

编辑：Jan 2018-7-8

在 MATLAB Online 中打开

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

7 个评论
显示 5更早的评论隐藏 5更早的评论

Jan 2018-7-8

编辑：Jan 2018-7-8

在 MATLAB Online 中打开

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))

Jan 2018-7-9

@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 2018-7-9