To code based on percent in a matrix.

a=[0 0 0 0 1 1 0 0; 0 0 0 0 0 0 0 0; 0 0 0 0 0 1 1 1; 1 0 0 0 0 0 0 0]

I need to replace zero value in a m x n matrix with value based on below function. For example, the matrix is as matrix a above. How can I code this problem to replace the matrix by values of this function? I just know the code to search zero values.

% code to search zero values

iszero=cellfun(@(p) isequal(p,0),dataShift);

          % This function will randomly set a rules and save it in mxn matrix without disturbed the zero values

          % case 1
          % +----------------------+
          % |  Value    |  Percent |
          % |11         |    0.5   |
          % |22         |    0.3   |
          % |33         |   0.175  |
          % |44         |   0.025  |
          % +----------------------+
          % case 2
          % +----------------------+
          % |  Value    |  Percent |
          % |11         |   0.45   |
          % |22         |   0.33   |
          % |33         |   0.195  |
          % |44         |   0.025  |
          % +----------------------+
          % case 3
          % +----------------------+
          % |  Value    |  Percent |
          % |11         |   0.40   |
          % |22         |   0.36   |
          % |33         |   0.215  |
          % |44         |   0.025  |
          % +----------------------+

update 4.29pm p.m'sia:

*algorithm required as below:

*1. In first row of matrix, case selected randomly is case 3. Therefore, in row 1, 40% zero values will assign as 11, 36% zero values will assign as 22, 21.5% zero values will assign as 33 and 2.5% zero values will assign as 44. The position in zero values available for 11, 22, 33 and 44, the selection will do randomly in that row.

2. Then, the next row will repeat process 1, but the case select is based on random.*

[David Young]

Please explain the function more clearly. If a(p,q) is zero, what value should be assigned to a(p,q)? For example, what value should be assigned to a(2,3), and how does that relate to the table you display?

[yue ishida]

Ok, the algorithm as below:

1. In first row of matrix, case selected randomly is case 3. Therefore, in row 1, 40% zero values will assign as 11, 36% zero values will assign as 22, 21.5% zero values will assign as 33 and 2.5% zero values will assign as 44. The position in zero values available for 11, 22, 33 and 44, the selection will do randomly in that row.

2. Then, the next row will repeat process 1, but the case select is based on random.

[David Young]

That's clearer, thank you.

There's still a problem in knowing how to apply the probabilities. In row 1, there are 6 zeros. 40% of 6 is 2.4, 36% of 6 is 2.16, and so on. So exactly how many values should be set to 11, how many to 22 etc? Or do you want to set each zero element to a value chosen randomly from the distribution given (which would result in an unpredictable number of 11s in the row on any given run)?

It would help if you explain what the application is, or whether this is an exercise aimed at learning about programming.

Incidentally, your code that calls cellfun seems irrelevant. There are no cells involved in this task.

[yue ishida]

this is a problem given by my instructor. she want me to study how to code based on problem given

Answer 1

[Andrei Bobrov]

try this is code:

a=[0 0 0 0 1 1 0 0;
   0 0 0 0 0 0 0 0; 
   0 0 0 0 0 1 1 1; 
   1 0 0 0 0 0 0 0];
cases = {[0.5 0.3   0.175  0.025],...
         [ 0.45  0.33    0.195   0.025],...
         [ 0.40  0.36   0.215   0.025]};
a2 = cellfun(@(x)cumsum([0 x]),cases,'un',0);
out = a;
for j1 = 1:size(a,1)
    id = find(~a(j1,:));
    n = numel(id);
    [~,b] = histc(linspace(0,1-1e2*eps,n),a2{randi(3)});
    b = b*11;
    out(j1,id) = b(randperm(n));
end