How many times each element of an array have the 1 value after n times assigned different logical values?

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Thao Linh Tran 2018-9-21

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评论： Thao Linh Tran 2018-9-21

I have n same size matrices containing different logical values such as A1, A2, A3 matrices. It also can be considered that a matrix_A was assigned n times with different logical values 1 and 0. Please help to point out how to count the number of times each elements of this A matrix having 1 value. Any suggestion would be much appreciated!

A1 = 1 0 0 1 0 1 0
1 0 1 0 0 0
1 1 0 0 1 0
1 1 0 1 0 1
A2 = 1 0 0 0 0 1 0
1 0 0 0 0 0
1 1 1 0 1 0
1 1 1 1 0 1
A3 = 0 0 0 0 0 1 1
1 1 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 1

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Stephen23 2018-9-21

编辑：Stephen23 2018-9-21

Numbered variables are a sign that you are doing something wrong. It is much more efficient to use one array (which could be numeric, cell, struct, table, etc) and indexing.

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

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Answer 1

Stephen23 2018-9-21

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编辑：Stephen23 2018-9-21

在 MATLAB Online 中打开

Simple: use one 3D array:

A = cat(3,...
    [1 0 0 1 0 1 0
     0 1 0 1 0 0 0
     0 1 1 0 0 1 0
     0 1 1 0 1 0 1],...
    [1 0 0 0 0 1 0
     0 1 0 0 0 0 0
     0 1 1 1 0 1 0
     0 1 1 1 1 0 1],...
    [0 0 0 0 0 1 1
     0 1 1 1 1 0 0
     0 0 0 0 0 0 0
     0 0 0 0 0 0 1]);

And then:

>> sum(A,3)
ans =
   2   0   0   1   0   3   1
   0   3   1   2   1   0   0
   0   2   2   1   0   2   0
   0   2   2   1   2   0   3

Note: if your arrays are in one cell array C then you can easily conver this into a 3D array:

A = cat(3,C{:});

If you have lots of separate variables, then you need to fix your code design.

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Thao Linh Tran 2018-9-21

Thanks a lot for prompt reply with effective solution!

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Answer 2

Fangjun Jiang 2018-9-21

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在 MATLAB Online 中打开

Constructing your data as 3d matrix will make it easy.

A(1,:,:) =[ 1 0 0 1 0 1 0
     0 1 0 1 0 0 0
     0 1 1 0 0 1 0
     0 1 1 0 1 0 1];
 A(2,:,:) = [1 0 0 0 0 1 0
     0 1 0 0 0 0 0
     0 1 1 1 0 1 0
     0 1 1 1 1 0 1];
 A(3,:,:) = [0 0 0 0 0 1 1
     0 1 1 1 1 0 0
     0 0 0 0 0 0 0
     0 0 0 0 0 0 1];
   m=4;
   n=5;
   sum(A(:,m,n))