Could anyone help me to obtain new_C for the below mentioned code

Question

Prabha Kumaresan 2018-1-3

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https://ww2.mathworks.cn/matlabcentral/answers/375307-could-anyone-help-me-to-obtain-new_c-for-the-below-mentioned-code

编辑： Rik 2018-1-5

If
C=[0 0 41 0 0;
    45 0 0 0 0;
    0 43 0 0 0;
    0 0 0 42 0;
    0 0 0 0 44]
non_0=  sum(C);
non_0=repmat(non_0,2,1);
A_part=[ 1 2 3 4 5;
    11 12 13 14 15;
    6 7 8 9 10;
    21 22 23 24 25;
    26 27 28 29 30]
new_C=C;
new_C(non_0 & C==0)=A_part(non_0 & C==0)

but i want to have new_C in

     [ 0     0    41     4     0;
      45    12     0     0    15;
       6    43     0     0    10;
       0     0    23    42     0;
      26    27     0     0    44]

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Prabha Kumaresan 2018-1-3

编辑：Cedric 2018-1-3

In new C row 1 and row 4 are sharing with each other.Thats why row 1 of new C has 41 from C and 4 from A_part.Similarly row 4 of new C has 23 from A_part and 42 from C.Since row 1 and row 4 has values on the third and fourth place the remaining places first,second and fifth are zeros.The same method applies to row2,3,and 5.

Geoff Hayes 2018-1-3

I'm sorry, I don't understand what is meant by In new C row 1 and row 4 are sharing with each other. What are they sharing? What is common between row 1 and 4?

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Answer 1

Rik 2018-1-5

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https://ww2.mathworks.cn/matlabcentral/answers/375307-could-anyone-help-me-to-obtain-new_c-for-the-below-mentioned-code#answer_298835

在 MATLAB Online 中打开

Let me help you out here, before everybody jumps out of the window.

You have already had some help about how to get a vector with randperm that should denote which rows are clustered ( here is one example I found). In this post I'm going to skip that step. If you have trouble putting that step back into this code, DON'T ASK A NEW QUESTION. Reply here instead.

function C=cluster_rows(A,B,rows)
%extract the parts of the matrices
A_part=A(rows,:);
B_part=B(rows,:);
%sum along the the vertical axis to get indices for the non-0 columns
non_0=sum(B_part);
%Repeat the vector back to the same size as the partial matrix. It will be
%converted to a logical later on.
non_0=repmat(non_0,length(rows),1);
%create a copy of B, as that is the basis for C
C=B;C_part=B_part;
%for all non-zero columns, replace the 0 positions with the values from A
C_part(non_0 & C_part==0)=A_part(non_0 & C_part==0);
%paste the edited partial matrix back
C(rows,:)=C_part;
end

You can call this function like this:

A=[ 1 2 3 4 5;
    11 12 13 14 15;
    6 7 8 9 10;
    21 22 23 24 25;
    26 27 28 29 30];
B=[0 0 41 0 0;
    45 0 0 0 0;
    0 43 0 0 0;
    0 0 0 42 0;
    0 0 0 0 44];
C_after_step_1=cluster_rows(A,B,[1 4]);
C=cluster_rows(A,C_after_step_1,[2 3 5]);

And, no, there isn't an easy way to do this without calling the function twice or repeating the code.

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jaah navi 2018-1-5

Thanks.could you please help me if the size of the matrix A and B are (n,m) then how it can be done.

Rik 2018-1-5

编辑：Rik 2018-1-5

The function works on matrices of arbitrary sizes. As long as you provide row indices that are still inside the matrix dimensions, this will work. (so don't try to get row 1001 if your A and B have only 1000 rows)

The function processes the entire row at once, so the number of columns doesn't matter.

I would also be interested to know if you ( Jaah Navi and Prabha Kumaresan) are indeed working together on the same project.

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