How to replace unique interval of columns in each row of a matrix with zero?

I have a matrix consisting of ones and zeros and have to replace the entries of each row up to a unique column number for each row with a zero. A for-loop is not acceptable.

Exaple:

A1= [1 1 1 1 1 1 1 0 0 1 1 0 0 1

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

1 1 1 1 1 1 1 0 1 1 1 1 1 0]

idx_column = [5

1

3]

A2= [0 0 0 0 0 1 1 0 0 1 1 0 0 1

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

0 0 0 1 1 1 1 0 1 1 1 1 1 0]

I tried the following without success:

A2 = A1;

A2(:,[1;1;1]:[5;1;3]) = 0;

Is there a possibility to replace a unique interval in each row of a matrix efficiently? A for loop is not acceptable.

Thank you very much!

Best regards,

Michael

[Bruno Luong]

"A for loop is not acceptable."

Don't be silly.

Answer 1

[Bruno Luong]

A1= [1    1    1    1    1    1    1    0    0    1    1    0    0    1
    1    1    1    1    1    1    1    1    0    1    1    0    0    1
    1    1    1    1    1    1    1    0    1    1    1    1    1    0]

A1 = 3×14

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

idx_column = [5
    1
    3]

idx_column = 3×1

5 1 3

A2 = A1 .* ((1:size(A1,2))>idx_column)

A2 = 3×14

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

Answer 2

[Abderrahim. B]

Not a smart way but still do the job:

A1 = [1    1    1    1    1    1    1    0    0    1    1    0    0    1
    1    1    1    1    1    1    1    1    0    1    1    0    0    1
    1    1    1    1    1    1    1    0    1    1    1    1    1    0] ;

A2 = A1 ;
A2(1, 1:5) = 0; A2(2, 1) = 0; A2(3, 1:3) = 0;
A2

A2 = 3×14

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

In case you have a huge matrix and you want replace multiple elements, use for loop if you see that it is not fast try parfor. Note that parfor requires to have Parallel Computing Toolbox license.

[Michael König]

Thank you for your answer. I will check whether a simple for loop is actually faster - but the method of your solution is very helpful for different kinds of problems in any way.

Answer 3

[David Hill]

Why is a for-loop not acceptable?

for k=1:length(idx_column)
   A1(k,1:idx_column(k))=0;
end

[Michael König]

I have huge matrices and am searching for a more efficient method in order to reduce computing times. Is there a more direct way?

I found https://de.mathworks.com/matlabcentral/answers/164993-how-to-use-a-vector-as-indices-for-matrix - but in my problem there can be multiple rows with a number, so that solution is not fitting for me.

[Bruno Luong]

For-loop is the most straighforward and possible the fastest method.