How to quickly find the first non-zero element without iterations in all columns in a sparse matrix?

Do you think I can use your code to do the same thing for the rows in a sparse matrix? I want to find the index of the LAST non-zero element in each row in A.

For example, I have matrix A = [2 0 0;0 5 0;0 0 1;0 3 0;-1 0 -4]. My code gives the following result:

reorderRow = [1 2 3 2 3];

Thanks a lot.

Benson

Matt J 2020-3-19

Yes, it's a simple modification. See here,

https://www.mathworks.com/matlabcentral/answers/511800-how-to-quickly-find-the-column-index-of-the-last-non-zero-element-in-all-rows-in-a-sparse-matrix#answer_420999

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Answer 2

KSSV 2020-3-18

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在 MATLAB Online 中打开

[val,id] = min(abs(A)) ;

Use the minimum function.

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Benson Gou 2020-3-18

Hi, KSSV, thanks for your reply. I am sorry for the unclarity of my statement. I am looking for the position of the first non-zero element in each column, not the value of the first non-zero element. Thanks!

Benson

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Answer 3

Les Beckham 2020-3-18

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编辑：Les Beckham 2020-3-18

在 MATLAB Online 中打开

Try this. In my test with a 1000x1000 random sparse 0 or 1 matrix (A = sparse(randi([0 1], 1000, 1000));) it is about 4 times faster.

i = find(A(:) ~= 0, 1, 'first');
ij = ind2sub(size(A), i);

For a 5000x5000 A it is actually slower, however (about 83% as fast).

Note that I am using Octave as I don't currently have access to Matlab.

Iteration is not always necessary, or desirable, to avoid. In fact, many of the Matlab 'tricks' are just iteration in disguise.

I hope this helps.

Perhaps your results will be different using Matlab vs. Octave. Let me know.

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Benson Gou 2020-3-19

Hi, Les,

I tried your code. I found it only gives me one integer which is the first non-zero lement in the first column in A. How can I apply for all the columns in A? Thanks.

Benson

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