Retrieve all non-zeros indices except the first one for each row

Hi, I have the following question. How can I retrieve all nonzero indices of each row except the first nonzero index in a matrix? Suppose I have the following matrix:

A = [2 0 1 0 0; 4 -9 0 5 0;0 0 2 0 -3;0 -3 2 0 0;1 2 0 0 6];

I would like the output to be: 11, 0, 7, 17, 0, 23, 0, 14, 0, 10, 25, 0 as the linear indices.

If there are no more non-zeros in a row after the last one, then the index is zero.

-Thanks in advance-

Romeo

Answer 1

[Image Analyst]

Zero cannot be a linear index. This code will work:

A = [2 0 1 0 0; 4 -9 0 5 0;0 0 2 0 -3;0 -3 2 0 0;1 2 0 0 6]
linearIndexes = [];
for row = 1 : size(A, 1)
	% Find columns where A is nonzero
	indexes = find(A(row,:) ~= 0);
	% Tack on the second and other elements more to the right in this row.
	for k = 2 : length(indexes)
		% Get the linear index for this (row, column) location.
		linearIndex = sub2ind(size(A), row, indexes(k));
		% Tack it on to our accumulating/growing output variable.
		linearIndexes(end+1) = linearIndex;
	end
end
linearIndexes % Echo to command window.

It shows

A =

2 0 1 0 0

4 -9 0 5 0

0 0 2 0 -3

0 -3 2 0 0

1 2 0 0 6

linearIndexes =

11 7 17 23 14 10 25

Same as you except for no zeros.

[Romeo Tahal]

Thank you, sir

The solution looks great, but I still have 2 more questions regarding this.

Q1.The linearIndexes are actually the next non-zeros after the first non-zeros per row. After the last column, there are no more non-zeros, so I would like to insert a zero (0) for that position. How can I include that in the code?

Q2. Can I use the same approach for the non-zeros in the columns?

[Image Analyst]

Q1: Yes, to append a zero:

linearIndexes(end+1) = 0;

Q2: the for loop should be over columns, not rows. I'm sure you've already made the obvious modifications by now, and got something like this:

A = [2 0 1 0 0; 4 -9 0 5 0;0 0 2 0 -3;0 -3 2 0 0;1 2 0 0 6]
linearIndexes = [];
for col = 1 : size(A, 1)
	% Find rows where A is nonzero
	indexes = find(A(:, col) ~= 0);
	% Tack on the second and other elements more to the right in this row.
	for k = 2 : length(indexes)
		% Get the linear index for this (row, column) location.
		linearIndex = sub2ind(size(A), indexes(k), col);
		% Tack it on to our accumulating/growing output variable.
		linearIndexes(end+1) = linearIndex;
	end
end
linearIndexes % Echo to command window.

[Romeo Tahal]

Once again, thank you very much, sir

The codes work perfectly. Now I only need to fill in the zeros after the last col index and last row index of mon-zeros.

kind regards,

Romeo

[Image Analyst]

What does "fill in the zeros mean"? You mean like

linearIndexes(end:numel(A)) = 0;

??? That would append zeros to linearIndexes. Or do you mean that you want to "fill in" / replace the zeros in A with some other value? If so what value? And why do you need to do this unusual/quirky thing?

[Romeo Tahal]

Maybe I can explain what I'm trying to accomplish.

I'm trying to store a sparse matrix (of any size) in a tabular format. The miminum requirements are to have at least 3 arrays, one will store the row index, one will store the column index, and one will store the actual value. That is not a problem; I already have that part.

Then two more arrays are needed: one will store the indices of the first non-zero element per row, and one will store the indices of the first non-zero element per column. That is also not a problem; I already covered that part. These two arrays will act as pointers, sort of.

Then two more arrays are needed: one will store all the indices of the next non-zero elements per row, and will will do the same for each column. If the end of the row or column is reached, there is no other non-zero element to point at, so instead 'null' is used to indicate the end of each row or each column. This was the part I was struggling with. In my example matrix there should be 12 non-zero elements in total. The next-in-row array and the next-in-column array should also have 12 elements. You've helped me with the code to accomplish the last part of the storage. Based on your solution, there are 7 non-zeros for the rows and 7 for the columns. So, the zeros will be at the end for each row (5 in total) and for each column (5 in total).

I hope this will make more sense of what I'm trying to do.

[Image Analyst]

Seems complicated. MATLAB has capability to deal with sparse matrices. Why not just use it and not have this complicated scheme? Are the matrices so gigantic, even when sparse, that you're running out of memory? Well anyway, good luck.

[Romeo Tahal]

Well sir,

There is an issue with the sequence of the indices. Using the code, I get the following sequence for the row indices:

11 0 7 17 0 23 0 14 0 10 25 0

This is because the indices for the rows are row-based, but matlab uses column-wise ordering. The correct sequence should be: 11 7 10 17 14 25 0 23 0 0 0 0

How can I get this ordering? What change(s) do I have to make in the code for the row indices?

Thanks in advance.