Is there a method to linear index to a transpose matrix without taking the transpose?

Is there a method to linear index to a transpose matrix without taking the transpose?

[James Tursa]

What do you mean precisely? Do you want to traverse the entire array by rows and want the indexing for that? Or do you have a linear index for the transpose and want to know where this is in the un-transposed matrix? Or ...?

[James Tursa]

2nd question, what are you doing with this linear indexing downstream in your code? It might be faster to actually transpose first and then do linear indexing on that rather than linear indexing into rows and jumping around in memory a lot.

[Douglas Farinelli]

I didn't write the original code. But the indexing is not sequential in either direction. I can time the operation and determine which is faster. Thanks!

[Image Analyst]

What code is that? Where did you attach it? Who wrote it?

And what is wrong with using the transpose operator '? Why do you need a different way?

[Douglas Farinelli]

The matrix is rather large ... did not desire a copy of it.

Answer 1

[Douglas Farinelli]

Yes want to index the matrix by rows vs columns. I suppose i can use the function ind2sub and switch the row and col.

[Douglas Farinelli]

The matrix is not square so the ind2sub would not work.

[Walter Roberson]

Yes.

Note: for a 2D matrix, ind2sub is (column_number - 1) * number_of_rows + row_number .

You can therefore optimize the access (row_number, column_number) to the transpose (column_number, row_number) as

(row_number - 1) * number_of_rows + column_number