Is there a method to linear index to a transpose matrix without taking the transpose?
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Douglas Farinelli
am 18 Jun. 2019
Kommentiert: Douglas Farinelli
am 19 Jun. 2019
Is there a method to linear index to a transpose matrix without taking the transpose?
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Image Analyst
am 19 Jun. 2019
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?
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Douglas Farinelli
am 18 Jun. 2019
2 Kommentare
Walter Roberson
am 18 Jun. 2019
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
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