The array tmp is used to index into prowB and elemsB when adding a new element and then is updated accordingly. Then elemsB and prowB can be populated as follows for j = 1.,n Now suppose that tmp has been overwritten by its cumulative sum. Hence, the cumulative sum of tmp is the same as pcolB. Then it follows that tmp(i + 1) is the number of elements in column i of B. Let tmp be an array such that tmp(1) = 0 and tmp(i + 1) is the number of elements in row i of A for i = 1.,m. To do so, we use the fact that the rows of A form the columns of B. If B denotes the transpose of A, then our goal is to define the analogous arrays elemsB, prowB, pcolB.

pcolA, of length n + 1, such that all nonzero elements in column j of A are indexed by the range [pcolA(j), pcolA(j + 1)).

prowA, of length nnz(A), that stores the row indices of nonzero elements in A.

elemsA, of length nnz(A), that stores the nonzero elements in A.

Suppose that A is an m x n matrix stored in CSC format, i.e., A is defined by three arrays: Transposing a sparse matrix is actually a straightforward task that can be accomplished in time proportional to the number of nonzero elements in the input matrix. I realize I am bit late to the game, but I thought I could help shed some light on this question. So the only reason for the fast speed of column scaling of X' should be that X is actually transposed to X' instead of setting a flag.Īlso, if every sparse matrix is stored in CSC format, simply setting a flag cannot make X' in CSC format. It makes sense because sparse matrices are stored column by column. This result means that accessing column by column is faster than accessing row by row. T = 138.3586 seconds % scaling each row by transposing X and transforming back I (1 0) J (0 -1) (0 1) (1 0) and notice that the transpose of J ( JT) is just equal to -J. Consider a matrix representation of complex numbers. For example, I have a sparse matrix X with size m*n=7984*12411, and I want to scale each column and each row: % scaling each column Actually I'd argue that there are deep reasons why the transpose IS the conjugate. But it appears to me that Matlab does 'actual' transposition. To do this (accessing row by row), you can transpose the matrix, perform operations on the columns, and then retranspose the result … The time required to transpose the matrix is negligible.įollow-up (modified as suggested by agree with and that setting a flag is enough for many operations such as the BLAS or sparse BLAS operations in terms of their interfaces. I've been wondering about this question for quite a while but cannot find a reference: How does Matlab transpose a sparse matrix so fast, given that it is stored in CSC (compressed sparse column) format?Īlso its documentation verifies the efficiency of sparse matrix transposition:

Matlab transpose matrix