[SciPy-user] Sparse matrix: division by vector

Dinesh B Vadhia dineshbvadhia@hotmail....
Tue Apr 29 21:04:03 CDT 2008

Does this work for row sums?  For example, assuming A is in CSR and rowSum[i] stores the i-th row sum can you do

> A.data /= rowSum[A.indices]


From: Nathan Bell <wnbell <at> gmail.com>
Subject: Re: Sparse matrix: division by vector
Newsgroups: gmane.comp.python.scientific.user
Date: 2008-04-29 18:16:19 GMT (7 hours and 44 minutes ago)

On Tue, Apr 29, 2008 at 12:16 PM, Dinesh B Vadhia
<dineshbvadhia <at> hotmail.com> wrote:
> Sparse matrix A is in csr_matrix format and I want to divide each column
> element j of A by A's column j sum (where colSum is a numpy vector) ie.
> > A[:,j] = A[:,j]/colSum[j]
> What is the most efficient way to achieve this apart from brute force ie.
> > A[i,j] = A[i,j]/colSum[j] ?
> I can initially create A in a different c**_matrix method if that helps but
> the final A has to be in csr_matrix form.

Assuing A is in CSR and colSum[j] stores the j-th column sum you can do

A.data /= colSum[A.indices]

Nathan Bell wnbell <at> gmail.com
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