[petsc-users] [Ext] change matrix
Matthew Knepley
knepley at gmail.com
Mon Aug 31 14:20:12 CDT 2020
On Mon, Aug 31, 2020 at 3:12 PM Barry Smith <bsmith at petsc.dev> wrote:
> On Aug 31, 2020, at 2:08 PM, Kun Jiao <KJiao at slb.com> wrote:
>
> If I am correct, to do this, it will double the peak memory usage.
>
>
> Yes
>
>
> Is there any way no to double the peak memory usage?
>
>
> The only way would be to destroy the old matrix, allocate a new one and
> recompute the entries.
>
> Barry
>
> Depending on the application etc the extra memory for storing two copies
> of the matrix may not be a fundamental problem
>
The other thing is to look at why you are adding rows. If you know how many
will eventually show up you can allocate them, but fill
with zeros, etc. until you get the values.
Thanks,
Matt
>
>
> Schlumberger-Private
> *From:* Barry Smith <bsmith at petsc.dev>
> *Sent:* Monday, August 31, 2020 1:56 PM
> *To:* Kun Jiao <KJiao at slb.com>
> *Cc:* petsc-users <petsc-users at mcs.anl.gov>
> *Subject:* [Ext] Re: [petsc-users] change matrix
>
>
> Kun,
>
> This is not possible, PETSc matrices have a static size (resizing in
> parallel is tricky so we don't support it).
>
> If it is more efficient to reuse the matrix entries than recompute them
> you can create a larger matrix and then loop over the old matrix calling
> MatGetRow() and then call MatSetValues() to copy that row into the new
> matrix.
>
> Barry
>
>
>
>
> On Aug 31, 2020, at 1:51 PM, Kun Jiao via petsc-users <
> petsc-users at mcs.anl.gov> wrote:
>
> Hi Petsc Experts,
>
> Trying to do something like appending some rows (~100 rows) to an already
> created matrix, but could not find any document about it.
>
> Could anyone provide some information about it?
>
> Regards,
> Kun
>
> Schlumberger-Private
>
>
>
--
What most experimenters take for granted before they begin their
experiments is infinitely more interesting than any results to which their
experiments lead.
-- Norbert Wiener
https://www.cse.buffalo.edu/~knepley/ <http://www.cse.buffalo.edu/~knepley/>
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