|
From: | James Sherman Jr. |
Subject: | Re: Gallery (or circulant matrices) |
Date: | Fri, 7 Jun 2013 23:29:45 -0400 |
Thanks, Markus. I'm afraid this didn't work. You can't subtract (1:length(v)) from (1:length(v))'. Did you mean to repmat them, somehow?On Fri, Jun 7, 2013 at 9:13 PM, Markus Appel [via Octave] <[hidden email]> wrote:
[hidden email]On 06/08/2013 02:26 AM, Jeffrey wrote:> Converting from Matlab, the "gallery" command to create circulant
> matrices is not implemented. Is there an Octave equivalent to
>
> K=gallery('circul',[4 -1 zeros(1,N-3) -1]);
>
> How does one find octave equivalents to Matlab statements? Thanks.I can't answer the general question of how to find equivalents, but in
the special case of this matrix (and if i understood the 'circul'
gallery function right) you should be able to substitute
K=gallery('circul',v)
by
K = v( 1 + mod( (1:length(v))' - (1:length(v)), length(v)) )
The _expression_ is basically creating an index matrix with shifted
columns and picking the elements out of your input vector v in the right
order.
Hope it helps,
Markus
_______________________________________________
Help-octave mailing list
https://mailman.cae.wisc.edu/listinfo/help-octave
If you reply to this email, your message will be added to the discussion below:http://octave.1599824.n4.nabble.com/Gallery-or-circulant-matrices-tp4653859p4653860.htmlTo start a new topic under Octave - General, email [hidden email]
To unsubscribe from Octave, click here.
NAML
View this message in context: Re: Gallery (or circulant matrices)
_______________________________________________
Help-octave mailing list
address@hidden
https://mailman.cae.wisc.edu/listinfo/help-octave
[Prev in Thread] | Current Thread | [Next in Thread] |