Re: 2D pattern matching
- To: mathgroup at smc.vnet.net
- Subject: [mg77232] Re: 2D pattern matching
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Wed, 6 Jun 2007 06:46:47 -0400 (EDT)
- Organization: Uni Leipzig
- References: <f43dm5$f8$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
Hi,
a = {{0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 1, 0, 1, 1, 0}, {0,
0, 0, 1, 1, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}}
and
a /. {top___, {a___, 1, 1, b___}, {c___, 1, 1, d___}, bottom___} /;
Length[{a}] === Length[{c}] :>
{top, {a, this, is, b}, {c, "a 2d", pattern, d}, bottom}
will do it.
Regards
Jens
alexxx.magni at gmail.com wrote:
> hi everybody,
>
> do you know how to do a pattern match involving 2D matrices?
> If e.g. you have a list
>
> a = {{0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 1, 0, 1, 1, 0}, {0,
> 0, 0, 1, 1, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}}
>
> i.e.
> 0 0 0 0 0 0
> 0 1 0 0 0 0
> 0 1 0 1 1 0
> 0 0 0 1 1 0
> 0 0 0 0 0 0
>
> is it possible for me to write a rule by which I am able to find the
> 2x2 square of 1's, and replace it with something else?
>
> I studied all I could on patterns, but didnt find an answer...
>
> thanks for ANY help
>
> Alessandro Magni
>
>