Re: ReplacePart in an If[] construct?
- To: mathgroup at smc.vnet.net
- Subject: [mg67823] Re: ReplacePart in an If[] construct?
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Sun, 9 Jul 2006 04:50:47 -0400 (EDT)
- References: <e8nttf$kcf$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
because you discard your first
ReplacePart[#1, replaced!, 3];
ReplacePart[#1, replaced!, 2]
but you mean
tmp=ReplacePart[#1, replaced!, 3];
ReplacePart[tmp, replaced!, 2]
Regards
Jens
gardyloo wrote:
> Hi, all,
>
> It's probably that I'm too tired, but I have a question about using
> ReplacePart in an If[] construct. I've made up a (very) minimal example:
>
> In[1]:=
> testList = {{3, an, example, list},
> {4, another, example, list}}
>
> Out[1]=
> {{3, an, example, list}, {4, another, example, list}}
>
> In[2]:=
> (If[ (ListQ[#1] && First[#1] == 4),
> ReplacePart[#1, replaced!, 3];
> ReplacePart[#1, replaced!, 2],
> #1 (*otherwise*)
> ] & ) /@ testList
>
> Out[2]=
> {{3, an, example, list}, {4, replaced!, example, list}}
>
>
> Can someone tell me why BOTH the second and third positions in
> the second element in testList aren't turned to "replaced!" ? I have
> plenty of other ways of doing this, but this seemed the most
> straightforward, and I can't wrap my head around it, for some reason.
>
>
>