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. > > >