Re: easier method for Flatten[Position[list2,x_x...??
- To: mathgroup at smc.vnet.net
- Subject: [mg94023] Re: easier method for Flatten[Position[list2,x_x...??
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Tue, 2 Dec 2008 00:42:26 -0500 (EST)
- Organization: Uni Leipzig
- References: <gh0jlg$5mu$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
Hi,
list1 = {a, b, b, e};
list2 = {{a, aP}, {b, bP}, {c, cP}, {d, dP}, {e, eP}, {f, fP}};
list1 /. Rule @@@ Select[list2, MemberQ[list1, First[#]] &]
??
Regards
Jens
Van Der Burgt, Maarten wrote:
> Dear all,
>
> Below I have two lists.
>
> The elements x in list1 all appear again in the 2nd level of list2
> together with an associated parameter xP.
>
> I want to extract the xP as illustrated below.
>
> I have the feeling it can be done in an easier way.
>
>
>
> Does anyone have an idea how?
>
>
>
> Thanks for your help,
>
>
>
> Maarten
>
>
>
>
>
> list1={a,b,b,e};
>
>
>
> list2={{a,aP},{b,bP},{c,cP},{d,dP},{e,eP},{f,fP}};
>
>
>
>
>
> pos = Flatten[Position[list2,x_List/;x[[1]]==#]&/@list1];
>
>
>
> list2[[pos,2]]
>
>
>
> Out-> {aP,bP,bP,eP}
>
>
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