Re: easier method for Flatten[Position[list2,x_x...??
- To: mathgroup at smc.vnet.net
- Subject: [mg94024] Re: easier method for Flatten[Position[list2,x_x...??
- From: Raffy <raffy at mac.com>
- Date: Tue, 2 Dec 2008 00:42:37 -0500 (EST)
- References: <gh0jlg$5mu$1@smc.vnet.net>
On Dec 1, 7:00 am, "Van Der Burgt, Maarten"
<Maarten.VanDerBu... at icos.be> 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}
A simple replacement will work:
Replace[ list1, Rule @@@ list2, {1} ]
Additionally, you can tag unmapped keys:
Replace[ list1, Join[ Rule @@@ list2, { x_ :> unknown[x] } ], {1} ]
Or, remove unmapped keys:
Replace[list1, Join[ Rule @@@ list2, { _ -> Sequence[] } ], {1} ]