Re: easier method for Flatten[Position[list2,x_x...??
- To: mathgroup at smc.vnet.net
- Subject: [mg94032] Re: [mg93990] easier method for Flatten[Position[list2,x_x...??
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Tue, 2 Dec 2008 00:44:03 -0500 (EST)
- Reply-to: hanlonr at cox.net
list1 = {a, b, b, e};
list2 = {{a, aP}, {b, bP}, {c, cP}, {d, dP}, {e, eP}, {f, fP}};
Flatten[Cases[list2, {#, y_} :> y] & /@ list1]
{aP,bP,bP,eP}
Bob Hanlon
---- "Van Der Burgt 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}
--
Bob Hanlon