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Re: easier method for Flatten[Position[list2,x_x...??

• To: mathgroup at smc.vnet.net
• Subject: [mg94012] Re: easier method for Flatten[Position[list2,x_x...??
• From: Szabolcs Horvát <szhorvat at gmail.com>
• Date: Tue, 2 Dec 2008 00:40:27 -0500 (EST)
• Organization: University of Bergen
• References: <gh0jlg$5mu$1@smc.vnet.net>

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?

Like this:

In[1]:=
list1 = {a, b, b, e};
list2 = {{a, aP}, {b, bP}, {c, cP}, {d, dP}, {e, eP}, {f, fP}};

In[3]:= list1 /. Rule @@@ list2
Out[3]= {aP, bP, bP, eP}



> 
> 
> 
> 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}
> 
> 

• Follow-Ups:
  • Re: easier method for Flatten[Position[list2,x_x...??
    • From: Raffy <raffy@mac.com>
  • Re: easier method for Flatten[Position[list2,x_x...??
    • From: Curtis Osterhoudt <cfo@lanl.gov>