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

Hallo,

Thanks to everybody helping me finding a simpler solution.
The 'Rule@@@list' solution is indeed very elegant and clever.
It also works on more complicated lists:
	In: {a, b, b, d} /.  Rule @@@ {{a, {a1, a2}}, {b, {b1, b2}}, {c,
c1},{ d, d1}}
=09
	Out: {{a1, a2}, {b1, b2}, {b1, b2}, d1}

I often use the Rule[lhs, rhs] in the form of Thread[lhlist -> rhlist],
which is the same as Thread[Rule[lhlist, rhlist].

Finding back the @@@ is less trivia: pressing F1 for help on @@@ bring
you to Apply (@@) where @@@ is explained under the Scope/Level
Specifications/ section.
In the virtual book @@@ is found in
tutorial/ApplyingFunctionsToListsAndOtherExpressions
but it is also not trivial to find back.

I am not sure if the list with all the shorthand notations like @@, //,
;; etc. we had in 5.2 or older versions is still there in 6 or 7.0..

regards,

Maarten




-----Original Message-----
From: Curtis Osterhoudt [mailto:cfo at lanl.gov]
Sent: 03 December 2008 11:42
To: mathgroup at smc.vnet.net
Subject: [mg94091] [mg94042] Re: easier method for Flatten[Position[list2,x_x...??


  These methods based on the replacement list are very clever! I have
long
gotten used to the replacements notation, but didn't know it could be
used
like this.
   I tried to see if there was any documentation provided with
Mathematica
(v. 7, here) illustrating anything like that. However, the only
documentation I can find on Rule gives the shorthand right-arrow
notation "->" and say
s NOTHING about the Rule[lhs, rhs] possibility (I had to deduce that
from the solutions posted here, and look at the FullForm of some
replacements). Can anyone point me to a spot in the documentation where
an example using the "Rule[lhs, rhs]" notation is used or explained? If
not, it would be a nice improvement in future versions.

          Regards,
                       C.O.

On Monday 01 December 2008 22:40:27 Szabolcs Horv=C3=A1t wrote:
> 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}
> >
> >
>
>



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Curtis Osterhoudt
cfo at remove_this.lanl.and_this.gov
PGP Key ID: 0x4DCA2A10
Please avoid sending me Word or PowerPoint attachments
See http://www.gnu.org/philosophy/no-word-attachments.html
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