Re: Pattern Matching Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg43434] Re: Pattern Matching Problem
- From: Dr Reinhard Simonovits <Reinhard.Simonovits at uni-graz.at>
- Date: Wed, 17 Sep 2003 07:58:50 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Dear Ted,
One method is the "1:1 Translation" trick, from words directly into the
Mathematica syntax
In words,
Take at first the terms without f, Cases[expr,a_/;FreeQ[a,_f] ]
then those f 2 terms, Cases[expr,f[_,2]]
then those f 3 terms, Cases[expr,f[_,3]]
Put it all together:
In[4]:=
Clear[a,b,c,x,f,g,y,z];
expr =a+b+c+f[w,2]+f[w,3]+x+f[x,2]+f[x,3]+y+f[y,2]+f[z,2];
Plus@@@{ Cases[expr, a_/;FreeQ[a,_f]], Cases[expr,f[_,2]], Cases[expr,f[_,3]] }
Out[6]=
{a+b+c+x+y, f[w,2]+f[x,2]+f[y,2]+f[z,2], f[w,3]+f[x,3] }
or the (slightly improved) "marker method" from Paul Abbot:
In[7]:=
CoefficientList[Collect[expr /. c:f[_, n_] :> c g^n , g], g] /.(
0->Sequence[] )
Out[7]=
{a+b+c+x+y, f[w,2]+f[x,2]+f[y,2]+f[z,2], f[w,3]+f[x,3] }
Actually you are collecting the coefficients of the powers of g:
In[9]:=
Collect[expr /. c:f[_, n_] :> g^n c, g ]//InputForm
Out[9]//InputForm=
a + b + c + x + y + g^3*( f[w, 3] + f[x, 3] ) + g^2*( f[w, 2] + f[x, 2]
+ f[y, 2] + f[z, 2] )
>"Ersek, Ted R" <ErsekTR at navair.navy.mil> wrote:
>
> > Consider the following:
> >
> > In[1]:=
> > ClearAll[f,a,b,c,w,x,y,z];
> > expr=a+b+c+f[w,2]+f[w,3]+x+f[x,2]+f[x,3]+y+f[y,2]+f[z,2];
> >
> >
> > Can somebody suggest a general way to seperate the terms above into like
> > groups. By "like" I mean having the same second argument for (f). So for
> > this example I want to get
> >
> > {a+b+c+x+y, f[w,2]+f[x,2]+f[y,2]+f[z,2], f[w,3]+f[x,3]}
********************************************
Dr. Reinhard Simonovits
Handelsakademie | Karl Franzens University
Math Department | Inst. of Th. Physics
Grazbachgasse 71 | Universitaetsplatz 5
A-8010 Graz, Austria
Email: Reinhard.Simonovits at uni-graz.at
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