Re: ReplaceList -- Unexpected Answer
- To: mathgroup at smc.vnet.net
- Subject: [mg46815] Re: ReplaceList -- Unexpected Answer
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Tue, 9 Mar 2004 04:31:01 -0500 (EST)
- Organization: Universitaet Leipzig
- References: <c2hdqc$aeq$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
because ReplaceAll[] or /. is different from Replace[] and Replace[]
return your expression unchanged.
In[]:=Replace[{f[h[4], h[4]], f[h[4], h[5]]}, f[x : h[_], x_] -> r[x]]
Out[]={f[h[4], h[4]], f[h[4], h[5]]}
and that's why ReplaceList[] return {}
Regards
Jens
Harold Noffke wrote:
>
> MathGroup:
>
> In The Mathematica 5 Book, Section 2.3.3 Naming Pieces of Patterns, we
> find the following pattern matching exercise ...
>
> Now both arguments of f are constrained to be the same, and only the
> first case matches.
>
> In[5]:=
> {f[h[4], h[4]], f[h[4], h[5]]} /. f[x:h[_], x_] -> r[x]
> Out[5]=
> {r[h[4]],f[h[4],h[5]]}
>
> Now, let's use ReplaceList to get more insight into this matching
> process ...
>
> In[6]:=
> ReplaceList[ {f[h[4],h[4]],f[h[4],h[5]]},f[x:h[_],x_] -> r[x] ]
> Out[6]=
> {}
>
> I do not understand why ReplaceList returns {} instead of { r[h[4]] }.
>
> Regards,
> Harold