Re: ReplaceList -- Unexpected Answer
- To: mathgroup at smc.vnet.net
- Subject: [mg46822] Re: ReplaceList -- Unexpected Answer
- From: drbob at bigfoot.com (Bobby R. Treat)
- Date: Tue, 9 Mar 2004 04:31:12 -0500 (EST)
- References: <c2hdqc$aeq$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
ReplaceList (contrary to the documentation in my opinion) tries to
transform the expression at its top level only; it doesn't apply rules
to subexpressions.
{f[h[4],h[4]],f[h[4],h[5]]};
ReplaceList[# ,f[x:h[_],x_] -> r[x] ]&/@%
{{r[h[4]]},{}}
or
g@f[x : h[_], x_] := r[x]
g@a_ := a
g //@ {f[h[4], h[4]], f[h[4], h[5]]}
{r[h[4]], f[h[4], h[5]]}
Bobby
Harold.Noffke at wpafb.af.mil (Harold Noffke) wrote in message news:<c2hdqc$aeq$1 at smc.vnet.net>...
> 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