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Re: Pattern Matching
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23907] Re: Pattern Matching
*From*: Hartmut Wolf <hwolf at debis.com>
*Date*: Fri, 16 Jun 2000 00:56:48 -0400 (EDT)
*Organization*: debis Systemhaus
*References*: <8hsu3g$dh4@smc.vnet.net> <8i1suj$jls@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
My comments below at >>>>>>>>
Allan Hayes schrieb:
>
> Johannes,
>
> Re your first question.
>
> We get
>
> x[1] b[1] + x[2] b[2] /.
> Plus[Times[x[_], b[_]] ..] -> z
>
> b[1] x[1] + b[2] x[2]
>
> possibly because, before the matching and replacement, the left side of the
> rule is evaluated thus:
>
> Plus[Times[x[_], b[_]] ..]
>
> b[_] x[_] ..
>
> So we end up with
>
> x[1] b[1] + x[2] b[2] /. (*1*)
> (x[_] b[_]) .. -> z
>
> b[1] x[1] + b[2] x[2]
>
> We can avoid this by using HoldPattern:
>
> x[1] b[1] + x[2] b[2] /.
> HoldPattern[Plus[Times[x[_], b[_]] ..]] -> z
>
> z
>
> However it does seem odd that, in spite of (*1*), we get
>
> x * b[1] + x* b[2] /.
> (x* b[_]) .. -> z
>
> 2 z
>
> --
> Allan
> ---------------------
> Allan Hayes
> Mathematica Training and Consulting
> Leicester UK
> www.haystack.demon.co.uk
> hay at haystack.demon.co.uk
> Voice: +44 (0)116 271 4198
> Fax: +44 (0)870 164 0565
>
>>>>>>>>
Dear Allan,
as so often you come up with the right questions. You observed (example
simplified)
In[35]:= h[x, x] /. x .. -> z (*2*)
Out[35]= h[z, z]
whereas (*1*) doesn't substitute. Before coming back to that, why isn't Out[35]
== h[z]?
Time to go back to the Book. Although I didn't find it stated, and although we
wouldn't get an error message, all examples given in the book indicate that the
use of Repeated in (*1*) and (*2*) isn't legal. Since Repeat[...] is equivalent
to a Sequence the matched object must be enclosed in a head anyways, so
In[38]:= h[x, x] /. head_[x ..] -> head[z]
Out[38]= h[z]
works right, as well as does
In[26]:= h[x[1] b[1], x[2] b[2]] /. head_[x[_] b[_] ..] -> head[z]
Out[26]= h[z]
(And of course the HoldPattern version suggested by you and Andrzej corresponds
to this syntax.)
However this doesn't yet solve the problem as to replace only for equal indices.
In[47]:= x[1] b[1] + x[2] b[2] /. HoldPattern[Plus[Times[x[n_], b[n_]] ..]] -> z
Out[47]= z + b[2] x[2]
gives only one substitution. I would have liked to restrict the scope of the
pattern variable n_ only to Times[x[n_], b[n_]] and not to the whole sequence,
but I found no way to do that. David Park, and similarily Wagner Truppel,
proposed
In[80]:= Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /. b[n_]x[n_] -> z /.
n_Integer z -> z
Out[80]= z + b[2] x[1] + b[1] x[2]
using iterated replacement (which is fine!). But let's try to do that within one
stroke. Since it is easy to simply remove all elements x[n_] b[n_]
In[88]:= Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
p : b[n_]x[n_] :> Sequence[]
Out[88]= b[2] x[1] + b[1] x[2]
we may try just to not remove the first occurance:
In[125]:= Module[{virgo = True},
Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
b[n_]x[n_] :>
RuleCondition[ If[virgo, virgo = False; z, Unevaluated[Sequence[]]] ]]
Out[125]=
z + b[2] x[1] + b[1] x[2]
RuleCondition is not striktly neccessary in this case, but would give a more
pleasing result if the lhs in ReplaceAll is held: then the rhs of RuleDelayed
will be evaluated *after* pattern matching and before substitution (the
Trott-Strzebonski-Hayes method; else the rhsides will be evaluated later when
the result of ReplaceAll is evaluated as the final step, which of course will
not occur if held).
A less obfuscated method to do the same would be
In[127]:= Module[{virgo = True},
Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
{b[n_]x[n_] /; First[{virgo, virgo = False}] -> z,
b[n_] x[n_] -> Sequence[]}]
Out[127]=
z + b[2] x[1] + b[1] x[2]
where the condition is evaluated at the lhs (of Rule).
Finally a remark addressed to Johannes: I certainly would not try to program
that way; it is much easier (and clearer) just to seperate "diagonal" and
"off-diagonal" elements of your sum
In[90]:=
Plus[x[1] b[1], x[1]b[2], x[2] b[1],
x[2] b[2]] /. {{b[n_]x[n_] -> Sequence[]}, {b[n_]x[m_] /; n =!= m ->
Sequence[]}}
Out[90]=
{b[2] x[1] + b[1] x[2], b[1] x[1] + b[2] x[2]}
and then proceed with the parts as you like.
Kind regards to everyone,
Hartmut Wolf
> "Johannes Ludsteck" <ludsteck at zew.de> wrote in message
> news:8hsu3g$dh4 at smc.vnet.net...
> > Dear Group Members,
> > I would like to "find" and replace expressions with the simple
> > structure x[1] b[1]+x[2] b2]+...+x[n] b[n]
> > I tried to use the following replacement rule
> > In[27]:= x[1] b[1] + x[2] b[2] /. Plus[Times[x[_], b[_]] ..] -> z
> >
> > Out[27]= b[1] x[1] + b[2] x[2] + b[3] x[3]
> > Which didn't work (Out[27] should be z).
> > Why?
> > The following FullForm seems to give exactly the structure I used
> > in my replacement rule.
> >
> > In[17]:=
> > FullForm[x[1] b[1] + x[2] b[2] + x[3] b[3]]
> > Out[17]//FullForm=
> > Plus[Times[b[1], x[1]], Times[b[2], x[2]], Times[b[3], x[3]]]
> >
> > Even if this worked, my pattern wouldn't account for equal indices,
> > i.e. it would match x[1] b[500]+x[12] b[3], even if it shouldn't.
> >
> > Any suggestions?
> > Thanks,
> > Johannes Ludsteck
> >
> >
> > Johannes Ludsteck
> > Centre for European Economic Research (ZEW)
> > Department of Labour Economics,
> > Human Resources and Social Policy
> > Phone (+49)(0)621/1235-157
> > Fax (+49)(0)621/1235-225
> >
> > P.O.Box 103443
> > D-68034 Mannheim
> > GERMANY
> >
> > Email: ludsteck at zew.de
> >
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