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MathGroup Archive 1997

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Re: Bug in pattern matching?

  • To: mathgroup at
  • Subject: [mg8726] Re: [mg8722] Bug in pattern matching?
  • From: jpk at
  • Date: Sat, 20 Sep 1997 22:27:56 -0400
  • Sender: owner-wri-mathgroup at

Hi Peter,

the explanation is quite easy, Your rule says

ls/.{first_, middle___, last_} -> {{first},C {middle},{last}}

{{first},C {middle},{last}}

is to evaluated (You use Rule[] instead RuleDelayed[])
since Times[] has the attribute Listable  the evaluation yields

{{first} {Times[C,middle]},{last}}

than the transformation is applied to ls, middle is bounded to


and You get Times[C,Sequence[a[1],a[2],a[3],a[4]]] and furture

Your mistake is to use Rule[] instead of RuleDelayed[] with

ls/.{first_, middle___, last_} :> {{first},C {middle},{last}}

{{a[0]},{C a[1],C a[2],C a[3],C a[4]},{a[5]}}

all works as You want.

Hope that helps

> Hi, I've posted this problem to Wolfram support but received no answer =
so I
> thought I might just post it here aswell.
> I'm trying to "deconstruct" a list, perform an operation to a pert of =
> list and then recombine the result. When I did this I found out that
> multiplication in pattern matching performs differently than =
> Check version:
> In[1]:=
> $Version
> Out[1]=
> "Microsoft Windows 3.0 (October 6, 1996)"
> Define list:
> In[2]:=
> ls=Table[a[j], {j,0,5}]
> Out[2]=
> {a[0],a[1],a[2],a[3],a[4],a[5]}
> I want to split this list into the following three sublists:
> {a[0],a[1],a[2],a[3],a[4],a[5]} -> {a[0]}, {a[1], a[2], a[3], a[4]}, =
> Use pattern matching:
> In[3]:=
> ls/.{first_, middle___, last_}\[Rule] {{first},{middle},{last}}
> Out[3]=
> {{a[0]},{a[1],a[2],a[3],a[4]},{a[5]}}
> This works fine. Now I want to multiply the middle list by a factor C.
> Ordinary multiplication works like this:
> In[4]:=
> C ls
> Out[4]=
> {C a[0],C a[1],C a[2],C a[3],C a[4],C a[5]}
> Doing this in a pattern matching statement produces a totally =
> result:
> In[5]:=
> ls/.{first_, middle___, last_}\[Rule] {{first},C {middle},{last}}
> Out[5]=
> {{a[0]},{C a[1] a[2] a[3] a[4]},{a[5]}}
> The middle list now contains a completely different result than =
> Can anyone please explain this to me?
> /Peter Str=F6mbeck

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