Re: Sum pattern
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- Subject: [mg128289] Re: Sum pattern
- From: Fred Simons <f.h.simons at tue.nl>
- Date: Fri, 5 Oct 2012 02:48:27 -0400 (EDT)
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On Oct 3, 12:12 am, "Dave Snead"<dsne... at charter.net> wrote: > Hi, > > I'm trying to put together a rule whose left hand side is a sum of arbitrary > length whose elements all have the same head f. > > For example: > > In[4]:= x = f[a1, s] + f[a2, s] + f[a3, s] > > Out[4]= f[a1, s] + f[a2, s] + f[a3, s] > > In[6]:= y = f[First /@ x, s] > > Out[6]= f[a1 + a2 + a3, s] > > which is what I want. > > However when I turn this into a rule > > In[7]:= z = x /. (p : Plus[__f]) -> f[First /@ p, s] > > Out[7]= f[f[a1, s], s] + f[f[a2, s], s] + f[f[a3, s], s] > > Why isn't z equal to y? > How can I make this rule work? > > Thanks in advance, > Dave Snead Your solution is almost correct, but fails since you did not take into account that both sides of your rule are evaluated before it is applied. In[1]:= (p:Plus[__f])->f[First/@p,s] Out[1]= p__f->f[p,s] Your rule wraps every expression with head f in another f, as you see in your output. To prevent the evaluation, use RuleDelayed for keeping the right-hand side and HoldPattern for the left-hand side: In[2]:= x=f[a1,s]+f[a2,s]+f[a3,s] x /. (p:HoldPattern[Plus[__f]]):>f[First/@p,s] Out[2]= f[a1,s]+f[a2,s]+f[a3,s] Out[3]= f[a1+a2+a3,s] Fred Simons Eindhoven University of Technology
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