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Topic: Sum pattern
Replies: 1   Last Post: Oct 6, 2012 4:13 AM

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Dave Snead

Posts: 52
Registered: 12/13/04
Re: Sum pattern
Posted: Oct 6, 2012 4:13 AM
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Thanks everyone,

I've concluded that the shortest way to do this
(in terms of the number of characters I need to type)
which works is:

In[1]:= x = f[a1, s] + f[a2, s] + f[a3, s]

Out[1]= f[a1, s] + f[a2, s] + f[a3, s]

In[3]:= z = x /. (p : Plus[_f, __f]) :> f[First /@ p, s]

Out[3]= f[a1 + a2 + a3, s]

-- Dave Snead

-----Original Message-----
From: Fred Simons
Sent: Thursday, October 04, 2012 11:48 PM
Subject: Re: Sum pattern

On Oct 3, 12:12 am, "Dave Snead"<> 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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