Re: Sum pattern
- To: mathgroup at smc.vnet.net
- Subject: [mg128278] Re: Sum pattern
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Wed, 3 Oct 2012 23:38:55 -0400 (EDT)
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Clear[f]
x = f[a1, s] + f[a2, s] + f[a3, s];
y = f[First /@ x, s]
f[a1 + a2 + a3, s]
To see why your rule didn't work, take it one step at a time
x /. (p : Plus[__f]) -> p
f[a1, s] + f[a2, s] + f[a3, s]
x /. (p : Plus[__f]) -> (First /@ p)
f[a1, s] + f[a2, s] + f[a3, s]
This is where it broke.
Alternatives:
y === (x /. (p : Plus[__f]) ->
f[Map[First, x, {1}], s]/3)
True
y === f[x[[All, 1]], s]
True
y === f[x /. f[a_, s] -> a, s]
True
y === f[Total@Cases[x, f[a_, s] -> a], s]
True
y === f[Plus @@ Cases[x, f[a_, s] -> a], s]
True
y === (x //. f[a_, s] + f[b_, s] -> f[a + b, s])
True
Using UpValues
f /: f[a_, s] + f[b_, s] := f[a + b, s];
y === x
True
Bob Hanlon
On Wed, Oct 3, 2012 at 3:07 AM, Dave Snead <dsnead6 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
>
>
>