Comment #6 on issue 1577 by smichr: match doesn't put every wild as a key
http://code.google.com/p/sympy/issues/detail?id=1577
Regarding the comment that the replacement dictionary (if present) should
allow the
pattern to rebuild the expression is exactly the problem with the return
dictionary
given in issue 1826:
d=oo
a, b, c, p = map(Wild, 'abcp')
r=d.match(a+b*c**p)
r
{c_: nan, a_: oo, p_: 1}
(a+b*c**p).subs(r) <<<< it doesn't rebuild the expression
nan
r={a:oo,b:0,c:1,d:1} <<< this does
(a+b*c**p).subs(r)
oo
An approach like as_independent would be useful where, if and Add is being
analyzed
and there is not independent term you get a 0. And a Mul generates a 1. It
seems to
me that if a term doesn't match in a pattern then it should get a value
that will
make it disappear: if only 'a' matches in an expression below, then
for a+b*c**p, b=0 shuts off the other term
for a+c**p, c=0, p=1 shuts it off
for a*b**p, b=1 and p=0 tell you there was no power term
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