Comment #39 on issue 2607 by [email protected]: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
By keeping track of the terms internally rather than feeding it back to
Mul, some significant improvements can be made:
>>> from timeit import timeit as T
>>> T('Add(*[1/Dummy() for i in xrange(500)]).as_numer_denom6()','from
sympy imp
ort Dummy, Add', number=1)
1.4353856028742573
>>> T('Add(*[1/Dummy() for i in xrange(500)]).as_numer_denom4()','from
sympy imp
ort Dummy, Add', number=1)
21.225003537452913
>>> T('Add(*[Dummy() for i in xrange(1000)]).as_numer_denom6()','from
sympy impo
rt Dummy, Add', number=1)
0.016849042016737314
>>> T('Add(*[Dummy()/i for i in
xrange(1,1000)]).as_numer_denom6()','from sympy
import Dummy, Add', number=1)
0.9006821644700267
This handles the constant, too:
>>> (x/y+x/2+3/x).as_numer_denom6()
(x**2*y + 2*x**2 + 6*y, 2*x*y)
>>> (x/y+x/2+3/x).as_numer_denom()
(x**2*y + 2*x**2 + 6*y, 2*x*y)
>>> (x/1+y/2+z/3).as_numer_denom6()
(6*x + 3*y + 2*z, 6)
>>> (x/1+y/2+z/3).as_numer_denom()
(6*x + 3*y + 2*z, 6)
>>> (x/2 + 1/(5*(x + 1))).as_numer_denom6()
(x*(5*x + 5) + 2, 10*x + 10)
>>> (x/2 + 1/(5*(x + 1))).as_numer_denom()
(x*(5*x + 5) + 2, 10*x + 10)
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