Comment #19 on issue 2607 by asmeurer: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
Sorry, that's as_numer_denom4(). And here are some timings:
In [12]: numers, denoms = zip(*((Symbol('n%d'%i),Symbol('d%d'%i)) for i in
xrange(1000)))
In [13]: a = Add(*(n/d for n, d in zip(numers, denoms)))
In [14]: %timeit a.as_numer_denom4()
1 loops, best of 3: 134 ms per loop
Compare the above to timings 18-21 in comment 15. And for all denoms == to
1:
In [1]: a = Add(*(x**i for i in xrange(1000)))
In [2]: %timeit a.as_numer_denom()
1 loops, best of 3: 7.21 s per loop
In [3]: %timeit a.as_numer_denom1()
1 loops, best of 3: 17.2 s per loop
In [4]: %timeit a.as_numer_denom2()
1 loops, best of 3: 8.42 s per loop
In [5]: %timeit a.as_numer_denom3()
1 loops, best of 3: 135 ms per loop
In [6]: %timeit a.as_numer_denom4()
1 loops, best of 3: 124 ms per loop
Based on what I've tried so far, we must reject as_numer_denom0-3 (I'll
leave them around for reference). If you can break 4, then we will have to
try harder, but so far, it seems to work and work fast. My main concern
with it is not speed that but that canceling won't work correctly (see the
above comments).
Thanks for trying sifting. I figured that no matter what optimization we
do, that will only make it fast, so I wan't going to implement it until the
end.
Other kinds of expressions to try:
- Add(random_poly/random_poly)
- Add(exp(i*x) for i in xrange(whatever))
- Add(n/d for n in [n1, n2] for d in [d1, d2, ..., dn]) # Many denoms with
some common denoms
- Add(n/d for n in [n1, n2, ...., nn] for d in [d1, d2]) # Few denoms with
many common denoms
- See if you can get automatic distribution of numbers through Adds to mess
up the cancelation from as_numer_denom4().
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