Comment #13 on issue 2607 by asmeurer: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
So this is not working so well. The problem is that the algorithm relies
on *= type multiplication being fast, but currently it is not:
In [1]: n = numbered_symbols()
Comment #14 on issue 2607 by asmeurer: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
Another way would be to not distribute the denominator in each step, so
compute
(a1*d2 + a2*d1)/(d1*d2)
((a1*d2 + a2*d1)*d3 + a3*d1*d2)/(d1*d2*d3)
(((a1*d2 + a2*d1)*d3
Comment #15 on issue 2607 by asmeurer: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
Here's what I've got so far:
def as_numer_denom(self):
return self.as_numer_denom_orig()
def as_numer_denom_orig(self):
numers, denoms = [],[]
for n,d in
Comment #16 on issue 2607 by asmeurer: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
By the way, if anyone wants to play around with these, I've pushed them up
to https://github.com/asmeurer/sympy/tree/as_numer_denom.
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Labels: Type-Defect Priority-Medium
New issue 2624 by alex.ebe...@gmail.com: Sympy 0.7.1 can't integrate
Gaussians
http://code.google.com/p/sympy/issues/detail?id=2624
Integration of a basic Gaussian works:
In [9]: integrate(exp(-x**2), (x,-oo,+oo))
Out[9]:
⎽⎽⎽
╲╱
Updates:
Status: Started
Owner: ness...@googlemail.com
Labels: Integration
Comment #1 on issue 2624 by matt...@gmail.com: Sympy 0.7.1 can't integrate
Gaussians
http://code.google.com/p/sympy/issues/detail?id=2624
This works in one of our development branches
Comment #5 on issue 2618 by mrock...@gmail.com: Solve fails on expressions
containing finite symbols
http://code.google.com/p/sympy/issues/detail?id=2618
Ok, so apparently is_finite is operating correctly, given the definition
that
finite = bounded and not infinitesimal (whether or not
Comment #6 on issue 2618 by mrock...@gmail.com: Solve fails on expressions
containing finite symbols
http://code.google.com/p/sympy/issues/detail?id=2618
This commit is option #1
It works at least for the example above
Comment #7 on issue 2618 by nicolas@gmail.com: Solve fails on
expressions containing finite symbols
http://code.google.com/p/sympy/issues/detail?id=2618
If y = Symbol('y', bounded=True), then ask(Q.bounded(y)) should return True
indeed.
The problem is that the old assumption system
Comment #8 on issue 2618 by nicolas@gmail.com: Solve fails on
expressions containing finite symbols
http://code.google.com/p/sympy/issues/detail?id=2618
Ok, so apparently is_finite is operating correctly, given the definition
that
finite = bounded and not infinitesimal (whether or not
Comment #9 on issue 2618 by nicolas@gmail.com: Solve fails on
expressions containing finite symbols
http://code.google.com/p/sympy/issues/detail?id=2618
A temporary workaround might be to make check_assumptions() first check old
assumptions, then new ones, so that global assumptions
Updates:
Labels: W
Comment #10 on issue 2618 by asmeurer: Solve fails on expressions
containing finite symbols
http://code.google.com/p/sympy/issues/detail?id=2618
If ask(Q.is_bounded(y)) returns False, then I'd say it's a wrong result. It
should either know about the old
Updates:
Labels: -W WrongResult
Comment #11 on issue 2618 by asmeurer: Solve fails on expressions
containing finite symbols
http://code.google.com/p/sympy/issues/detail?id=2618
(No comment was entered for this change.)
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Comment #2 on issue 2624 by asmeurer: Sympy 0.7.1 can't integrate Gaussians
http://code.google.com/p/sympy/issues/detail?id=2624
Yes, Tom is working with Matthew to make sure that all the common integrals
from statistics work.
This will likely be in the next release, unless someone wants to
Comment #3 on issue 2624 by asmeurer: Sympy 0.7.1 can't integrate Gaussians
http://code.google.com/p/sympy/issues/detail?id=2624
And another issue of poor simplification:
In [2]: integrate(exp(-x**2 + 2*x - 1), (x, -oo, oo))
Out[2]:
⎽⎽⎽ ⎽⎽⎽
╲╱ π ⋅(-erf(1) + 1) ╲╱ π
Status: New
Owner:
Labels: Type-Defect Priority-Medium
New issue 2625 by arthur.n...@gmail.com: Imaginary unit in R, ordering of
complex numbers
http://code.google.com/p/sympy/issues/detail?id=2625
Sorry if this has already been reported; I searched open issues but didn't
see this
Updates:
Status: Accepted
Labels: WrongResult Assumptions
Comment #1 on issue 2625 by asmeurer: Imaginary unit in R, ordering of
complex numbers
http://code.google.com/p/sympy/issues/detail?id=2625
This is because we have in the class Infinity:
def __le__(a, b):
Comment #2 on issue 2625 by asmeurer: Imaginary unit in R, ordering of
complex numbers
http://code.google.com/p/sympy/issues/detail?id=2625
I suppose there is a natural ordering of the purely imaginary numbers but
statements such as '5*I 6*I' and 'I*4 I' do not return Boolean values.
Comment #17 on issue 2607 by smi...@gmail.com: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
I see two simple limiting cases: all denoms the same and all denoms
having the same base raised to different powers. Your code 3 handles
the latter well and only one
Comment #18 on issue 2607 by asmeurer: as_numer_denom() is too slow
http://code.google.com/p/sympy/issues/detail?id=2607
There is a fifth method, which is to cancel stuff, as discussed in comments
4 and 5. I've pushed that as as_numer_denom5() in my branch. For the
example expressions
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 =
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