On 27 October 2011 23:22, Satrajit Ghosh <[email protected]> wrote:
> hi robert,
>
>
>> I had a way of doing it that would be really robust, but really slow.
>> Expand each of the factorials on the denominator and numerator separately
>> and store as separate lists. Then remove all terms common to both lists and
>> multiply the results. However, this heavily uses append and would be too
>> slow for large value of N.
>>
>
> another exact solution would be to do an accumulator trick (treating every
> factorial as a factor):
>
> def run_accumulate(all_factors):
> a = np.zeros((all_factors.max()+1))
> b = np.arange(all_factors.max()+1)
> for factor in all_factors:
> a[:(abs(factor)+1)] += np.sign(factor)
> return np.exp(np.dot(a[2:], np.log(b[2:])))
>
> seems to run pretty quickly
>
> %timeit run_accumulate(np.array([50, 10, -30, -20])) # numerator factors
> are +ve, denominator is -ve
>
> 10000 loops, best of 3: 98.3 us per loop
>
> cheers,
>
> satra
>
>
>
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I haven't seen this pattern before - that should probably work well.
I'll have some time this weekend, so I think I'll be implementing three
different versions!
- Robert
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