[SciPy-User] scipy.stats.kendalltau bug?

Jeffrey zfyuan@mail.ustc.edu...
Sun Jul 29 06:25:51 CDT 2012


On 07/29/2012 06:30 PM, Nathaniel Smith wrote:
> On Sun, Jul 29, 2012 at 10:42 AM, Jeffrey <zfyuan@mail.ustc.edu.cn> wrote:
>> On 07/29/2012 03:47 PM, Nathaniel Smith wrote:
>>> On Sun, Jul 29, 2012 at 8:27 AM, Jeffrey <zfyuan@mail.ustc.edu.cn> wrote:
>>>> Thanks eat. I found the reason is that numpy.sqrt cannot deal with too large
>>>> number. When calculating kendalltau, assume n=len(x),then the total pair
>>>> number is 'tot' below:
>>>>
>>>>       tot=(n-1)*n//2
>>>>
>>>> when calculating tau, the de-numerator is as below:
>>>>
>>>>       np.sqrt((tot-u)*(tot-v))
>>>>
>>>> u and v stands for ties in x[] and y[perm[]], which is zero if the two array
>>>> sample from continuous dist. Hence (tot-u)*(tot-v) may be out of range for
>>>> the C written ufunc 'np.sqrt', and an Error is then raised.
>>>>
>>>> What about using math.sqrt here, or multiply two np.sqrt in the
>>>> de-numerator? Since big data sets are often seen these days.
>>> It seems like the bug is that np.sqrt is raising an AttributeError on
>>> valid input... can you give an example of a value that np.sqrt fails
>>> on? Like
>> Assume the input array x and y has n=100000 length, which is common
>> seen, and assume there is no tie in both x and y, hence u=0, v=0 and t=0
>> in the scipy.stats.kendalltau subroutine. Hence the de-numerator of
>> expression for calculating tau would be as follows:
>>
>>       np.sqrt( (tot-u) * (tot-v) )
>>
>> Here above, tot= n * (n-1) //2=499950000, and (tot-u) * (tot-v)= tot*tot
>> = 24999500002500000000L, this long int will raise Error when np.sqrt is
>> applied. I think type convert, like 'float()' should be done before
>> np.sqrt, or write like np.sqrt(tot-u) * np.sqrt(tot-v) to avoid long
>> integer.
>>
>> Thanks a lot : )
> Thanks, that clarifies things: https://github.com/numpy/numpy/issues/368
>
> For now, yeah, some sort of workaround makes sense, though... in
> addition to the ones you mention, I noticed that this also seems to
> work:
>
> np.sqrt(bignum, dtype=float)
>
> You should submit a pull request :-).
>
> -n
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:-) thanks for your pull request. My English is a little pool, and I'm 
new to Python.

-- 

Jeffrey




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