On 07/29/2012 01:06 AM, eat wrote:

Hi,

On Sat, Jul 28, 2012 at 7:48 PM, Jeffrey <zfyuan@mail.ustc.edu.cn <mailto:zfyuan@mail.ustc.edu.cn>> wrote:

On 07/29/2012 12:23 AM, Jeffrey wrote: > Dear all, > > The sentences bellow will always raise an Error or Exception just > as follows, which is a little anomaly. Is this a bug? > > >>> u1=numpy.random.rand(100000) > >>> u2=numpy.random.rand(100000) > >>> scipy.stats.kendalltau(u1,u2) > --------------------------------------------------------------------------- > > AttributeError Traceback (most recent call > last) > /home/zfyuan/phd/paper1/pyvine_lap/<ipython-input-28-98f367090ed1> in > <module>() > ----> 1 sp.stats.kendalltau(u1,u2) > > /usr/lib64/python2.7/site-packages/scipy/stats/stats.pyc in > kendalltau(x, y, initial_lexsort) > 2673 > 2674 tau = ((tot - (v + u - t)) - 2.0 * exchanges) / \ > -> 2675 np.sqrt((tot - u) * (tot - v)) > 2676 > 2677 # what follows reproduces the ending of Gary Strangman's > original > > > AttributeError: sqrt >

Sorry, I didn't describe this bug with details. What I mean is that when the two array have larger length, for example with length 100000, then it is more possible that the Error would occur.

My scipy version is 0.9.0 and numpy is 1.6.2.

Thanks a lot for your answering.

I can confirm this, like In []: os.sys.version Out[]: '2.7.2 (default, Jun 12 2011, 15:08:59) [MSC v.1500 32 bit (Intel)]' In []: np.version.version Out[]: '1.6.0' In []: sp.version.version Out[]: '0.9.0'

In []: stats.kendalltau(rand(77929), rand(77929)) Out[]: (0.0060807135427758865, 0.010891543687108114) In []: stats.kendalltau(rand(77939), rand(77939)) ------------------------------------------------------------ Traceback (most recent call last): File "<ipython console>", line 1, in <module> File "C:\Python27\lib\site-packages\scipy\stats\stats.py", line 2675, in kendalltau np.sqrt((tot - u) * (tot - v)) AttributeError: sqrt

There really seems to be odd problem above a certain length of arrays.

My 2 cents, -eat

--

Jeffrey

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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. Thanks a lot ! -- Jeffrey

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 : )

...

...

...

np.sqrt(<something>) AttributeError

-n _______________________________________________ SciPy-User mailing list SciPy-User@scipy.org http://mail.scipy.org/mailman/listinfo/scipy-user

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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 _______________________________________________ SciPy-User mailing list SciPy-User@scipy.org http://mail.scipy.org/mailman/listinfo/scipy-user

:-) thanks for your pull request. My English is a little pool, and I'm new to Python. -- Jeffrey