[SciPy-User] multivariate empirical distribution function, avoid double loop ?

[David Baddeley]

Sounds like it could be a case for scipy.spatial.kdtree. 

Cheers, David

On Thu, 25 Aug 2011 06:59 NZST josef.pktd at gmail.com wrote:

>On Wed, Aug 24, 2011 at 2:27 PM, Alan G Isaac <alan.isaac at gmail.com> wrote:
>> On 8/24/2011 10:23 AM, josef.pktd at gmail.com wrote:
>>> Does anyone know whether there is an algorithm that avoids the double
>>> loop to get a multivariate empirical distribution function?
>>
>> I think that is pretty standard.
>> I'll attach something posted awhile ago.
>> It seemed right at the time, but I did
>> not test it.  Once upon a time it was at
>> http://svn.scipy.org/svn/scipy/trunk/scipy/sandbox/dhuard/stats.py
>>
>> Cheers,
>> Alan
>>
>>
>> def empiricalcdf(data, method='Hazen'):
>>     """Return the empirical cdf.
>>
>>     Methods available (here i goes from 1 to N)
>>         Hazen:       (i-0.5)/N
>>         Weibull:     i/(N+1)
>>         Chegodayev:  (i-.3)/(N+.4)
>>         Cunnane:     (i-.4)/(N+.2)
>>         Gringorten:  (i-.44)/(N+.12)
>>         California:  (i-1)/N
>>
>>     :author: David Huard
>>     """
>>     i = np.argsort(np.argsort(data)) + 1.
>>     nobs = len(data)
>>     method = method.lower()
>>     if method == 'hazen':
>>         cdf = (i-0.5)/nobs
>>     elif method == 'weibull':
>>         cdf = i/(nobs+1.)
>>     elif method == 'california':
>>         cdf = (i-1.)/nobs
>>     elif method == 'chegodayev':
>>         cdf = (i-.3)/(nobs+.4)
>>     elif method == 'cunnane':
>>         cdf = (i-.4)/(nobs+.2)
>>     elif method == 'gringorten':
>>         cdf = (i-.44)/(nobs+.12)
>>     else:
>>         raise 'Unknown method. Choose among Weibull, Hazen, Chegodayev, Cunnane, Gringorten and California.'
>>     return cdf
>
>
>Unfortunately it's 1d only, and I am working on multivariate, at least
>bivariate.
>
>Pierre has a 1d version similar to this in scipy.stats.mstats and a,
>so far unused, copy is in statsmodels.
>
>Thanks,
>Josef
>
>
>>
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