[SciPy-User] generic expectation operator
josef.pktd@gmai...
josef.pktd@gmai...
Thu Aug 23 12:30:00 CDT 2012
On Thu, Aug 23, 2012 at 1:24 PM, <josef.pktd@gmail.com> wrote:
> On Thu, Aug 23, 2012 at 1:11 PM, nicky van foreest <vanforeest@gmail.com> wrote:
>> Hi,
>>
>> I noticed that rv_frozen does not have an expect attribute. Is there a
>> design reason for this? It feels somewhat unnatural to me that frozen
>> distributions cannot be used directly to compute expectations.
>
> oversight to add it, I guess.
>
> can be added and it can delegate like in the other methods. PR welcome.
> frozen_dist.dist.expect(..)
>
> (I almost never work with frozen distribution and didn't see this.)
>>> miss = set(dir(stats.norm)) - set(dir(stats.norm(loc=5)))
>>> [i for i in miss if not i[0]=='_']
['moment_type', 'xb', 'xa', 'freeze', 'badvalue', 'expect', 'xtol',
'vecentropy', 'fit_loc_scale', 'expandarr', 'fit', 'vecfunc',
'est_loc_scale', 'logsf', 'logcdf', 'logpdf', 'generic_moment',
'numargs', 'name', 'a', 'nnlf', 'b', 'm', 'shapes', 'extradoc',
'veccdf']
>>> import scipy
>>> scipy.__version__
'0.9.0'
'logsf', 'logcdf', 'logpdf', are also missing (in my version of scipy)
Josef
>
> Josef
>
>
>>
>> bye
>>
>> Nicky
>>
>>
>>
>> On 23 August 2012 18:36, nicky van foreest <vanforeest@gmail.com> wrote:
>>> Hi Josef,
>>>
>>> Thanks for your answers.
>>>
>>>>>> return scipy.integrate.quad(lambda x: x*g(x), X.dist.a, X.dist.b)
>>>
>>>>>> X = stats.norm(3,sqrt(3))
>>>>>> print E(sqrt, X)
>>>
>>>> I don't know why this works, sqrt of a normal distributed random
>>>> variable has lots of nans (or complex numbers)
>>>
>>> You are right, it shouldn't work. The point is that my code above is
>>> this: lambda x: x*g(x), while it should have been lambda x: f(x)*g(x).
>>>
>>> I'll give expect a try.
>>>
>>> Nicky
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