On Mon, Feb 1, 2010 at 10:02 PM, <josef.p...@gmail.com> wrote: > On Mon, Feb 1, 2010 at 11:45 PM, Charles R Harris > <charlesr.har...@gmail.com> wrote: > > > > > > On Mon, Feb 1, 2010 at 9:36 PM, David Cournapeau <courn...@gmail.com> > wrote: > >> > >> On Tue, Feb 2, 2010 at 1:05 PM, <josef.p...@gmail.com> wrote: > >> > >> > I think this could be considered as a correct answer, the count of any > >> > integer is zero. > >> > >> Maybe, but this shape is random - it would be different in different > >> conditions, as the length of the returned array is just some random > >> memory location. > >> > >> > > >> > Returning an array with one zero, or the empty array or raising an > >> > exception? I don't see much of a pattern > >> > >> Since there is no obvious solution, the only rationale for not raising > >> an exception I could see is to accommodate often-encountered special > >> cases. I find returning [0] more confusing than returning empty > >> arrays, though - maybe there is a usecase I don't know about. > >> > > > > In this case I would expect an empty input to be a programming error and > > raising an error to be the right thing. > > Not necessarily, if you run the bincount over groups in a dataset and > your not sure if every group is actually observed. The main question, > is whether the user needs or wants to check for empty groups before or > after the loop over bincount. > > How would they know which bin to check? This seems like an unlikely way to check for an empty input.
> Like > >>> np.sum([]) > 0.0 > >>> sum([]) > 0 > the empty array or the array([0]) can be considered as the default > argument. In this case it is not really a programming error. > > I like that better than an empty array. > Since bincount usually returns redundant zero count unless > np.unique(data) = np.arange(data.max()+1), > array([0]) would also make sense as a minimum answer > >>> np.bincount([7,8,9]) > array([0, 0, 0, 0, 0, 0, 0, 1, 1, 1]) > > I use bincount quite a lot but only with fixed sized arrays, so I > never actually used it in this way (yet). > > Chuck
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