On Sat, Mar 3, 2012 at 12:50 PM, Olivier Delalleau <sh...@keba.be> wrote:
> >> Would it be helpful if I went ahead and submitted a pull request with the >> function in my original question called "isclose" (along with a complete >> docstring and a few tests)? >> >> One note: >> At the moment, it deliberately compares NaN's as equal. E.g. >> >> isclose([np.nan, np.nan], [np.nan, np.nan]) >> >> will return: >> >> [True, True] >> >> This obviously runs counter to the standard way NaN's are handled (and >> indeed the definition of NaN). >> >> However, in the context of a floating point "close to" function, I think >> it makes the most sense. >> >> I've had this sitting around in a small project for awhile now, and it's >> been more useful to have it compare NaN's as "approximately equal" than not >> for my purposes at least. >> >> Nonetheless, it's something that needs additional consideration. >> >> Thanks, >> -Joe >> > > It would be confusing if numpy.isclose().all() was different from > numpy.allclose(). That being said, I agree it's useful to have NaNs compare > equal in some cases, maybe it could be a new argument to the function? > Good point. I went ahead and added an "equal_nan" kwarg and removed the "check_invalid" kwarg I had in before. I also made it mimic what np.equal() does in the case of two scalars (return a scalar instead of an array). I went ahead an make a pull request: https://github.com/numpy/numpy/pull/224 Hope that's alright. Cheers, -Joe > > -=- Olivier > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > >
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