On Fri, Nov 19, 2010 at 10:59 PM, Keith Goodman <kwgood...@gmail.com> wrote: > On Fri, Nov 19, 2010 at 7:51 PM, <josef.p...@gmail.com> wrote: >> >> does this give you the correct answer? >> >>>>> 1>np.nan >> False >> >> What's the starting value for amax? -inf? > > Because "1 > np.nan" is False, the current running max does not get > updated, which is what we want. > >>> import nanny as ny >>> np.nanmax([1, np.nan]) > 1.0 >>> np.nanmax([np.nan, 1]) > 1.0 >>> np.nanmax([np.nan, 1, np.nan]) > 1.0 > > Starting value is -np.inf for floats and stuff like this for ints: > > cdef np.int32_t MININT32 = np.iinfo(np.int32).min > cdef np.int64_t MININT64 = np.iinfo(np.int64).min
That's what I thought halfway through typing the question. >>> -np.inf>-np.inf False If the only value is -np.inf, you will return nan, I guess. >>> np.nanmax([-np.inf, np.nan]) -inf Josef (being picky) > > Numpy does this: > >>> np.nanmax([]) > <snip> > ValueError: zero-size array to ufunc.reduce without identity > > Nanny does this: > >>> ny.nanmax([]) > nan > > So I haven't taken care of that corner case yet. I'll commit nanmax to > github in case anyone wants to give it a try. > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
