On Thu, Mar 31, 2011 at 10:00 AM, Ralf Gommers <ralf.gomm...@googlemail.com>wrote:
> On Wed, Mar 30, 2011 at 11:52 PM, Derek Homeier > <de...@astro.physik.uni-goettingen.de> wrote: > > > > On 30 Mar 2011, at 23:26, Benjamin Root wrote: > > > >> Ticket 301: 'Make power and divide return floats from int inputs (like > >> true_divide)' > >> http://projects.scipy.org/numpy/ticket/301 > >> Invalid because the output dtype is the same as the input dtype unless > >> you override using the dtype argument: > >> >>> np.power(3, 1, dtype=np.float128).dtype > >> dtype('float128') > >> Alternatively return a float and indicate in the docstring that the > >> output dtype can be changed. > >> > >> > >> FWIW, > >> > >> Just thought I'd note (on a python 2.6 system): > >> > >> >>> import numpy as np > >> >>> a = np.array([1, 2, 3, 4]) > >> >>> a.dtype > >> dtype('int32') > >> >>> 2 / a > >> array([2, 1, 0, 0]) > >> >>> from __future__ import division > >> >>> 2 / a > >> array([ 2. , 1. , 0.66666667, 0.5 ]) > >> > >> So, numpy already does this when true division is imported (and > >> therefore consistent with whatever the python environment does), and > >> python currently also returns integers for exponentials when both > >> inputs are integers. > > > > I'd agree, and in my view power(3, -1) is well defined as 1 / 3 - > > also, in future (or Python3) > > > > >>> a/2 > > array([ 0.5, 1. , 1.5, 2. ]) > > >>> a//2 > > array([0, 1, 1, 2], dtype=int32) > > The ticket is about the functions np.divide and np.power, not / and > **. This currently does not work the same, unlike what's said above: > > >>> from __future__ import division > >>> x = np.arange(4) + 1 > >>> 2 / x > array([ 2. , 1. , 0.66666667, 0.5 ]) > >>> np.divide(2, x) > array([2, 1, 0, 0]) > > The power and divide functions should not check the "from __future__ > import division", just return floats IMHO. This is what's expected by > most users, and much more useful than integer math. > > Hmm, I didn't notice that distinction. So long as there is a consistent difference between the function and the operator, I am fine with that. > > > so I think at least a**n should follow integer math rules; depends on > > whether we want > > np.power to behave differently from ** (if they are internally handled > > separately at all)... > > Not sure if I understand the overload suggestion in the ticket > > I don't understand that either, it'll just lead to more confusion. > > Did a little further investigation of this (I hadn't thought of negative powers), and now I am getting thoroughly confused... >>> import numpy as np >>> a = np.array([1, 2]) >>> a.dtype dtype('int32') >>> 1**2 1 >>> 2**2 4 >>> 1**-2 1.0 >>> 2**-2 0.25 >>> a**2 array([1, 4]) >>> a**-2 array([1, 0]) >>> np.power(a, 2) array([1, 4]) >>> np.power(a, -2) array([1, 0]) So, in this case, the ** operator and np.power are identical, but diverges from the behavior of python's operator. FWIW, Ben Root
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