[Numpy-discussion] question about creating numpy arrays

[josef.pktd at gmail.com]

On Thu, May 20, 2010 at 12:13 PM, Benjamin Root <ben.root at ou.edu> wrote:
>
>
> On Thu, May 20, 2010 at 10:30 AM, Ryan May <rmay31 at gmail.com> wrote:
>>
>> On Thu, May 20, 2010 at 9:44 AM, Benjamin Root <ben.root at ou.edu> wrote:
>> >> I gave two counterexamples of why.
>> >
>> > The examples you gave aren't counterexamples.  See below...
>> >
>> > On Wed, May 19, 2010 at 7:06 PM, Darren Dale <dsdale24 at gmail.com> wrote:
>> >>
>> >> On Wed, May 19, 2010 at 4:19 PM,  <josef.pktd at gmail.com> wrote:
>> >> > On Wed, May 19, 2010 at 4:08 PM, Darren Dale <dsdale24 at gmail.com>
>> >> > wrote:
>> >> >> I have a question about creation of numpy arrays from a list of
>> >> >> objects, which bears on the Quantities project and also on masked
>> >> >> arrays:
>> >> >>
>> >> >>>>> import quantities as pq
>> >> >>>>> import numpy as np
>> >> >>>>> a, b = 2*pq.m,1*pq.s
>> >> >>>>> np.array([a, b])
>> >> >> array([ 12.,   1.])
>> >> >>
>> >> >> Why doesn't that create an object array? Similarly:
>> >> >>
>> >
>> >
>> > Consider the use case of a person creating a 1-D numpy array:
>> >  > np.array([12.0, 1.0])
>> > array([ 12.,  1.])
>> >
>> > How is python supposed to tell the difference between
>> >  > np.array([a, b])
>> > and
>> >  > np.array([12.0, 1.0])
>> > ?
>> >
>> > It can't, and there are plenty of times when one wants to explicitly
>> > initialize a small numpy array with a few discrete variables.
>>
>> What do you mean it can't? 12.0 and 1.0 are floats, a and b are not.
>> While, yes, they can be coerced to floats, this is a *lossy*
>> transformation--it strips away information contained in the class, and
>> IMHO should not be the default behavior. If I want the objects, I can
>> force it:
>>
>> In [7]: np.array([a,b],dtype=np.object)
>> Out[7]: array([2.0 m, 1.0 s], dtype=object)
>>
>> This works fine, but feels ugly since I have to explicitly tell numpy
>> not to do something. It feels to me like it's violating the principle
>> of "in the face of ambiguity, resist the temptation to guess."
>
> I have thought about this further, and I think I am starting to see your
> point (from both of you).  Here are my thoughts:
>
> As I understand it, numpy.array() (rather, array_like()) essentially builds
> the dimensions of the array by first identifying if there is an iterable
> object, and then if the contents of the iterable is also iterable, until it
> reaches a non-iterable.
>
> Therefore, the question becomes, why is numpy.array() implicitly coercing
> the non-iterable type into a numeric?  Is there some reason that I am not
> seeing for why there is an implicit coercion?

I think because the dtype is numeric (float), otherwise it wouldn't
operate on numbers, and none of the other numerical functions might
work  (just a guess)

>>> a = np.array(['2.0', '1.0'], dtype=object)
>>> a
array([2.0, 1.0], dtype=object)
>>> np.sqrt(a)
Traceback (most recent call last):
  File "<pyshell#31>", line 1, in <module>
    np.sqrt(a)
AttributeError: sqrt
>>> np.array([a,a])
array([[2.0, 1.0],
       [2.0, 1.0]], dtype=object)
>>> 2*a
array([2.02.0, 1.01.0], dtype=object)


>>> b = np.array(['2.0', '1.0'])
>>> np.sqrt(b)
NotImplemented
>>> np.array([b,b])
array([['2.0', '1.0'],
       ['2.0', '1.0']],
      dtype='|S3')
>>> 2*b
Traceback (most recent call last):
  File "<pyshell#41>", line 1, in <module>
    2*b
TypeError: unsupported operand type(s) for *: 'int' and 'numpy.ndarray'


Josef

>
> At first glance, I did not see a problem with this behavior, and I have come
> to expect it (hence my original reply). But now, I am not quite so sure.
>
>>
>> Ryan
>>
>> --
>> Ryan May
>> Graduate Research Assistant
>> School of Meteorology
>> University of Oklahoma
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