[Numpy-discussion] Determining the 'viewness' of an array, and flags.owndata confusion

Val Kalatsky kalatsky@gmail....
Wed Feb 29 00:38:38 CST 2012


Viewness is in the eyes of the beholder.
You have to use indirect methods to figure it out.
Probably the most robust approach is to go up the base chain until you get
None.

In [71]: c1=np.arange(16)
In [72]: c2=c1[::2]
In [73]: c4=c2[::2]
In [74]: c8=c4[::2]
In [75]: id(c8.base)==id(c4)
Out[75]: True
In [76]: id(c8.base.base.base)==id(c1)
Out[76]: True

The graph of dependencies is only a tree, so sooner or later you'll get to
the root.

Val

On Tue, Feb 28, 2012 at 8:06 PM, Jonathan Rocher <jrocher@enthought.com>wrote:

> Thank you all for your answers. Kurt and I were training new developers on
> numpy and telling them that
> - fancy indexing creates a copy
> - owndata was a good way to know if an array is a view or a copy.
> It turns out that these were both correct statements but we didn't think
> that the fancy indexing would sometimes return a view after making a copy
> (BTW is that necessary? It sounds to me like it is complicating things for
> no real benefit.)
>
> So a more precise question then is "is a_fancy a view on a's data or does
> it copy the data and its data is independent?" and the answer is probably:
> since "a_fancy1.base is a" is false then the 2 array's data are
> independent from each other.
>
> Do I have it right now?
>
> Jonathan
>
>
> On Tue, Feb 28, 2012 at 5:31 PM, Nathaniel Smith <njs@pobox.com> wrote:
>
>> On Tue, Feb 28, 2012 at 11:01 PM, Kurt Smith <kwmsmith@gmail.com> wrote:
>> > For an arbitrary numpy array 'a', what does 'a.flags.owndata' indicate?
>>
>> I think what it really indicates is whether a's destructor should call
>> free() on a's data pointer.
>>
>> > I originally thought that owndata is False iff 'a' is a view.  But
>> > that is incorrect.
>> >
>> > Consider the following:
>> >
>> > In [119]: a = np.zeros((3,3))
>> >
>> > In [120]: a.flags.owndata  # should be True; zeros() creates and
>> > returns a non-view array.
>> > Out[120]: True
>> >
>> > In [121]: a_view1 = a[2:, :]
>> >
>> > In [122]: a_view1.flags.owndata   # expected to be False
>> > Out[122]: False
>> >
>> > In [123]: a_fancy1 = a[[0,1], :]
>> >
>> > In [124]: a_fancy1.flags.owndata  # expected to be True, a_fancy1 is a
>> > fancy-indexed array.
>> > Out[124]: True
>> >
>> > In [125]: a_fancy2 = a[:, [0,1]]
>> >
>> > In [126]: a_fancy2.flags.owndata # expected to be True, a_fancy2 is a
>> > fancy-indexed array.
>> > Out[126]: False
>> >
>> > So when I query an array's flags.owndata, what is it telling me?  What
>> > I want to know is whether an array is a view or not.  If flags.owndata
>> > has nothing to do with the 'viewness' of an array, how would I
>> > determine if an array is a view?
>> >
>> > In the previous example, a_fancy2 does not own its data, as indicated
>> > by 'owndata' being False.  But when I modify a_fancy2, 'a' is not
>> > modified, as expected, but contrary to what 'owndata' would seem to
>> > indicate.
>>
>> It looks like the fancy indexing code in this case actually creates a
>> new array, and then instead of returning it directly, returns a
>> (reshaped) view on this new array. If you look at a_fancy2.base,
>> you'll see this new array.
>>
>> So: a_fancy2 *is* a view... it's just not a view of 'a'. It's a view
>> of this other array.
>>
>> > If I cannot use flags.owndata, what is a reliable way to determine
>> > whether or not an array is a view?
>>
>> owndata *is* a reliable way to determine whether or not an array is a
>> view; it just turns out that this is not a very useful question to
>> ask.
>>
>> What are you actually trying to do? There's probably another way to
>> accomplish it.
>>
>> -- Nathaniel
>> _______________________________________________
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>> NumPy-Discussion@scipy.org
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>>
>
>
>
> --
> Jonathan Rocher, PhD
> Scientific software developer
> Enthought, Inc.
> jrocher@enthought.com
> 1-512-536-1057
> http://www.enthought.com
>
>
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
>
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