On 06/05/2015 06:39 AM, Todd wrote:
On Fri, Jun 5, 2015 at 3:23 PM, Gary Herron
<gary.her...@islandtraining.com
<mailto:gary.her...@islandtraining.com>> wrote:
On 06/05/2015 06:11 AM, Paul Appleby wrote:
On Fri, 05 Jun 2015 14:55:11 +0200, Todd wrote:
Numpy arrays are not lists, they are numpy arrays. They
are two
different data types with different behaviors. In lists,
slicing is a
copy. In numpy arrays, it is a view (a data structure
representing some
part of another data structure). You need to explicitly
copy the numpy
array using the "copy" method to get a copy rather than a
view:
OK, thanks. I see.
(I'd have thought that id(a[1]) and id(b[1]) would be the same
if they
were the same element via different "views", but the id's seem
to change
according to rules that I can't fathom.)
Nope. It's odder than that. a[1] is still a view into the
inderlying numpy array, and your id is the id of that view. Each
such index produces a new such view object. Check this out:
>>> import numpy
>>> a = numpy.array([1,2,3])
>>> id(a[1])
28392768
>>> id(a[1])
28409872
This produces two different view of the same underlying object.
a[1] and b[1] are not views:
>>> a[1].flags['OWNDATA']
True
>>> b[1].flags['OWNDATA']
True
>>> a[1:2].flags['OWNDATA']
False
Right. My bad. Each execution of a[1] creates a new numpy.int64 object
with the value from the array.
>>> type(a[1])
<class 'numpy.int64'>
Each execution of id(a[1]) creates an int64 object which is immediately
used and then deleted. Two successive executions of id(a[1]) may or may
not reuse the same piece of memory, depending on what else is going on
in memory. Indeed when I produced the above example with id(a[1]), a
third and fourth runs of id(a[1]) did indeed repeat 28409872, but they
are all new creations of an int64 object which happen to use the same
recently freed bit of memory.
Gary Herron
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