id() and is operator
Gary Herron
gherron at digipen.edu
Sun Feb 22 14:16:56 EST 2015
On 02/22/2015 09:53 AM, LJ wrote:
> Hi everyone. Quick question here. Lets suppose if have the following numpy array:
>
> b=np.array([[0]*2]*3)
>
> and then:
>
>>>> id(b[0])
> 45855552
>>>> id(b[1])
> 45857512
>>>> id(b[2])
> 45855552
>
> Please correct me if I am wrong, but according to this b[2] and b[0] are the same object. Now,
>
>>>> b[0] is b[2]
> False
>
>
> Any clarification is much appreciated.
>
> Cheers,
In fact, b[0] and b[2] are different objects as can be seen here:
>>> import numpy as np
>>> b=np.array([[0]*2]*3)
>>> b[0]=1 // broadcast into both ints in row 0
>>> b[1]=2 // ... row 1
>>> b[2]=3 // ... row 2
>>> b
array([[1, 1],
[2, 2],
[3, 3]])
When you extracted b[0], you got a newly created python/numpy object
(1x2 array of ints) briefly stored at location 45855552 but then
deleted immediately after that use. A little later, the extraction of
b[2] used the same bit of memory. The id of a temporarily created value
is meaningless, and apparently misleading.
As a separate issue, each of b, b[0], b[1], and b[2] do *all* refer to
the same underlying array of ints as can be seen here:
>>> r = b[0]
>>> r[0] = 123
>>> b
array([[123, 1],
[ 2, 2],
[ 3, 3]])
but the Python/numpy objects that wrap portions of that underlying array
of ints are all distinct.
Gary Herron
--
Dr. Gary Herron
Department of Computer Science
DigiPen Institute of Technology
(425) 895-4418
More information about the Python-list
mailing list