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array slicing in numpy

today I used the numpy array for some calculation and found a strange problem, for example, assume i already imported numpy.arange in Ipython, and I run some scripts as follows:

In [5]: foo = arange(10)                                                      

In [8]: foo1 = foo[arange(3)]                                                 

In [11]: foo1[:] = 0                                                          

In [12]: foo
Out[12]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In [16]: foo2 = foo[0:3]                                                      

In [19]: foo2[:]=0                                                            

In [21]: foo
Out[21]: array([0, 0, 0, 3, 4, 5, 6, 7, 8, 9])

above shows that when i slice the array by foo[arange(3)], i got a copy of the array slice, but when i slice the array by foo[0:3], i got a reference of the array slice, thus foo changes with foo2. Then I thought foo and foo2 should have the same id, but that seems is not true

In [59]: id(foo)
Out[59]: 27502608

In [60]: id(foo2)
Out[60]: 28866880

In [61]: id(foo[0])
Out[61]: 38796768

In [62]: id(foo2[0])
Out[62]: 38813248

...

even more strange, if I keep checking the id of foo and foo2, they are not constant, and sometimes, they did match each other!

In [65]: id(foo2[0])
Out[65]: 38928592

In [66]: id(foo[0])                                                          
Out[66]: 37111504

In [67]: id(foo[0])
Out[67]: 38928592

can anyone explain this a little bit? I am really confused by this dynamic feature of python

thanks alot

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shelper Avatar asked Oct 25 '13 23:10

shelper


1 Answers

foo[arange(3)]

is not a slice. The elements of arange(3) are used to select elements of foo to construct a new array. Since this can't efficiently return a view (every element of the view would have to be an independent reference, and operations on the view would require following far too many pointers), it returns a new array.

foo[0:3]

is a slice. This can be done efficiently as a view; it only requires adjusting some bounds. Thus, it returns a view.

id(foo[0])

foo[0] doesn't refer to a specific Python object. Keeping separate Python objects for every array element would be far too expensive, negating much of the benefit of numpy. Instead, when an indexing operation is performed on a numpy ndarray, numpy constructs a new object to return. You'll get a different object with a different ID every time.

like image 146
user2357112 supports Monica Avatar answered Sep 17 '22 17:09

user2357112 supports Monica