array assignment using slicing

Question

When I was using array assignment using slicing, there is some thing strange happened. The source code is below:

import numpy as np
a = np.array([1,2,3,4]).reshape(2,2)
b = np.array([5,6,7,8]).reshape(2,2)
print(id(a))
print(id(b))
b = a[:]
b[1,1] = 10
print(b is a)
print(id(a))
print(id(b))
print(a)
print(b)

The result is given as:

From the result, the id of b and a is different after array assignment, but when I change the value of b, the value in a also changed. Why is this?

Using Sublime Text, Python 3.4.3.

That link is about lists. This question is about numpy arrays. — hpaulj
– hpaulj, Commented Jun 17, 2016 at 14:33
@hpaulj Thank you. You are right. Numpy array is different from list. — Moon Zoe
– Moon Zoe, Commented Jun 17, 2016 at 14:57
I removed the duplicates because this is about an array view v copy, not about list copies and deep copies. stackoverflow.com/questions/2612802/…. numpy arrays are not lists — hpaulj
– hpaulj, Commented Jun 17, 2016 at 23:57

Community · Accepted Answer · 2017-05-23 11:58:47Z

1

I think you might have an issue with referencing (b=a[:]). Here is a previous answer that might help:

Python objects confusion: a=b, modify b and a changes!

edited May 23, 2017 at 11:58

CommunityBot

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answered Jun 17, 2016 at 14:43

janeDoe

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Comments

hpaulj · Accepted Answer · 2016-06-17 15:24:18Z

0

With lists, b=a[:] makes a copy of a. Changes to b will not affect a or its values.

But with an numpy array, this action makes view. b is a new object, but it shares the underlying data buffer. Changes to values in b will affect a.

Use b=b.copy() is you want a true copy.

https://docs.scipy.org/doc/numpy-dev/user/quickstart.html#copies-and-views

edited Jun 17, 2016 at 15:24

answered Jun 17, 2016 at 15:16

hpaulj

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1 Comment

Moon Zoe Over a year ago

This is the file I found, exactly what you said.docs.scipy.org/doc/numpy-dev/user/quickstart

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