Python: Assigning an initial numpy array to another variable, changing few elements of the new variable changes the initial numpy array

Question

I need to do calculations on an intial 3x3 array (say 'x') which I would need again later for further calculations. So, I have another variable ('y') to make a copy of 'x' (y = x), do calculations on 'y' and then use x for later purposes. But somehow address of 'y' changes to 'x' even if I assign it initially other than 'x'

import numpy as np

x = np.random.rand(3,3)
y = np.random.rand(3,3)

print 'y',id(y)

y = x
y[1][1] = -y[1][1]

print x[1][1] + y[1][1]  #This needs to be 0.
print 'x',id(x)
print 'y',id(y)

In the above code, I need line 9 ('x[1][1] + y[1][1]'), to print 0, but what is giving is 2 times -x[1][1]. What is the reasoning behind this problem and if you could suggest some method to avoid this?

When indexing a numpy array, get in the habit of using y[1, 1] instead of y[1][1]. Here they do the same thing, but y[:][1] is not the same as y[:,1]. — hpaulj
– hpaulj, Commented Jun 10, 2018 at 20:49

Ofer Sadan · Accepted Answer · 2018-06-10 21:35:23Z

1

Because y=x copies just the refrence, and does not create another copy of the array

You should replace that line with

y = x[:]

Otherwise changing x also changes y and vise versa.

However, this method is good only for a regular list and not numpy arrays. That is possible like this (also more explicit and readable):

y = np.copy(x)

If you want to check it yourself, you can print id(y) and print id(x) after that assignment, and see that in your case they are the same, while in a true copy they are different

edited Jun 10, 2018 at 21:35

answered Jun 10, 2018 at 18:31

Ofer Sadan

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2 Comments

juanpa.arrivillaga Over a year ago

This does not work. Slicing creates a view in numpy, and now x and y share the same underlying buffer. You have to use .copy

Ofer Sadan Over a year ago

@juanpa.arrivillaga i've updated the answer to reflect that

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