Different function outputs for integers and numpy arrays

Question

I am having trouble understanding, why I get different values in the following two cases:

-Case 1:

def myfunc(a,b,c):
    xx = a+b
    yy = b+c
    return xx, yy

q,w = myfunc(1,2,3)

print(q,w)

Output 1: 3 5

-Case 2:

import numpy as np

q=w=np.zeros(3)

def myfunc(a,b,c):
    xx = a+b
    yy = b+c
    return xx, yy


for i in range(3):
    q[i],w[i] = myfunc(1,2,3)

print(q,w)

Output 2: [5. 5. 5.] [5. 5. 5.]

In the second case, both arrays have their entries equal to 5. Could someone explain, why?

Anass ABEA · Accepted Answer · 2022-01-06 22:25:40Z

1

I won't talk about the first case because it's simple and clear. For the second case, you have defined the variables q and w as following

q=w=np.zeros(3)

In this case, what ever changes you make in q they will be applied to w because they q and w have the same address.

When you run this:

q[i],w[i] = myfunc(1,2,3)

q[i] gets the value 3 and w[i] gets the value 5 and since q and w have the same address, q[i] will get the value 5 as well. That explains why you have 5 every time.

If you want to solve it, change the variable definition line from:

q=w=np.zeros(3)

to :

w=np.zeros(3)
q=np.zeros(3)

answered Jan 6, 2022 at 22:25

Anass ABEA

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4 Comments

chepner Over a year ago

More precisely, q and w are both references to the same object created by np.zeros(3). (There is no notion of "address" in Python.)

numpy Over a year ago

Thank you, that helps. I thought I would save space by using q=w=np.zeros(3) instead of w=np.zeros(3) q=np.zeros(3), but did not know they would be assigned to the same memory address.

numpy Over a year ago

@chepner, yes, that makes sense! Thank you!

chepner Over a year ago

@numpy There is no "they"; you only created one object, then defined two references two it. Simple assignment never creates a copy in Python. Be sure to read nedbatchelder.com/text/names.html.

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