Question

I came across the fact that numpy arrays are passed by reference at multiple places, but then when I execute the following code, why is there a difference between the behavior of foo and bar

import numpy as np

def foo(arr):
   arr = arr - 3

def bar(arr):
   arr -= 3

a = np.array([3, 4, 5])
foo(a)
print a # prints [3, 4, 5]

bar(a)
print a # prints [0, 1, 2]

I’m using python 2.7 and numpy version 1.6.1

Answer 1

In Python, all variable names are references to values.

When Python evaluates an assignment, the right-hand side is evaluated before the left-hand side. arr - 3 creates a new array; it does not modify arr in-place.

arr = arr - 3 makes the local variable arr reference this new array. It does not modify the value originally referenced by arr which was passed to foo. The variable name arr simply gets bound to the new array, arr - 3. Moreover, arr is local variable name in the scope of the foo function. Once the foo function completes, there is no more reference to arr and Python is free to garbage collect the value it references. As Reti43 points out, in order for arr‘s value to affect a, foo must return arr and a must be assigned to that value:

def foo(arr):
    arr = arr - 3
    return arr
    # or simply combine both lines into `return arr - 3`

a = foo(a)

In contrast, arr -= 3, which Python translates into a call to the __iadd__ special method, does modify the array referenced by arr in-place.