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pythonarraysnumpypass-by-reference
Vincent
Vincent

Reputation: 60371

Modifying a numpy array inside a function?

I have a problem with the simple program below :

def my_function(my_array = np.zeros(0)):
    my_array = [1, 2, 3]

my_array = np.zeros(0)
my_function(my_array)
print my_array

It prints an empty array as if my_array was passed by copy and not by reference inside the function. How to correct that ?

Upvotes: 6

Views: 28674

Answers (3)

YXD
YXD

Reputation: 32511

np.zeros(0) gives you an empty numpy array. The reference inside your function now points to a new Python list but you haven't actually modified your empty numpy array, so that's still what you're printing out.

Recommend to read this answer to clear up some concepts.

Upvotes: 6

Ramón J Romero y Vigil
Ramón J Romero y Vigil

Reputation: 17933

The pass-by-reference model is more of a pass-by-value of a pointer. So inside your my_function you have a copy of a pointer to your original my_array. If you were to use that pointer to manipulate the inputed array directly that would make a change, but a re-assignment of the copied pointer won't affect the original array.

As an example:

def my_func(a):
    a[1] = 2.0

ar = np.zeros(4)
my_func(ar)
print ar

The above code will change the internal value of ar.

Upvotes: 18

mgilson
mgilson

Reputation: 309881

You can use slice assignment here just like you would a list:

def func(my_array):
    my_array[:3] = [1,2,3]

Note that this still requires that my_array has at least 3 elements in it ... Example usage:

>>> def func(my_array):
...     my_array[:3] = [1,2,3]
... 
>>> a = np.zeros(4)
>>> a
array([ 0.,  0.,  0.,  0.])
>>> func(a)
>>> a
array([ 1.,  2.,  3.,  0.])

The thing you're missing is how python deals with references. When you enter my_function, you have a reference to the original ndarray object bound to the name my_array. However, as soon as you assign something new to that name, you lose your original reference and replace it with a reference to a new object (in this case, a list).

Note that having a default argument which is a mutable object can often lead to surprises

Upvotes: 9

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