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Why does Python's `random.shuffle()` return `None` instead of the shuffled list?

Dec 07, 2024 pm 03:49 PM

Why does Python's `random.shuffle()` return `None` instead of the shuffled list?

Shuffling a List of Objects

A common task in programming is to shuffle a list of objects. In Python, there's a built-in function for this: random.shuffle(). However, its behavior can be unexpected if you're not familiar with its nuances.

Question

Why does random.shuffle(), when called on a list of objects, return None instead of the shuffled list?

Answer

random.shuffle() modifies its input list in place, meaning it doesn't create a new list but rather shuffles the elements of the existing list. As a result, it doesn't have any value to return, hence the None result. This behavior is consistent with Python's convention for mutable objects, where functions that modify them usually return None.

Example

To demonstrate this, consider the following code:

from random import shuffle

l = [[i] for i in range(10)]
shuffle(l)
print(l)
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The code creates a list of lists and shuffles it using shuffle(). Printing the resulting list shows that it has been successfully rearranged:

[[9], [2], [7], [0], [4], [5], [3], [1], [8], [6]]
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In this example, random.shuffle() indeed modified the original list in place, returning None.

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