Why Does Python\'s `random.shuffle` Return `None`?
Python's Random Shuffle: Why the Non-Apparent Return?
When using Python's random.shuffle function, one might be perplexed to see it return None. This behavior stems from its in-place nature.
Unlike functions that return the transformed data structure after modifications, random.shuffle directly alters the input list. Therefore, it returns None to indicate that no explicit value is being returned.
To illustrate, consider the following code:
Upon executing this code, the list x will be shuffled in-place, but the random.shuffle function itself will not return any value. Consequently, the print statement will output the now-shuffled list.
If you desire a new, shuffled list without modifying the original, you can utilize alternative methods:
- random.sample(x, len(x)): Creates a new randomly-shuffled list by sampling values from the original list.
- shuffled = sorted(x, key=lambda k: random.random()): Uses sorting with a randomized key to generate a new, shuffled list.
Understand that while random.sample has an O(N) complexity, sorted has an O(N log N) complexity due to the underlying sorting operation.
By comprehending random.shuffle's in-place behavior and utilizing alternative methods as necessary, you can effectively control how you handle list shuffling operations in Python.
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