When delving into the depths of Python's random module, one may stumble upon the perplexing behavior of random.shuffle(). In defiance of expectations, this function appears to return the ethereal None value after ostensibly shuffling a list. To unravel this enigma, let's embark on a journey into the inner workings of random.shuffle.
Contrary to the convention of returning the modified data structure, random.shuffle() chooses a more enigmatic path. It transforms the list it is applied to silently, without any tangible evidence of its actions. This subtle alteration, known as "in-place modification," is a hallmark of many Python API methods that modify data structures. Hence, random.shuffle() returns None, leaving behind a shuffled list without any explicit confirmation.
To witness the elusive work of random.shuffle(), consider the following code:
x = ['foo', 'bar', 'black', 'sheep'] from random import shuffle random.shuffle(x) print(x)
Executing this code will print the shuffled list, revealing the fruits of random.shuffle()'s invisible labor.
However, if your desire is not merely to witness a list's transformation but to create a new shuffled list, other avenues await you. Random.sample() stands ready to serve, providing a way to generate a shuffled copy of an existing list while leaving the original untouched:
shuffled_x = random.sample(x, len(x))
Alternatively, sorted() combined with random.random() offers another route to a shuffled list, albeit with a computational cost:
shuffled_x = sorted(x, key=lambda k: random.random())
Both approaches yield a shuffled copy of your list, allowing you to manipulate the new list without disrupting the original.
The random.shuffle() function's peculiar behavior may initially raise an eyebrow, but understanding its in-place modification nature and exploring alternative approaches for creating new shuffled lists empowers you to harness its randomizing magic effectively.
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