python - Why does random.shuffle return None?

Question

Why is random.shuffle returning None in Python?

>>> x = ['foo','bar','black','sheep']
>>> from random import shuffle
>>> print shuffle(x)
None

How do I get the shuffled value instead of None?

Answer 1

random.shuffle() changes the x list in place.

Python API methods that alter a structure in-place generally return None, not the modified data structure.

>>> x = ['foo', 'bar', 'black', 'sheep']
>>> random.shuffle(x)
>>> x
['black', 'bar', 'sheep', 'foo']

---

If you wanted to create a new randomly-shuffled list based on an existing one, where the existing list is kept in order, you could use random.sample() with the full length of the input:

random.sample(x, len(x))     

You could also use sorted() with random.random() for a sorting key:

shuffled = sorted(x, key=lambda k: random.random())

but this invokes sorting (an O(N log N) operation), while sampling to the input length only takes O(N) operations (the same process as random.shuffle() is used, swapping out random values from a shrinking pool).

Demo:

>>> import random
>>> x = ['foo', 'bar', 'black', 'sheep']
>>> random.sample(x, len(x))
['bar', 'sheep', 'black', 'foo']
>>> sorted(x, key=lambda k: random.random())
['sheep', 'foo', 'black', 'bar']
>>> x
['foo', 'bar', 'black', 'sheep']