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?
Answers:
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']
According to docs:
Shuffle the sequence x in place. The optional argument random is a
0-argument function returning a random float in [0.0, 1.0); by
default, this is the function random().
>>> x = ['foo','bar','black','sheep']
>>> from random import shuffle
>>> shuffle(x)
>>> x
['bar', 'black', 'sheep', 'foo']
Why, really?
1. Efficiency
shuffle modifies the list in place.
This is nice, because copying a large list would be pure overhead if you do not need the original list anymore.
2. Pythonic style
According to the “explicit is better than implicit” principle of pythonic style, returning the list would be a bad idea, because then one might think it is a new one although in fact it is not.
But I don’t like it like this!
If you do need a fresh list, you will have to write something like
new_x = list(x) # make a copy
random.shuffle(new_x)
which is nicely explicit.
If you need this idiom frequently, wrap it in a function shuffled (see sorted) that returns new_x.
This method works too.
import random
shuffled = random.sample(original, len(original))
I had my aha moment with this concept like this:
from random import shuffle
x = ['foo','black','sheep'] #original list
y = list(x) # an independent copy of the original
for i in range(5):
print shuffle(y) # shuffles the original "in place" prints "None" return
print x,y #prints original, and shuffled independent copy
>>>
None
['foo', 'black', 'sheep'] ['foo', 'black', 'sheep']
None
['foo', 'black', 'sheep'] ['black', 'foo', 'sheep']
None
['foo', 'black', 'sheep'] ['sheep', 'black', 'foo']
None
['foo', 'black', 'sheep'] ['black', 'foo', 'sheep']
None
['foo', 'black', 'sheep'] ['sheep', 'black', 'foo']
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?
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']
According to docs:
Shuffle the sequence x in place. The optional argument random is a
0-argument function returning a random float in [0.0, 1.0); by
default, this is the function random().
>>> x = ['foo','bar','black','sheep']
>>> from random import shuffle
>>> shuffle(x)
>>> x
['bar', 'black', 'sheep', 'foo']
Why, really?
1. Efficiency
shuffle modifies the list in place.
This is nice, because copying a large list would be pure overhead if you do not need the original list anymore.
2. Pythonic style
According to the “explicit is better than implicit” principle of pythonic style, returning the list would be a bad idea, because then one might think it is a new one although in fact it is not.
But I don’t like it like this!
If you do need a fresh list, you will have to write something like
new_x = list(x) # make a copy
random.shuffle(new_x)
which is nicely explicit.
If you need this idiom frequently, wrap it in a function shuffled (see sorted) that returns new_x.
This method works too.
import random
shuffled = random.sample(original, len(original))
I had my aha moment with this concept like this:
from random import shuffle
x = ['foo','black','sheep'] #original list
y = list(x) # an independent copy of the original
for i in range(5):
print shuffle(y) # shuffles the original "in place" prints "None" return
print x,y #prints original, and shuffled independent copy
>>>
None
['foo', 'black', 'sheep'] ['foo', 'black', 'sheep']
None
['foo', 'black', 'sheep'] ['black', 'foo', 'sheep']
None
['foo', 'black', 'sheep'] ['sheep', 'black', 'foo']
None
['foo', 'black', 'sheep'] ['black', 'foo', 'sheep']
None
['foo', 'black', 'sheep'] ['sheep', 'black', 'foo']