How to use random.shuffle() on a generator? python

Question

How do I use random.shuffle() on a generator without initializing a list from the generator? Is that even possible? if not, how else should I use random.shuffle() on my list?

>>> import random >>> random.seed(2) >>> x = [1,2,3,4,5,6,7,8,9] >>> def yielding(ls): ...     for i in ls: ...             yield i ...  >>> for i in random.shuffle(yielding(x)): ...     print i ...  Traceback (most recent call last):   File "<stdin>", line 1, in <module>   File "/usr/lib/python2.7/random.py", line 287, in shuffle     for i in reversed(xrange(1, len(x))): TypeError: object of type 'generator' has no len() 

Note: random.seed() was designed such that it returns the same output after each script run?

Answer 1

In order to shuffle the sequence uniformly, random.shuffle() needs to know how long the input is. A generator cannot provide this; you have to materialize it into a list:

lst = list(yielding(x)) random.shuffle(lst) for i in lst:     print i 

You could, instead, use sorted() with random.random() as the key:

for i in sorted(yielding(x), key=lambda k: random.random()):     print i 

but since this also produces a list, there is little point in going this route.

Demo:

>>> import random >>> x = [1,2,3,4,5,6,7,8,9] >>> sorted(iter(x), key=lambda k: random.random()) [9, 7, 3, 2, 5, 4, 6, 1, 8] 

Answer 2

It's not possible to randomize the yield of a generator without temporarily saving all the elements somewhere. Luckily, this is pretty easy in Python:

tmp = list(yielding(x)) random.shuffle(tmp) for i in tmp:     print i 

Note the call to list() which will read all items and put them into a list.

If you don't want to or can't store all elements, you will need to change the generator to yield in a random order.