Python: .pop() on unordered sets

问题

I have a question regarding the .pop() method. In the documentation, it states that on sets:

Remove and return an arbitrary element from the set.

So what exactly does "arbitrary" mean? Like, for example: If I wanted to build a queue from a list of users where position is determined at random, could I enter in all users into a set, and then build a randomly assigned queue using .pop()?

So the code would look something like

queue_set = {'bob','rachel','sara','david'}
queue = dequeue([])
while queue_set:
        queue.append(queue_set.pop())

Would that be a reasonable way to randomly assign members to an index? Or is the .pop() reliant on some way the data is entered or something?

Thanks!

回答1:

No.

"Arbitrary" means the implementation returns whichever element is most convenient. You are not guaranteed randomness. Use random.sample() or random.shuffle() if you need random selection.

回答2:

I think the best way would be to use random.shuffle

>>> import random
>>> li = ["bob", "rachel", "sara", "david"]
>>> random.shuffle(li)
>>> li
['sara', 'bob', 'david', 'rachel']

If you want to get the elements one by one, you can then use li.pop().

来源：https://stackoverflow.com/questions/27952058/python-pop-on-unordered-sets