Python random.choice() function - how to never hav

问题:

Let's say I have

mychoice = random.choice(['this is random response 1','this is random response 2', 
'this is random response 3', 'and 4', 'and so on'])

How can I avoid having the same choice being repeated more than once in a row? Or how can I can I set a condition to make a particular choice only appear after a certain number of other choices have been chosen? Or is there a module better suited to my needs in this regard?

回答1:

the simplest solution would probably be to construct a usedQueue of length k (where k is the number of selections before a choice is allowed to repeat.) When you select a choice, remove it from your original list and place it in usedQueue. Then, if usedQueue.length > k, pop one back onto your array.

As already stated, this significantly reduces the randomness of your algorithm. That said, it does have practical uses (take a look at iTunes.)

回答2:

Example implementation ensuring a minimum distance between two occurrences of the same item:

def choice_gen(choices, min_dist):
    last_choices = collections.deque(maxlen=min_dist)
    choices = set(choices)
    while 1:
        c = random.choice(list(choices - set(last_choices)))
        last_choices.append(c)
        yield c

回答3:

This routine selects and prints random choices from mylist without replacement, until all choices have been exhausted.

index = range(len(mylist))
while len(index) > 0:
    i = random.choice(index)
    print mylist[i]
    index.pop(i)