如何从Python的Counter类中获得加权随机选择？ [英] How can I get a weighted random pick from Python's Counter class?

问题描述

我有一个程序，我正在使用 collections.Counter 跟踪各种事物的成功 - 事物递增相应的计数器：

I have a program where I'm keeping track of the success of various things using collections.Counter — each success of a thing increments the corresponding counter:

import collections
scoreboard = collections.Counter()

if test(thing):
    scoreboard[thing]+ = 1

然后，为了将来的测试，我想倾向于那些已经取得最大成功的 thing 。 Counter.elements（）似乎是理想的，因为它返回的元素（以任意顺序）重复多次等于计数。所以我想我可以做：

Then, for future tests, I want to skew towards things which have generated the most success. Counter.elements() seemed ideal for this, since it returns the elements (in arbitrary order) repeated a number of times equal to the count. So I figured I could just do:

import random
nextthing=random.choice(scoreboard.elements())

但不，这会引发 TypeError：'itertools.chain'类型的对象没有len （）即可。好的，所以 random.choice 无法与迭代器一起使用。但是，在这种情况下，长度是已知的（或可知） - 它是 sum（scoreboard.values（））。

But no, that raises TypeError: object of type 'itertools.chain' has no len(). Okay, so random.choice can't work with iterators. But, in this case, the length is known (or knowable) — it's sum(scoreboard.values()).

我知道迭代未知长度列表并随机选择元素的基本算法，但我怀疑有更优雅的东西。我应该在这做什么？

I know the basic algorithm for iterating through a list of unknown length and fairly picking an element at random, but I suspect that there's something more elegant. What should I be doing here?

推荐答案

您可以使用 itertools.islice 获取可迭代的第N项：

You can do this rather easily by using itertools.islice to get the Nth item of an iterable:

>>> import random
>>> import itertools
>>> import collections
>>> c = collections.Counter({'a': 2, 'b': 1})
>>> i = random.randrange(sum(c.values()))
>>> next(itertools.islice(c.elements(), i, None))
'a'

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