Pythonic方法以递减的顺序迭代collections.Counter（）实例？ [英] Pythonic way to iterate over a collections.Counter() instance in descending order?

问题描述

在Python 2（2.7，更准确地说）中，我想以递减计数顺序迭代collections.Counter实例。

In Python 2 (2.7, to be more precise), I want to iterate over a collections.Counter instance in descending count order.

>>> import collections
>>> c = collections.Counter()
>>> c['a'] = 1
>>> c['b'] = 999
>>> c
Counter({'b': 999, 'a': 1})
>>> for x in c:
        print x
a
b

在上面的示例，似乎元素按照它们添加到Counter实例的顺序进行迭代。

In the example above, it appears that the elements are iterated in the order they were added to the Counter instance.

我想从最高到最低迭代列表。我看到Counter的字符串表示就是这样，只是想知道是否有推荐的方法来做它。

I'd like to iterate over the list from highest to lowest. I see that the string representation of Counter does this, just wondering if there's a recommended way to do it.

推荐答案

你可以迭代超过 c.most_common（）以获得所需订单的商品。另请参阅 Counter.most_common（）的文档 。

You can iterate over c.most_common() to get the items in the desired order. See also the documentation of Counter.most_common().

示例：

>>> c = collections.Counter(a=1, b=999)
>>> c.most_common()
[('b', 999), ('a', 1)]

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