Python：如何获得列表中项目的排序计数？
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Python: how to get sorted count of items in a list?
提问by AP257
In Python, I've got a list of items like:
在 Python 中，我有一个项目列表，例如：
mylist = [a, a, a, a, b, b, b, d, d, d, c, c, e]
And I'd like to output something like:
我想输出如下内容：
a (4)
b (3)
d (3)
c (2)
e (1)
How can I output a count and leaderboard of items in a list? I'm not too bothered about efficiency, just any way that works :)
如何输出列表中项目的计数和排行榜？我不太关心效率，只是任何有效的方式:)
Thanks!
谢谢！
采纳答案by Eli Courtwright
from collections import defaultdict
def leaders(xs, top=10):
counts = defaultdict(int)
for x in xs:
counts[x] += 1
return sorted(counts.items(), reverse=True, key=lambda tup: tup[1])[:top]
So this function uses a defaultdictto count the number of each entry in our list. We then take each pair of the entry and its count and sort it in descending order according to the count. We then take the topnumber of entries and return that.
所以这个函数使用 adefaultdict来计算我们列表中每个条目的数量。然后我们取出每对条目及其计数，并根据计数将其按降序排序。然后我们获取top条目的数量并返回它。
So now we can say
所以现在我们可以说
>>> xs = list("jkl;fpfmklmcvuioqwerklmwqpmksdvjioh0-45mkofwk903rmiok0fmdfjsd")
>>> print leaders(xs)
[('k', 7), ('m', 7), ('f', 5), ('o', 4), ('0', 3), ('d', 3), ('i', 3), ('j', 3), ('l', 3), ('w', 3)]
回答by AXO
I'm surprised that no one has mentioned collections.Counter. Assuming
我很惊讶没有人提到collections.Counter。假设
import collections
mylist = ['a', 'a', 'a', 'a', 'b', 'b', 'b', 'd', 'd', 'd', 'c', 'c', 'e']
it's just a one liner:
这只是一个班轮：
print(collections.Counter(mylist).most_common())
which will print:
这将打印：
[('a', 4), ('b', 3), ('d', 3), ('c', 2), ('e', 1)]
Note that Counteris a subclass of dictwith some useful methods for counting objects. Refer to the documentationfor more info.
请注意，它Counter是dict具有一些用于计数对象的有用方法的子类。有关更多信息，请参阅文档。
回答by Otto Allmendinger
A two-liner:
两线：
for count, elem in sorted(((mylist.count(e), e) for e in set(mylist)), reverse=True):
print '%s (%d)' % (elem, count)
