按列表中出现的频率对列表进行排序 [英] Sorting a List by frequency of occurrence in a list

问题描述

我有一个整数列表(或者甚至可以是字符串)，我想按Python中出现的频率排序，例如:

I have a list of integers(or could be even strings), which I would like to sort by the frequency of occurrences in Python, for instance:

a = [1, 1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5]

元素5在列表中出现4次，4在列表中出现3次.因此，输出的排序列表将是:

Here the element 5 appears 4 times in the list, 4 appears 3 times. So the output sorted list would be :

result = [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

我尝试使用a.count()，但是它给出了元素出现的次数. 我想对它进行排序.知道怎么做吗?

I tried using a.count(), but it gives the number of occurrence of the element. I would like to sort it. Any idea how to do it ?

谢谢

推荐答案

from collections import Counter
print [item for items, c in Counter(a).most_common() for item in [items] * c]
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

甚至更好(高效)的实施方式

Or even better (efficient) implementation

from collections import Counter
from itertools import repeat, chain
print list(chain.from_iterable(repeat(i, c) for i,c in Counter(a).most_common()))
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

或

from collections import Counter
print sorted(a, key=Counter(a).get, reverse=True)
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

如果您喜欢就地排序

a.sort(key=Counter(a).get, reverse=True)

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