python - Python：给定一个列表列表，创建一个按内部列表中出现次数排序的列表

Question

我有一个列表列表：

[['a','b','c'], ['a'], ['a','b']]

我想对其进行排序并返回一个列表，因此输出如下所示：

['a', 'b', 'c']

即按每个元素出现的次数排序。a 出现 3 次，b 出现两次，c 出现一次。

我该怎么做呢？

Answer 1

使用collections.Counter,itertools.chain.from_iterable和列表推导：

from itertools import chain
from collections import Counter

data = [['a', 'b', 'c'], ['a'], ['a', 'b']]

d = Counter(chain.from_iterable(data))
print([i for i, c in d.most_common()])

输出：

['a', 'b', 'c']

_{注意：当你想计算列表中某些项目的频率时，记得使用Counter，它真的很有帮助。}

Answer 2

用于itertools.chain.from_iterable()首先展平列表，然后collections.Counter()对元素进行计数。

>>> from collections import Counter
>>> from itertools import chain
>>> [x[0] for x in Counter(chain.from_iterable(mylist)).most_common())
['a', 'b', 'c']

Answer 3

collections.Counter应该这样做：

>>> from collections import Counter
>>> lol = [['a','b','c'], ['a'], ['a','b']]
>>> c = Counter(elem for sublist in lol for elem in sublist)
>>> c
Counter({'a': 3, 'b': 2, 'c': 1})

elem for sublist..bit 只是一个扁平化的习语。你可以使用

>>> Counter(chain.from_iterable(lol))
Counter({'a': 3, 'b': 2, 'c': 1})

相反，在from itertools import chain. 然后你可以得到最常见的：

>>> c.most_common()
[('a', 3), ('b', 2), ('c', 1)]

然后提取您喜欢的密钥：

>>> [x[0] for x in c.most_common()]
['a', 'b', 'c']
>>> zip(*c.most_common())[0]
('a', 'b', 'c')

Answer 4

这有效：

LoL= [['a','b','c'], ['a'], ['a','b']]     

d={}
for e in [i for sub in LoL for i in sub]:
    d[e]=d.setdefault(e,0)+1

print sorted(d, key=d.get, reverse=True)  

印刷：

['a', 'b', 'c']

Answer 5

您可以使用collections.Counter为每个子列表中的每个元素保留一个计数器。像这样的东西，

>>> from collections import Counter
>>> lst = [['a','b','c'], ['a'], ['a','b']]
>>> counts = Counter()
>>> for sublst in lst:
...     for ele in sublst:
...         counts[ele] += 1
... 
>>> [ele for ele, _ in counts.most_common()]
['a', 'b', 'c']

Answer 6

from collections import Counter
import operator
mlist= [['a','b','c'], ['a'], ['a','b']]
elist= [x for x,y in sorted(
    Counter([l for s in mlist for l in s]).iteritems(),
    key=operator.itemgetter(1),reverse=True)]
print elist