我使用计数器在1行代码中从文本文件单词生成一个frequency . dict

def _fileIndex(fh):
''' create a dict using Counter of a
flat list of words (re.findall(re.compile(r"[a-zA-Z]+"), lines)) in (lines in file->for lines in fh)
'''
return Counter(
    [wrd.lower() for wrdList in
     [words for words in
      [re.findall(re.compile(r'[a-zA-Z]+'), lines) for lines in fh]]
     for wrd in wrdList])

from collections import OrderedDict
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
def get_count(lists):
    dictionary = OrderedDict()
    for val in lists:
        dictionary.setdefault(val,[]).append(1)
    return [sum(val) for val in dictionary.values()]
print(get_count(a))
>>>[4, 4, 2, 1, 2]

删除副本并维持秩序:

list(dict.fromkeys(get_count(a)))
>>>[4, 2, 1]

seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.

对于一个无序列表，你应该使用:

[a.count(el) for el in set(a)]

输出为

[4, 4, 2, 1, 2]

通过遍历列表并计算它们，手动计算出现的数量，使用collections.defaultdict跟踪到目前为止看到的内容:

from collections import defaultdict

appearances = defaultdict(int)

for curr in a:
    appearances[curr] += 1

我使用计数器在1行代码中从文本文件单词生成一个frequency . dict

def _fileIndex(fh):
''' create a dict using Counter of a
flat list of words (re.findall(re.compile(r"[a-zA-Z]+"), lines)) in (lines in file->for lines in fh)
'''
return Counter(
    [wrd.lower() for wrdList in
     [words for words in
      [re.findall(re.compile(r'[a-zA-Z]+'), lines) for lines in fh]]
     for wrd in wrdList])