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

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]

计算元素的频率可能最好使用字典:

b = {}
for item in a:
    b[item] = b.get(item, 0) + 1

要删除重复项，请使用set:

a = list(set(a))

另一种方法是使用较重但功能强大的库——NLTK。

import nltk

fdist = nltk.FreqDist(a)
fdist.values()
fdist.most_common()

还有另一种不使用集合的算法解决方案:

def countFreq(A):
   n=len(A)
   count=[0]*n                     # Create a new list initialized with '0'
   for i in range(n):
      count[A[i]]+= 1              # increase occurrence for value A[i]
   return [x for x in count if x]  # return non-zero count

from collections import Counter
a=["E","D","C","G","B","A","B","F","D","D","C","A","G","A","C","B","F","C","B"]

counter=Counter(a)

kk=[list(counter.keys()),list(counter.values())]

pd.DataFrame(np.array(kk).T, columns=['Letter','Count'])