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]

这个答案更加明确

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

d = {}
for item in a:
    if item in d:
        d[item] = d.get(item)+1
    else:
        d[item] = 1

for k,v in d.items():
    print(str(k)+':'+str(v))

# output
#1:4
#2:4
#3:3
#4:2

#remove dups
d = set(a)
print(d)
#{1, 2, 3, 4}

对于您的第一个问题，迭代列表并使用字典跟踪元素的存在。

对于你的第二个问题，只需使用集合操作符。

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'])

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]

假设我们有一个列表:

fruits = ['banana', 'banana', 'apple', 'banana']

我们可以在列表中找出每种水果的数量，像这样:

import numpy as np    
(unique, counts) = np.unique(fruits, return_counts=True)
{x:y for x,y in zip(unique, counts)}

结果:

{'banana': 3, 'apple': 1}