郑重声明，一个实用的答案:

>>> L = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> import functools
>>> >>> functools.reduce(lambda acc, e: [v+(i==e) for i, v in enumerate(acc,1)] if e<=len(acc) else acc+[0 for _ in range(e-len(acc)-1)]+[1], L, [])
[4, 4, 2, 1, 2]

如果你把0也算进去，那就更简洁了:

>>> functools.reduce(lambda acc, e: [v+(i==e) for i, v in enumerate(acc)] if e<len(acc) else acc+[0 for _ in range(e-len(acc))]+[1], L, [])
[0, 4, 4, 2, 1, 2]

一个解释:

我们从一个空的acc列表开始; 如果L的下一个元素e小于acc的大小，我们只需更新这个元素:如果acc的索引i是当前元素e，则v+(i==e)表示v+1，否则为之前的值v; 如果L的下一个元素e大于或等于acc的大小，我们必须展开acc以容纳新的1。

元素不必排序(itertools.groupby)。如果是负数，结果会很奇怪。

如果您不想使用任何库并保持简单和简短，可以尝试这种方法!

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
marked = []
b = [(a.count(i), marked.append(i))[0] for i in a if i not in marked]
print(b)

o/p

[4, 4, 2, 1, 2]

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

# 1. Get counts and store in another list
output = []
for i in set(a):
    output.append(a.count(i))
print(output)

# 2. Remove duplicates using set constructor
a = list(set(a))
print(a)

Set集合不允许重复，将一个列表传递给Set()构造函数将给出一个完全唯一对象的可迭代对象。Count()函数在传递列表中的对象时返回一个整数计数。这样，对唯一对象进行计数，每个计数值通过附加到空列表输出来存储 List()构造函数用于将set(a)转换为List，并由相同的变量a引用

输出

D:\MLrec\venv\Scripts\python.exe D:/MLrec/listgroup.py
[4, 4, 2, 1, 2]
[1, 2, 3, 4, 5]

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}

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'])

num=[3,2,3,5,5,3,7,6,4,6,7,2]
print ('\nelements are:\t',num)
count_dict={}
for elements in num:
    count_dict[elements]=num.count(elements)
print ('\nfrequency:\t',count_dict)