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

我迟到了，但这也有用，也会帮助到其他人:

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
freq_list = []
a_l = list(set(a))

for x in a_l:
    freq_list.append(a.count(x))


print 'Freq',freq_list
print 'number',a_l

会产生这个…

Freq  [4, 4, 2, 1, 2]
number[1, 2, 3, 4, 5]

def frequencyDistribution(data):
    return {i: data.count(i) for i in data}   

print frequencyDistribution([1,2,3,4])

...

 {1: 1, 2: 1, 3: 1, 4: 1}   # originalNumber: count

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}

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

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)