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]

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

str1='the cat sat on the hat hat'
list1=str1.split();
list2=str1.split();

count=0;
m=[];

for i in range(len(list1)):
    t=list1.pop(0);
    print t
    for j in range(len(list2)):
        if(t==list2[j]):
            count=count+1;
            print count
    m.append(count)
    print m
    count=0;
#print m

我将简单地以以下方式使用scipy.stats.itemfreq:

from scipy.stats import itemfreq

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

freq = itemfreq(a)

a = freq[:,0]
b = freq[:,1]

您可以在这里查看文档:http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.stats.itemfreq.html

简单的解决方法就是用字典。

def frequency(l):
     d = {}
     for i in l:
        if i in d.keys():
           d[i] += 1
        else:
           d[i] = 1

     for k, v in d.iteritems():
        if v ==max (d.values()):
           return k,d.keys()

print(frequency([10,10,10,10,20,20,20,20,40,40,50,50,30]))

还有一种方法是使用字典和列表。数数，下面一种幼稚的做法。

dicio = dict()

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

b = list()

c = list()

for i in a:

   if i in dicio: continue 

   else:

      dicio[i] = a.count(i)

      b.append(a.count(i))

      c.append(i)

print (b)

print (c)