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()

您可以使用python中提供的内置函数

l.count(l[i])


  d=[]
  for i in range(len(l)):
        if l[i] not in d:
             d.append(l[i])
             print(l.count(l[i])

上面的代码自动删除列表中的重复项，并打印原始列表和没有重复项的列表中每个元素的频率。

一枪双鸟!X维

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)

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'])

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))

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}