给定一个无序的值列表，比如
a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
我怎样才能得到出现在列表中的每个值的频率，就像这样?
# `a` has 4 instances of `1`, 4 of `2`, 2 of `3`, 1 of `4,` 2 of `5`
b = [4, 4, 2, 1, 2] # expected output
当前回答
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
其他回答
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))
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
Python 2.7+引入了字典理解。从列表中构建字典将获得计数并去除重复项。
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> d = {x:a.count(x) for x in a}
>>> d
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
>>> a, b = d.keys(), d.values()
>>> a
[1, 2, 3, 4, 5]
>>> b
[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}
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'])
