[PY3]——找出一个序列中出现次数最多的元素/collections.Counter 类的用法

问题

怎样找出一个序列中出现次数最多的元素呢？

解决方案

collections.Counter 类就是专门为这类问题而设计的， 它甚至有一个有用的 most_common() 方法直接给了你答案

collections.Counter 类

1. most_common(n)统计top_n

from collections import Counter
words = [
    ‘look‘, ‘into‘, ‘my‘, ‘eyes‘, ‘look‘, ‘into‘, ‘my‘, ‘eyes‘,
    ‘the‘, ‘eyes‘, ‘the‘, ‘eyes‘, ‘the‘, ‘eyes‘, ‘not‘, ‘around‘, ‘the‘,
    ‘eyes‘, "don‘t", ‘look‘, ‘around‘, ‘the‘, ‘eyes‘, ‘look‘, ‘into‘,
    ‘my‘, ‘eyes‘, "you‘re", ‘under‘
]
#创建Counter对象
word_counts=Counter(words)

#most_common(n)函数可直接统计top-n
top_three=word_counts.most_common(3)

print(type(top_three))
<class ‘list‘>

print(top_three)
[(‘eyes‘, 8), (‘the‘, 5), (‘look‘, 4)]

2. 统计任意元素出现的次数

Counter 对象可以接受任意的由可哈希（hashable）元素构成的序列对象

在底层实现上，一个 Counter 对象就是一个字典，将元素映射到它出现的次数上

print(word_counts[‘look‘])
4
print(word_counts["you‘re"])
1

3. 两个Counter对象之间可以互相使用数学运算符，从而叠加/叠减统计

word1 = [
    ‘look‘, ‘into‘, ‘my‘, ‘eyes‘, ‘look‘, ‘into‘, ‘my‘, ‘eyes‘,
    ‘the‘, ‘eyes‘, ‘the‘, ‘eyes‘, ‘the‘, ‘eyes‘, ‘not‘, ‘around‘, ‘the‘,
    ‘eyes‘, "don‘t", ‘look‘, ‘around‘, ‘the‘, ‘eyes‘, ‘look‘, ‘into‘,
    ‘my‘, ‘eyes‘, "you‘re", ‘under‘
]
word2 = [‘why‘,‘are‘,‘you‘,‘not‘,‘looking‘,‘in‘,‘my‘,‘eyes‘]

a=Counter(word1)
b=Counter(word2)
print(a)
Counter({‘eyes‘: 8, ‘the‘: 5, ‘look‘: 4, ‘into‘: 3, ‘my‘: 3, ‘around‘: 2, ‘not‘: 1, ‘under‘: 1, "you‘re": 1, "don‘t": 1})

print(b)
Counter({‘eyes‘: 1, ‘not‘: 1, ‘are‘: 1, ‘you‘: 1, ‘in‘: 1, ‘why‘: 1, ‘my‘: 1, ‘looking‘: 1})

print(a+b)
Counter({‘eyes‘: 9, ‘the‘: 5, ‘my‘: 4, ‘look‘: 4, ‘into‘: 3, ‘around‘: 2, ‘not‘: 2, ‘under‘: 1, ‘looking‘: 1, ‘you‘: 1, ‘why‘: 1, ‘in‘: 1, ‘are‘: 1, "you‘re": 1, "don‘t": 1})

print(a-b)
Counter({‘eyes‘: 7, ‘the‘: 5, ‘look‘: 4, ‘into‘: 3, ‘around‘: 2, ‘my‘: 2, ‘under‘: 1, "you‘re": 1, "don‘t": 1})