Python cookbook(数据结构与算法)找出序列中出现次数最多的元素算法示例

# Determine the most common words in a list
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'
]
from collections import Counter
word_counts = Counter(words)
top_three = word_counts.most_common(3)
print(top_three)
# outputs [('eyes', 8), ('the', 5), ('look', 4)]
# Example of merging in more words
morewords = ['why','are','you','not','looking','in','my','eyes']
word_counts.update(morewords) #使用update()增加计数
print(word_counts.most_common(3))

>>> ================================ RESTART ================================
>>>
[('eyes', 8), ('the', 5), ('look', 4)]
[('eyes', 9), ('the', 5), ('my', 4)]
>>>

>>> word_counts
Counter({'eyes': 9, 'the': 5, 'my': 4, 'look': 4, 'into': 3, 'around': 2, 'not': 2, "don't": 1, 'under': 1, 'are': 1, 'looking': 1, "you're": 1, 'you': 1, 'why': 1, 'in': 1})
>>> word_counts.most_common(3) #top_three
[('eyes', 9), ('the', 5), ('my', 4)]
>>> word_counts['not']
2
>>> word_counts['eyes']
9
>>> word_counts['eyes']+1
10
>>> word_counts
Counter({'eyes': 9, 'the': 5, 'my': 4, 'look': 4, 'into': 3, 'around': 2, 'not': 2, "don't": 1, 'under': 1, 'are': 1, 'looking': 1, "you're": 1, 'you': 1, 'why': 1, 'in': 1})
>>> word_counts['eyes']=word_counts['eyes']+1 #手动增加元素计数
>>> word_counts
Counter({'eyes': 10, 'the': 5, 'my': 4, 'look': 4, 'into': 3, 'around': 2, 'not': 2, "don't": 1, 'under': 1, 'are': 1, 'looking': 1, "you're": 1, 'you': 1, 'why': 1, 'in': 1})
>>>

>>> a=Counter(words)
>>> a
Counter({'eyes': 8, 'the': 5, 'look': 4, 'my': 3, 'into': 3, 'around': 2, 'under': 1, "you're": 1, 'not': 1, "don't": 1})
>>> b=Counter(morewords)
>>> b
Counter({'not': 1, 'my': 1, 'in': 1, 'you': 1, 'looking': 1, 'are': 1, 'eyes': 1, 'why': 1})
>>> c=a+b
>>> c
Counter({'eyes': 9, 'the': 5, 'my': 4, 'look': 4, 'into': 3, 'around': 2, 'not': 2, "don't": 1, 'under': 1, 'are': 1, 'looking': 1, "you're": 1, 'you': 1, 'in': 1, 'why': 1})
>>> # substract counts
>>> d=a-b
>>> d
Counter({'eyes': 7, 'the': 5, 'look': 4, 'into': 3, 'my': 2, 'around': 2, 'under': 1, "you're": 1, "don't": 1})
>>>