百木园问题描述
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。
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\']
word_counts = Counter(words)
top_three = word_counts.most_common(3) # [(\'eyes\', 8), (\'the\', 5), (\'look\', 4)]
讨论
Counter对象在几乎所有需要制表或者计数数据的场合是非常有用的工具。在解决这类问题时应该有限选择它，而不是手动的利用字典去实现。
Counter对象可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就是一个字典，将元素映射到它出现的次数上。
word_counts[\'not\'] # 1
word_counts[\'eyes\'] # 8
如果想手动增加计数，可以简单的使用加法：
morewords = [\'why\', \'are\', \'you\', \'not\', \'looking\', \'in\', \'my\', \'eyes\']
for word in morewords:
word_counts[word] += 1
\"\"\"
word_counts[\'eyes\'] = 9
\"\"\"
或者用update()方法：
word_counts.update(morewords)
Counter实例还可以跟数学运算操作相结合，比如：
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\']
morewords = [\'why\', \'are\', \'you\', \'not\', \'looking\', \'in\', \'my\', \'eyes\']
a = Counter(words)
b = Counter(morewords)
c = a + b # c = Counter({\'eyes\': 9, \'the\': 5, \'look\': 4, \'my\': 4, \'into\': 3, \'not\': 2, \'around\': 2, \"don\'t\": 1, \"you\'re\": 1, \'under\': 1, \'why\': 1, \'are\': 1, \'you\': 1, \'looking\': 1, \'in\': 1})
d = a - b # d = Counter({\'eyes\': 7, \'the\': 5, \'look\': 4, \'into\': 3, \'my\': 2, \'around\': 2, \"don\'t\": 1, \"you\'re\": 1, \'under\': 1})
来源：https://www.cnblogs.com/L999C/p/15732550.html
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