cookbook 1.12

大马强

找出序列中出现次数最多的元素

问题描述：
        我们有有一个序列，想找出里面出现次数最多的元素

一、from collections import Counter

1. from collections import Counter
   
2. 
   
3. words = ["if", "you", "and", "me", "if", "I",
   
4.          "love", "cat", "dog", "if", "love", "me", "you", "if", "me", "dog"]
   
5. 
   
6. word_counts = Counter(words)
   
7. temp = Counter(words)
   
8. # 找出最多的三个
   
9. top_three = word_counts.most_common(4)  # 参数
   
10. print(word_counts)
    
11. print(top_three)

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Counter({'if': 4, 'me': 3, 'dog': 2, 'you': 2, 'love': 2, 'and': 1, 'I': 1, 'cat': 1})
[('if', 4), ('me', 3), ('dog', 2), ('you', 2)]

字典的顺序和其在原列表顺序一致

二、字典中数据更新

1. new_word = ["if", "you", "and", "I", "love"]
   
2. # 数据的更新
   
3. # 迭代法
   
4. for word in new_word:
   
5.     word_counts[word] += 1
   
6. print(word_counts)
   
7. # 内置方法
   
8. word_counts.update(new_word)
   
9. print(word_counts)
   

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Counter({'if': 5, 'you': 3, 'me': 3, 'love': 3, 'and': 2, 'I': 2, 'dog': 2, 'cat': 1})
Counter({'if': 6, 'you': 4, 'love': 4, 'and': 3, 'me': 3, 'I': 3, 'dog': 2, 'cat': 1})

三、两个Counter对象间的运算

1. temp["mq"] = 1
   
2. print(temp)
   
3. a = temp + word_counts
   
4. print(a)
   
5. b = temp - word_counts
   
6. print(b)
   
7. c = word_counts - temp
   
8. print(c)

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集合加减法
temp ({'if': 4, 'me': 3, 'you': 2, 'love': 2, 'dog': 2, 'and': 1, 'I': 1, 'cat': 1,'mq': 1})
Counter({'if': 10, 'you': 6, 'me': 6, 'love': 6, 'and': 4, 'I': 4, 'dog': 4, 'cat': 2})
Counter({'mq': 1})
Counter({'if': 2, 'you': 2, 'and': 2, 'I': 2, 'love': 2})