一、计数器（counter）
Counter是对字典类型的补充，用于追踪值的出现次数。
具备字典的所有功能 + 自己的功能。
1 import collections 2 aa = collections.Counter("sdfdsgsdf;sdfssfd") #把所有元素出现的次数统计下来了 3 print(aa) 4 5 输出结果： 6 Counter({'s': 6, 'd': 5, 'f': 4, ';': 1, 'g': 1})
部分源码分析：
1 def most_common(self, n=None):
2 '''List the n most common elements and their counts from the most
3 common to the least. If n is None, then list all element counts.
4
5 >>> Counter('abcdeabcdabcaba').most_common(3)
6 [('a', 5), ('b', 4), ('c', 3)]
7
8 '''
9 # Emulate Bag.sortedByCount from Smalltalk
10 if n is None:
11 return sorted(self.items(), key=_itemgetter(1), reverse=True)
12 return _heapq.nlargest(n, self.items(), key=_itemgetter(1))
1 #获取元素出现次数多的几个 2 bb = aa.most_common(3) #取元素次数最多的前3个 3 print(bb) 4 5 #执行结果： 6 [('s', 6), ('d', 5), ('f', 4)]
1 def elements(self):
2 '''Iterator over elements repeating each as many times as its count.
3
4 >>> c = Counter('ABCABC')
5 >>> sorted(c.elements())
6 ['A', 'A', 'B', 'B', 'C', 'C']
7
8 # Knuth's example for prime factors of 1836: 2**2 * 3**3 * 17**1
9 >>> prime_factors = Counter({2: 2, 3: 3, 17: 1})
10 >>> product = 1
11 >>> for factor in prime_factors.elements(): # loop over factors
12 ... product *= factor # and multiply them
13 >>> product
14 1836
15
16 Note, if an element's count has been set to zero or is a negative
17 number, elements() will ignore it.
18
19 '''
20 # Emulate Bag.do from Smalltalk and Multiset.begin from C++.
21 return _chain.from_iterable(_starmap(_repeat, self.items()))
22
23 # Override dict methods where necessary
1 #其他输出方式 2 for item in aa.elements(): #迭代输出元素（同一元素出现几次，就连续输出该元素几次），无序 3 print(item) 4 5 for k,v in aa.items(): #把所有元素拿出来，再统计该元素出现次数 6 print(k,v) 7 8 #执行结果： 9 ; 10 s 11 s 12 s 13 s 14 s 15 s 16 f 17 f 18 f 19 f 20 d 21 d 22 d 23 d 24 d 25 g 26 ; 1 27 s 6 28 f 4 29 d 5 30 g 1
counter和elements的区别：elements处理的是原生的值；而counter处理的是已经处理完的数据
1 def update(*args, **kwds):
2 '''Like dict.update() but add counts instead of replacing them.
3
4 Source can be an iterable, a dictionary, or another Counter instance.
1 #更新计数器， 2 import collections 3 aa = collections.Counter(["11","22","33","22"]) #把所有元素出现的次数统计下来了 4 print(aa) 5 aa.update(["ggg","11","11"]) 6 print(aa) 7 8 #执行结果: 9 Counter({'22': 2, '33': 1, '11': 1}) 10 Counter({'11': 3, '22': 2, 'ggg': 1, '33': 1})
1 def subtract(*args, **kwds):
2 '''Like dict.update() but subtracts counts instead of replacing them.
3 Counts can be reduced below zero. Both the inputs and outputs are
4 allowed to contain zero and negative counts.
5
6 Source can be an iterable, a dictionary, or another Counter instance.
1 #减少元素 2 import collections 3 aa = collections.Counter(["11","22","33","22"]) #把所有元素出现的次数统计下来了 4 print(aa) 5 aa.subtract(["ggg","11","11"]) #如果不存在，也减 6 print(aa) 7 8 #执行结果： 9 Counter({'22': 2, '11': 1, '33': 1}) 10 Counter({'22': 2, '33': 1, 'ggg': -1, '11': -1})