python的collection系列-counter

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。

import collections
aa = collections.Counter("sdfdsgsdf;sdfssfd")    #把所有元素出现的次数统计下来了
print(aa)

输出结果：
Counter({'s': 6, 'd': 5, 'f': 4, ';': 1, 'g': 1})

部分源码分析：

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 
 >>> Counter('abcdeabcdabcaba').most_common(3)  6  [('a', 5), ('b', 4), ('c', 3)]  7 
        '''
        # Emulate Bag.sortedByCount from Smalltalk
        if n is None: 11             return sorted(self.items(), key=_itemgetter(1), reverse=True) 12         return _heapq.nlargest(n, self.items(), key=_itemgetter(1))

#获取元素出现次数多的几个
bb = aa.most_common(3)     #取元素次数最多的前3个
print(bb)

#执行结果：
[('s', 6), ('d', 5), ('f', 4)]

def elements(self):  2         '''Iterator over elements repeating each as many times as its count.  3 
 >>> c = Counter('ABCABC')  5  >>> sorted(c.elements())  6  ['A', 'A', 'B', 'B', 'C', 'C']  7 
 # 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 
 Note, if an element's count has been set to zero or is a negative 17  number, elements() will ignore it. 18 
        '''
        # Emulate Bag.do from Smalltalk and Multiset.begin from C++.
        return _chain.from_iterable(_starmap(_repeat, self.items())) 22 
    # Override dict methods where necessary

#其他输出方式
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
    print(item)

for k,v in aa.items():     #把所有元素拿出来，再统计该元素出现次数
    print(k,v)

#执行结果：
;
s
s
s
s
s
s
f
f
f
f
d
d
d
d
d
g
; 1
s 6
f 4
d 5
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

def update(*args, **kwds): 2         '''Like dict.update() but add counts instead of replacing them. 3 
 Source can be an iterable, a dictionary, or another Counter instance.

#更新计数器，
import collections
aa = collections.Counter(["11","22","33","22"])    #把所有元素出现的次数统计下来了
print(aa)
aa.update(["ggg","11","11"])
print(aa)

#执行结果:
Counter({'22': 2, '33': 1, '11': 1})
Counter({'11': 3, '22': 2, 'ggg': 1, '33': 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 
 Source can be an iterable, a dictionary, or another Counter instance.

#减少元素
import collections
aa = collections.Counter(["11","22","33","22"])    #把所有元素出现的次数统计下来了
print(aa)
aa.subtract(["ggg","11","11"])   #如果不存在，也减
print(aa)

#执行结果：
Counter({'22': 2, '11': 1, '33': 1})
Counter({'22': 2, '33': 1, 'ggg': -1, '11': -1})