python的collection系列-counter

一、計數器（counter）

Counter是對字典類型的補充，用於追蹤值的出現次數。

具備字典的所有功能 + 自己的功能。

import collections
aa = collections.Counter("sdfdsgsdf;sdfssfd")    #把所有元素出現的次數統計下來了
print(aa)

輸出結果：
Counter({'s': 6, 'd': 5, 'f': 4, ';': 1, 'g': 1})

部分源碼分析：

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))

#獲取元素出現次數多的幾個
bb = aa.most_common(3)     #取元素次數最多的前3個
print(bb)

#執行結果：
[('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

#其他輸出方式
for item in aa.elements():   #迭代輸出元素（同一元素出現幾次，就連續輸出該元素幾次），無序
    print(item)

for k,v in aa.items():     #把所有元素拿出來，再統計該元素出現次數
    print(k,v)

#執行結果：
;
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的區別：elements處理的是原生的值；而counter處理的是已經處理完的數據

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.

#更新計數器，
import collections
aa = collections.Counter(["11","22","33","22"])    #把所有元素出現的次數統計下來了
print(aa)
aa.update(["ggg","11","11"])
print(aa)

#執行結果:
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.

#減少元素
import collections
aa = collections.Counter(["11","22","33","22"])    #把所有元素出現的次數統計下來了
print(aa)
aa.subtract(["ggg","11","11"])   #如果不存在，也減
print(aa)

#執行結果：
Counter({'22': 2, '11': 1, '33': 1})
Counter({'22': 2, '33': 1, 'ggg': -1, '11': -1})