雖然工作中常用Python，但都是些基本操作，對於這種高階的工具包，一直是只知道有那么個東西，沒調用過，每次都是自己造輪子。
人生苦短， 我用Python，為毛還重復造輪子，裝什么C呢。
看下collections的init
__all__ = ['deque', 'defaultdict', 'namedtuple', 'UserDict', 'UserList',
'UserString', 'Counter', 'OrderedDict', 'ChainMap']
挑個今天眼熱的模塊，開始鼓搗鼓搗。
一、Counter
猜名字，是跟計數有關的玩意兒
看源碼中類的介紹
1 class Counter(dict): 2 '''Dict subclass for counting hashable items. Sometimes called a bag 3 or multiset. Elements are stored as dictionary keys and their counts 4 are stored as dictionary values.'''
大概就是，字典的子類，為哈希元素提供計數功能，新生成的字典，元素為key，計數為values,按原來的key順序進行的排序。
import collections ret=collections.Counter("abbbbbccdeeeeeeeeeeeee") ret Out[12]: Counter({'a': 1, 'b': 5, 'c': 2, 'd': 1, 'e': 13})
看源碼中給的案例：
>>> c = Counter('abcdeabcdabcaba') # count elements from a string # 返回最多的3個key和value >>> c.most_common(3) # three most common elements [('a', 5), ('b', 4), ('c', 3)]
>>> sorted(c) # list all unique elements ['a', 'b', 'c', 'd', 'e'] >>> ''.join(sorted(c.elements())) # list elements with repetitions 'aaaaabbbbcccdde' >>> sum(c.values()) # total of all counts 15 >>> c['a'] # count of letter 'a' 5
#對元素的更新 >>> for elem in 'shazam': # update counts from an iterable ... c[elem] += 1 # by adding 1 to each element's count >>> c['a'] # now there are seven 'a' 7 >>> del c['b'] # remove all 'b' >>> c['b'] # now there are zero 'b' 0 >>> d = Counter('simsalabim') # make another counter >>> c.update(d) # add in the second counter >>> c['a'] # now there are nine 'a' 9 >>> c.clear() # empty the counter >>> c Counter() Note: If a count is set to zero or reduced to zero, it will remain in the counter until the entry is deleted or the counter is cleared: >>> c = Counter('aaabbc') >>> c['b'] -= 2 # reduce the count of 'b' by two >>> c.most_common() # 'b' is still in, but its count is zero [('a', 3), ('c', 1), ('b', 0)]
常用API:
1、most_common(num),返回計數最多的num個元素,如果不傳參數，則返回所以元素
def most_common(self, n=None): '''List the n most common elements and their counts from the most common to the least. If n is None, then list all element counts. >>> Counter('abcdeabcdabcaba').most_common(3) [('a', 5), ('b', 4), ('c', 3)] ''' # Emulate Bag.sortedByCount from Smalltalk if n is None: return sorted(self.items(), key=_itemgetter(1), reverse=True) return _heapq.nlargest(n, self.items(), key=_itemgetter(1))
2、elements
返回一個迭代器，迭代對象是所有的元素，只不過給你按原始數據的順排了一下序，一樣的給你放一起了
>>> c = Counter('ABCABC')
>>> c.elements() ==》<itertools.chain at 0x5f10828>
>>> list(c.elements())
>>> ['A', 'A', 'B', 'B', 'C', 'C'] #這里不是排序，是你恰好參數順序是ABC，官方給的這例子容易誤導人
d=Counter('ACBCDEFT')
list(d.elements())
['A','C','B','D','E','F','T']
所以，要想帶順序，自己在調用sorted一下
sorted(d.elements()) => ['A','B','C','D','E','F','T']
3、subtract，感覺在leetcode刷題時，會比較實用
def subtract(*args, **kwds): '''Like dict.update() but subtracts counts instead of replacing them. Counts can be reduced below zero. Both the inputs and outputs are allowed to contain zero and negative counts. Source can be an iterable, a dictionary, or another Counter instance.
啥意思，就是更新你的Counter對象，怎么更新，基於你傳入的參數，它給你做減法，參數是可迭代對象，字典，或者另一個Counter
看官方的例子
c=Counter("which")
out:>> Counter({'w': 1, 'h': 2, 'i': 1, 'c': 1}) c.subtract('witch') #傳入一個迭代對象，對迭代對象的每一個元素，對原對象進行減法，注意，t是原對象沒有的 out:>> Counter({'w': 0, 'h': 1, 'i': 0, 'c': 0, 't': -1}) c.subtract(Counter('watch')) #傳入另一個Counter對象 out:>> Counter({'w': -1, 'h': 0, 'i': 0, 'c': -1, 't': -2, 'a': -1})
c.subtract({'h':3,'q':5}) #傳入一個字典，value是個數 也就是減去多少個key
out:>> Counter({'w': -1, 'h': -3, 'i': 0, 'c': -1, 't': -2, 'a': -1, 'q': -5})
其他好像沒啥好玩的了，以為幾分鍾就搞定了Collections的所有模塊，想多了，先寫個Counter，后邊的幾個慢慢補。