Python高级数据结构-Collections模块
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在Python数据类型方法精心整理，不必死记硬背，看看源码一切都有了之中，认识了python基本的数据类型和数据结构，现在认识一个高级的：Collections
这个模块对上面的数据结构做了封装，增加了一些很酷的数据结构，比如：
a）Counter： 计数器，用于统计元素的数量
b）OrderDict：有序字典
c）defaultdict：值带有默认类型的字典
d）namedtuple：可命名元组，通过名字来访问元组元素
e）deque :双向队列，队列头尾都可以放，也都可以取（与单向队列对比，单向队列只能一头放，另一头取）
1. Counter
计数器，用于统计对象中每个元素出现的个数
按照老惯例，先看源码：
class Counter(dict): \'\'\'Dict subclass for counting hashable items. Sometimes called a bag or multiset. Elements are stored as dictionary keys and their counts are stored as dictionary values. >>> c = Counter(\'abcdeabcdabcaba\') # count elements from a string >>> 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 >>> c[\'a\'] # count of letter \'a\' >>> 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\' >>> del c[\'b\'] # remove all \'b\' >>> c[\'b\'] # now there are zero \'b\' >>> d = Counter(\'simsalabim\') # make another counter >>> c.update(d) # add in the second counter >>> c[\'a\'] # now there are nine \'a\' >>> 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)] \'\'\' # References: # http://en.wikipedia.org/wiki/Multiset # http://www.gnu.org/software/smalltalk/manual-base/html_node/Bag.html # http://www.demo2s.com/Tutorial/Cpp/0380__set-multiset/Catalog0380__set-multiset.htm # http://code.activestate.com/recipes/259174/ # Knuth, TAOCP Vol. II section 4.6.3 def __init__(self, iterable=None, **kwds): \'\'\'Create a new, empty Counter object. And if given, count elements from an input iterable. Or, initialize the count from another mapping of elements to their counts. >>> c = Counter() # a new, empty counter >>> c = Counter(\'gallahad\') # a new counter from an iterable >>> c = Counter({\'a\': 4, \'b\': 2}) # a new counter from a mapping >>> c = Counter(a=4, b=2) # a new counter from keyword args \'\'\' super(Counter, self).__init__() self.update(iterable, **kwds) def __missing__(self, key): """ 对于不存在的元素，返回计数器为0 """ \'The count of elements not in the Counter is zero.\' # Needed so that self[missing_item] does not raise KeyError return 0 def most_common(self, n=None): """ 数量大于等n的所有元素和计数器 """ \'\'\'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.iteritems(), key=_itemgetter(1), reverse=True) return _heapq.nlargest(n, self.iteritems(), key=_itemgetter(1)) def elements(self): """ 计数器中的所有元素，注：此处非所有元素集合，而是包含所有元素集合的迭代器 """ \'\'\'Iterator over elements repeating each as many times as its count. >>> c = Counter(\'ABCABC\') >>> sorted(c.elements()) [\'A\', \'A\', \'B\', \'B\', \'C\', \'C\'] # Knuth\'s example for prime factors of 1836: 2**2 * 3**3 * 17**1 >>> prime_factors = Counter({2: 2, 3: 3, 17: 1}) >>> product = 1 >>> for factor in prime_factors.elements(): # loop over factors ... product *= factor # and multiply them >>> product Note, if an element\'s count has been set to zero or is a negative number, elements() will ignore it. \'\'\' # Emulate Bag.do from Smalltalk and Multiset.begin from C++. return _chain.from_iterable(_starmap(_repeat, self.iteritems())) # Override dict methods where necessary @classmethod def fromkeys(cls, iterable, v=None): # There is no equivalent method for counters because setting v=1 # means that no element can have a count greater than one. raise NotImplementedError( \'Counter.fromkeys() is undefined. Use Counter(iterable) instead.\') def update(self, iterable=None, **kwds): """ 更新计数器，其实就是增加；如果原来没有，则新建，如果有则加一 """ \'\'\'Like dict.update() but add counts instead of replacing them. Source can be an iterable, a dictionary, or another Counter instance. >>> c = Counter(\'which\') >>> c.update(\'witch\') # add elements from another iterable >>> d = Counter(\'watch\') >>> c.update(d) # add elements from another counter >>> c[\'h\'] # four \'h\' in which, witch, and watch \'\'\' # The regular dict.update() operation makes no sense here because the # replace behavior results in the some of original untouched counts # being mixed-in with all of the other counts for a mismash that # doesn\'t have a straight-forward interpretation in most counting # contexts. Instead, we implement straight-addition. Both the inputs # and outputs are allowed to contain zero and negative counts. if iterable is not None: if isinstance(iterable, Mapping): if self: self_get = self.get for elem, count in iterable.iteritems(): self[elem] = self_get(elem, 0) + count else: super(Counter, self).update(iterable) # fast path when counter is empty else: self_get = self.get for elem in iterable: self[elem] = self_get(elem, 0) + 1 if kwds: self.update(kwds) def subtract(self, iterable=None, **kwds): """ 相减，原来的计数器中的每一个元素的数量减去后添加的元素的数量 """ \'\'\'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. >>> c = Counter(\'which\') >>> c.subtract(\'witch\') # subtract elements from another iterable >>> c.subtract(Counter(\'watch\')) # subtract elements from another counter >>> c[\'h\'] # 2 in which, minus 1 in witch, minus 1 in watch >>> c[\'w\'] # 1 in which, minus 1 in witch, minus 1 in watch -1 \'\'\' if iterable is not None: self_get = self.get if isinstance(iterable, Mapping): for elem, count in iterable.items(): self[elem] = self_get(elem, 0) - count else: for elem in iterable: self[elem] = self_get(elem, 0) - 1 if kwds: self.subtract(kwds) def copy(self): """ 拷贝 """ \'Return a shallow copy.\' return self.__class__(self) def __reduce__(self): """ 返回一个元组（类型，元组） """ return self.__class__, (dict(self),) def __delitem__(self, elem): """ 删除元素 """ \'Like dict.__delitem__() but does not raise KeyError for missing values.\' if elem in self: super(Counter, self).__delitem__(elem) def __repr__(self): if not self: return \'%s()\' % self.__class__.__name__ items = \', \'.join(map(\'%r: %r\'.__mod__, self.most_common())) return \'%s({%s})\' % (self.__class__.__name__, items) # Multiset-style mathematical operations discussed in: # Knuth TAOCP Volume II section 4.6.3 exercise 19 # and at http://en.wikipedia.org/wiki/Multiset # # Outputs guaranteed to only include positive counts. # # To strip negative and zero counts, add-in an empty counter: # c += Counter() def __add__(self, other): \'\'\'Add counts from two counters. >>> Counter(\'abbb\') + Counter(\'bcc\') Counter({\'b\': 4, \'c\': 2, \'a\': 1}) \'\'\' if not isinstance(other, Counter): return NotImplemented result = Counter() for elem, count in self.items(): newcount = count + other[elem] if newcount > 0: result[elem] = newcount for elem, count in other.items(): if elem not in self and count > 0: result[elem] = count return result def __sub__(self, other): \'\'\' Subtract count, but keep only results with positive counts. >>> Counter(\'abbbc\') - Counter(\'bccd\') Counter({\'b\': 2, \'a\': 1}) \'\'\' if not isinstance(other, Counter): return NotImplemented result = Counter() for elem, count in self.items(): newcount = count - other[elem] if newcount > 0: result[elem] = newcount for elem, count in other.items(): if elem not in self and count < 0: result[elem] = 0 - count return result def __or__(self, other): \'\'\'Union is the maximum of value in either of the input counters. >>> Counter(\'abbb\') | Counter(\'bcc\') Counter({\'b\': 3, \'c\': 2, \'a\': 1}) \'\'\' if not isinstance(other, Counter): return NotImplemented result = Counter() for elem, count in self.items(): other_count = other[elem] newcount = other_count if count < other_count else count if newcount > 0: result[elem] = newcount for elem, count in other.items(): if elem not in self and count > 0: result[elem] = count return result def __and__(self, other): \'\'\' Intersection is the minimum of corresponding counts. >>> Counter(\'abbb\') & Counter(\'bcc\') Counter({\'b\': 1}) \'\'\' if not isinstance(other, Counter): return NotImplemented result = Counter() for elem, count in self.items(): other_count = other[elem] newcount = count if count < other_count else other_count if newcount > 0: result[elem] = newcount return result def __pos__(self): \'Adds an empty counter, effectively stripping negative and zero counts\' result = Counter() for elem, count in self.items(): if count > 0: result[elem] = count return result def __neg__(self): \'\'\'Subtracts from an empty counter. Strips positive and zero counts, and flips the sign on negative counts. \'\'\' result = Counter() for elem, count in self.items(): if count < 0: result[elem] = 0 - count return result def _keep_positive(self): \'\'\'Internal method to strip elements with a negative or zero count\'\'\' nonpositive = [elem for elem, count in self.items() if not count > 0] for elem in nonpositive: del self[elem] return self def __iadd__(self, other): \'\'\'Inplace add from another counter, keeping only positive counts. >>> c = Counter(\'abbb\') >>> c += Counter(\'bcc\') >>> c Counter({\'b\': 4, \'c\': 2, \'a\': 1}) \'\'\' for elem, count in other.items(): self[elem] += count return self._keep_positive() def __isub__(self, other): \'\'\'Inplace subtract counter, but keep only results with positive counts. >>> c = Counter(\'abbbc\') >>> c -= Counter(\'bccd\') >>> c Counter({\'b\': 2, \'a\': 1}) \'\'\' for elem, count in other.items(): self[elem] -= count return self._keep_positive() def __ior__(self, other): \'\'\'Inplace union is the maximum of value from either counter. >>> c = Counter(\'abbb\') >>> c |= Counter(\'bcc\') >>> c Counter({\'b\': 3, \'c\': 2, \'a\': 1}) \'\'\' for elem, other_count in other.items(): count = self[elem] if other_count > count: self[elem] = other_count return self._keep_positive() def __iand__(self, other): \'\'\'Inplace intersection is the minimum of corresponding counts. >>> c = Counter(\'abbb\') >>> c &= Counter(\'bcc\') >>> c Counter({\'b\': 1}) \'\'\' for elem, count in self.items(): other_count = other[elem] if other_count < count: self[elem] = other_count return self._keep_positive()
实际上，Counter是dict的一个子类，实例：
#通过字典形式统计每个元素重复的次数传 res = collections.Counter(\'abcdabcaba\') print(res) #结果Counter({\'a\': 4, \'b\': 3, \'c\': 2, \'d\': 1}) #dict的子类，所以也可以以字典的形式取得键值对 for k in res: print(k, res[k], end=\' | \') #结果 a 4 | b 3 | c 2 | d 1 | for k, v in res.items(): print(k, v, end=\' | \') #结果 a 4 | b 3 | c 2 | d 1 | #通过most_common(n)，返回前n个重复次数最多的键值对 print(res.most_common()) #结果None print(res.most_common(2)) #结果[(\'a\', 4), (\'b\', 3)] #通过update来增加元素的重复次数，通过subtract来减少元素重复的次数 a = collections.Counter(\'abcde\') res.update(a) print(res) #结果Counter({\'a\': 5, \'b\': 4, \'c\': 3, \'d\': 2, \'e\': 1})，比原来的res增加了重复次数 b = collections.Counter(\'aaafff\') res.subtract(b) print(res) #结果Counter({\'b\': 4, \'c\': 3, \'a\': 2, \'d\': 2, \'e\': 1, \'f\': -3})，还有负值，要注意 #fromkeys功能还没实现，使用的话会报错
2. OrderDict
有序字典，数据结构字典Dict是无序的，有时使用起来不是很方便，Collections里提供一个有序字典OrderDict，用起来就很方便了
在介绍有序字典以前，用已知的知识其实可以自己实现一个有序字典，通过列表或者元祖来维护key，实现有序字典：
lst =[] dic = {} lst.append(\'name\') dic[\'name\'] = \'winter\' lst.append(\'age\') dic[\'age\'] = 18 for k in lst: print(k, dic[k])
实际上，OrderDict就是通过这种方式实现的
源代码：
class OrderedDict(dict): \'Dictionary that remembers insertion order\' # An inherited dict maps keys to values. # The inherited dict provides __getitem__, __len__, __contains__, and get. # The remaining methods are order-aware. # Big-O running times for all methods are the same as regular dictionaries. # The internal self.__map dict maps keys to links in a doubly linked list. # The circular doubly linked list starts and ends with a sentinel element. # The sentinel element never gets deleted (this simplifies the algorithm). # The sentinel is in self.__hardroot with a weakref proxy in self.__root. # The prev links are weakref proxies (to prevent circular references). # Individual links are kept alive by the hard reference in self.__map. # Those hard references disappear when a key is deleted from an OrderedDict. def __init__(*args, **kwds): \'\'\'Initialize an ordered dictionary. The signature is the same as regular dictionaries, but keyword arguments are not recommended because their insertion order is arbitrary. \'\'\' if not args: raise TypeError("descriptor \'__init__\' of \'OrderedDict\' object " "needs an argument") self, *args = args if len(args) > 1: raise TypeError(\'expected at most 1 arguments, got %d\' % len(args)) try: self.__root except AttributeError: self.__hardroot = _Link() self.__root = root = _proxy(self.__hardroot) root.prev = root.next = root self.__map = {} self.__update(*args, **kwds) def __setitem__(self, key, value, dict_setitem=dict.__setitem__, proxy=_proxy, Link=_Link): \'od.__setitem__(i, y) <==> od[i]=y\' # Setting a new item creates a new link at the end of the linked list, # and the inherited dictionary is updated with the new key/value pair. if key not in self: self.__map[key] = link = Link() root = self.__root last = root.prev link.prev, link.next, link.key = last, root, key last.next = link root.prev = proxy(link) dict_setitem(self, key, value) def __delitem__(self, key, dict_delitem=dict.__delitem__): \'od.__delitem__(y) <==> del od[y]\' # Deleting an existing item uses self.__map to find the link which gets # removed by updating the links in the predecessor and successor nodes. dict_delitem(self, key) link = self.__map.pop(key) link_prev = link.prev link_next = link.next link_prev.next = link_next link_next.prev = link_prev link.prev = None link.next = None def __iter__(self): \'od.__iter__() <==> iter(od)\' # Traverse the linked list in order. root = self.__root curr = root.next while curr is not root: yield curr.key curr = curr.next def __reversed__(self): \'od.__reversed__() <==> reversed(od)\' # Traverse the linked list in reverse order. root = self.__root curr = root.prev while curr is not root: yield curr.key curr = curr.prev def clear(self): \'od.clear() -> None. Remove all items from od.\' root = self.__root root.prev = root.next = root self.__map.clear() dict.clear(self) def popitem(self, last=True): \'\'\'od.popitem() -> (k, v), return and remove a (key, value) pair. Pairs are returned in LIFO order if last is true or FIFO order if false. \'\'\' if not self: raise KeyError(\'dictionary is empty\') root = self.__root if last: link = root.prev link_prev = link.prev link_prev.next = root root.prev = link_prev else: link = root.next link_next = link.next root.next = link_next link_next.prev = root key = link.key del self.__map[key] value = dict.pop(self, key) return key, value def move_to_end(self, key, last=True): \'\'\'Move an existing element to the end (or beginning if last==False). Raises KeyError if the element does not exist. When last=True, acts like a fast version of self[key]=self.pop(key). \'\'\' link = self.__map[key] link_prev = link.prev link_next = link.next soft_link = link_next.prev link_prev.next = link_next link_next.prev = link_prev root = self.__root if last: last = root.prev link.prev = last link.next = root root.prev = soft_link last.next = link else: first = root.next link.prev = root link.next = first first.prev = soft_link root.next = link def __sizeof__(self): sizeof = _sys.getsizeof n = len(self) + 1 # number of links including root size = sizeof(self.__dict__) # instance dictionary size += sizeof(self.__map) * 2 # internal dict and inherited dict size += sizeof(self.__hardroot) * n # link objects size += sizeof(self.__root) * n # proxy objects return size update = __update = MutableMapping.update def keys(self): "D.keys() -> a set-like object providing a view on D\'s keys" return _OrderedDictKeysView(self) def items(self): "D.items() -> a set-like object providing a view on D\'s items" return _OrderedDictItemsView(self) def values(self): "D.values() -> an object providing a view on D\'s values" return _OrderedDictValuesView(self) __ne__ = MutableMapping.__ne__ __marker = object() def pop(self, key, default=__marker): \'\'\'od.pop(k[,d]) -> v, remove specified key and return the corresponding value. If key is not found, d is returned if given, otherwise KeyError is raised. \'\'\' if key in self: result = self[key] del self[key] return result if default is self.__marker: raise KeyError(key) return default def setdefault(self, key, default=None): \'od.setdefault(k[,d]) -> od.get(k,d), also set od[k]=d if k not in od\' if key in self: return self[key] self[key] = default return default @_recursive_repr() def __repr__(self): \'od.__repr__() <==> repr(od)\' if not self: return \'%s()\' % (self.__class__.__name__,) return \'%s(%r)\' % (self.__class__.__name__, list(self.items())) def __reduce__(self): \'Return state information for pickling\' inst_dict = vars(self).copy() for k in vars(OrderedDict()): inst_dict.pop(k, None) return self.__class__, (), inst_dict Python collections模块总结阶段1 语言基础+高级_1-3-Java语言高级_04-集合_07 Collections工具类_1_Collections集合工具类的方法