Python collections queue

发布时间 2023-06-16 18:35:12作者: evescn

collections

一、计数器(counter)

  • Counter是对字典(无序)类型的补充,用于追踪值的出现次数。
  • 使用counter需要导入 collections 类

具备字典的所有功能 + 自己的功能

1、创建一个计数器

>>> import collections
>>> obj = collections.Counter('aaabbccsdfsdfdfsdfsdf')

2、查看计数器变量

>>> print(obj)
Counter({'d': 5, 'f': 5, 's': 4, 'a': 3, 'c': 2, 'b': 2})

3、查看计数器可使用的方法

>>> dir(obj)
['clear', 'copy', 'elements', 'fromkeys', 'get', 'items', 'keys', 'most_common', 'pop', 'popitem', 'setdefault', 'subtract', 'update', 'values']
collections.Counter 函数方法
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
    15

    >>> c['a']                          # count of letter 'a'
    5
    >>> 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)]

    '''
    # 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().__init__()
        self.update(iterable, **kwds)

    def __missing__(self, key):
        '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):
        '''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('abracadabra').most_common(3)
        [('a', 5), ('b', 2), ('r', 2)]

        '''
        # 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))

    def elements(self):
        '''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
        1836

        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.items()))

    # Override dict methods where necessary

    @classmethod
    def fromkeys(cls, iterable, v=None):
        # There is no equivalent method for counters because the semantics
        # would be ambiguous in cases such as Counter.fromkeys('aaabbc', v=2).
        # Initializing counters to zero values isn't necessary because zero
        # is already the default value for counter lookups.  Initializing
        # to one is easily accomplished with Counter(set(iterable)).  For
        # more exotic cases, create a dictionary first using a dictionary
        # comprehension or dict.fromkeys().
        raise NotImplementedError(
            'Counter.fromkeys() is undefined.  Use Counter(iterable) instead.')

    def update(self, iterable=None, /, **kwds):
        '''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
        4

        '''
        # 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, _collections_abc.Mapping):
                if self:
                    self_get = self.get
                    for elem, count in iterable.items():
                        self[elem] = count + self_get(elem, 0)
                else:
                    # fast path when counter is empty
                    super().update(iterable)
            else:
                _count_elements(self, iterable)
        if kwds:
            self.update(kwds)

    def subtract(self, iterable=None, /, **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.

        >>> 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
        0
        >>> 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, _collections_abc.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):
        'Return a shallow copy.'
        return self.__class__(self)

    def __reduce__(self):
        return self.__class__, (dict(self),)

    def __delitem__(self, elem):
        'Like dict.__delitem__() but does not raise KeyError for missing values.'
        if elem in self:
            super().__delitem__(elem)

    def __repr__(self):
        if not self:
            return f'{self.__class__.__name__}()'
        try:
            # dict() preserves the ordering returned by most_common()
            d = dict(self.most_common())
        except TypeError:
            # handle case where values are not orderable
            d = dict(self)
        return f'{self.__class__.__name__}({d!r})'

    # 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()
    #
    # Rich comparison operators for multiset subset and superset tests
    # are deliberately omitted due to semantic conflicts with the
    # existing inherited dict equality method.  Subset and superset
    # semantics ignore zero counts and require that p≤q ∧ p≥q → p=q;
    # however, that would not be the case for p=Counter(a=1, b=0)
    # and q=Counter(a=1) where the dictionaries are not equal.

    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()

4、常用的计数器操作

  • most_common 输出出现次数最多的字符
# most_common 输出出现次数最多的字符

>>> import collections
>>> obj = collections.Counter('aaabbccsdfsdfdfsdfsdf')
>>> print(obj.most_common(2))
[('d', 5), ('f', 5)]
  • items 计数器以k/v方式输出统计结果
# items 计数器以k/v方式输出统计结果

import collections

obj = collections.Counter('aaabbccsdfsdfdfsdfsdf')

for k,v in obj.items():
    print(k,v)

输出结果:
f 5
s 4
d 5
a 3
c 2
b 2
  • elements 把所有的元素拿到,并进行统计
# elements 把所有的元素拿到,并进行统计

>>> import collections
>>> obj = collections.Counter('aaabbccsdfsdfdfsdfsdf')
>>> print(obj.elements)
<bound method Counter.elements of Counter({'d': 5, 'f': 5, 's': 4, 'a': 3, 'c': 2, 'b': 2})>
  • update 更新计数器中的字符串或字符
# update 更新计数器中的字符串或字符

import collections

obj = collections.Counter(['11', '22', '33', '22'])
print(obj)
obj.update(['eric', '11', '11'])
print(obj)

输出结果:
Counter({'22': 2, '33': 1, '11': 1})
Counter({'11': 3, '22': 2, '33': 1, 'eric': 1})
  • subtract 删除计数器中的字符串
# subtract 删除计数器中的字符串

import collections

obj = collections.Counter(['11', '22', '33', '22'])
print(obj)
obj.update(['eric', '11', '11'])
print(obj)

obj.subtract(['eric', '11', '11', 'root'])
print(obj)

输出结果:
Counter({'22': 2, '11': 1, '33': 1})
Counter({'11': 3, '22': 2, '33': 1, 'eric': 1})
Counter({'22': 2, '11': 1, '33': 1, 'eric': 0, 'root': -1})

二、有序字典(orderedDict)

orderdDict是对字典类型的补充,他记住了字典元素添加的顺序

1、创建一个有序字典

>>> import collections

>>> dic = collections.OrderedDict()
>>> dic['k1'] = 'v1'
>>> dic['k2'] = 'v2'
>>> dic['k3'] = 'v3'

2、查看有序字典

>>> print(dic)
OrderedDict([('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3')])

3、对比有序字典和字典

import collections

dic1 = collections.OrderedDict()  # 有序字典
dic1['k1'] = 'v1'
dic1['k2'] = 'v2'
dic1['k3'] = 'v3'

dic2 = dict()     # 字典
dic2['k1'] = 'v1'
dic2['k2'] = 'v2'
dic2['k3'] = 'v3'

print(dic)

输出结果
OrderedDict([('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3')])   # 有序的
{'k2': 'v2', 'k1': 'v1', 'k3': 'v3'}    #无序的

4、查看有序字典的方法

>>> dir(dic)
['clear', 'copy', 'fromkeys', 'get', 'items', 'keys', 'move_to_end', 'pop', 'popitem', 'setdefault', 'update', 'values']
collections.Counter 函数方法
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__(self, other=(), /, **kwds):
        '''Initialize an ordered dictionary.  The signature is the same as
        regular dictionaries.  Keyword argument order is preserved.
        '''
        try:
            self.__root
        except AttributeError:
            self.__hardroot = _Link()
            self.__root = root = _proxy(self.__hardroot)
            root.prev = root.next = root
            self.__map = {}
        self.__update(other, **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):
        '''Remove and return a (key, value) pair from the dictionary.

        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 is false).

        Raise KeyError if the element does not exist.
        '''
        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 = _collections_abc.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__ = _collections_abc.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):
        '''Insert key with a value of default if key is not in the dictionary.

        Return the value for key if key is in the dictionary, else default.
        '''
        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 or None, None, iter(self.items())

    def copy(self):
        'od.copy() -> a shallow copy of od'
        return self.__class__(self)

    @classmethod
    def fromkeys(cls, iterable, value=None):
        '''Create a new ordered dictionary with keys from iterable and values set to value.
        '''
        self = cls()
        for key in iterable:
            self[key] = value
        return self

    def __eq__(self, other):
        '''od.__eq__(y) <==> od==y.  Comparison to another OD is order-sensitive
        while comparison to a regular mapping is order-insensitive.

        '''
        if isinstance(other, OrderedDict):
            return dict.__eq__(self, other) and all(map(_eq, self, other))
        return dict.__eq__(self, other)

    def __ior__(self, other):
        self.update(other)
        return self

    def __or__(self, other):
        if not isinstance(other, dict):
            return NotImplemented
        new = self.__class__(self)
        new.update(other)
        return new

    def __ror__(self, other):
        if not isinstance(other, dict):
            return NotImplemented
        new = self.__class__(other)
        new.update(self)
        return new

5、有序字典的常用操作方法

  • setdefault 设置新键的值,默认为None
# setdefault 设置新键的值,默认为None

import collections

dic = collections.OrderedDict()
dic['k1'] = 'v1'
dic['k2'] = 'v2'
dic['k3'] = 'v3'

dic['k4'] = None
dic.setdefault('k5')
dic.setdefault('k6', 'v6')

print(dic)
输出结果:
OrderedDict([('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3'), ('k4', None), ('k5', None), ('k6', 'v6')])
  • move_to_end 移动某个key到最后
# move_to_end 移动某个key到最后

import collections

dic = collections.OrderedDict()
dic['k1'] = 'v1'
dic['k2'] = 'v2'
dic['k3'] = 'v3'

print(dic)

dic.move_to_end('k1')

print(dic)

输出结果:
OrderedDict([('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3')])
OrderedDict([('k2', 'v2'), ('k3', 'v3'), ('k1', 'v1')])
  • pop 删除某个key
# pop 删除某个key

import collections

dic = collections.OrderedDict()
dic['k1'] = 'v1'
dic['k2'] = 'v2'
dic['k3'] = 'v3'

print(dic)

dic.pop('k2')
ret = dic.pop('k2')

print(dic)
print(ret)

输出结果:
OrderedDict([('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3')])
OrderedDict([('k1', 'v1'), ('k3', 'v3')])
v2
  • update 新增某个key或刷新某个key值
# update 新增某个key或刷新某个key值

import collections

dic = collections.OrderedDict()
dic['k1'] = 'v1'
dic['k2'] = 'v2'
dic['k3'] = 'v3'

print(dic)

dic.update({'k1':'v111','k10':'v10'})

print(dic)

输出结果:
OrderedDict([('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3')])
OrderedDict([('k1', 'v111'), ('k2', 'v2'), ('k3', 'v3'), ('k10', 'v10')])

三、默认字典(defaultdict)

defaultdict是对字典的类型的补充,它默认给字典的值设置了一个类型。

1、创建默认字典

import collections

dic = collections.defaultdict(list)    # 设置dic的值类型为list
dic['k1'].append('evescn')

2、查看默认字典

print(dic)

输出结果:
defaultdict(<class 'list'>, {'k1': ['evescn']})

3、查看默认字典的方法

>>> dir(dic)
['clear', 'copy', 'default_factory', 'fromkeys', 'get', 'items', 'keys', 'pop', 'popitem', 'setdefault', 'update', 'values']
collections.Counter 函数方法
class defaultdict(dict[_KT, _VT], Generic[_KT, _VT]):
    default_factory: Callable[[], _VT] | None
    @overload
    def __init__(self) -> None: ...
    @overload
    def __init__(self: defaultdict[str, _VT], **kwargs: _VT) -> None: ...
    @overload
    def __init__(self, __default_factory: Callable[[], _VT] | None) -> None: ...
    @overload
    def __init__(self: defaultdict[str, _VT], __default_factory: Callable[[], _VT] | None, **kwargs: _VT) -> None: ...
    @overload
    def __init__(self, __default_factory: Callable[[], _VT] | None, __map: SupportsKeysAndGetItem[_KT, _VT]) -> None: ...
    @overload
    def __init__(
        self: defaultdict[str, _VT],
        __default_factory: Callable[[], _VT] | None,
        __map: SupportsKeysAndGetItem[str, _VT],
        **kwargs: _VT,
    ) -> None: ...
    @overload
    def __init__(self, __default_factory: Callable[[], _VT] | None, __iterable: Iterable[tuple[_KT, _VT]]) -> None: ...
    @overload
    def __init__(
        self: defaultdict[str, _VT],
        __default_factory: Callable[[], _VT] | None,
        __iterable: Iterable[tuple[str, _VT]],
        **kwargs: _VT,
    ) -> None: ...
    def __missing__(self, __key: _KT) -> _VT: ...
    def __copy__(self) -> Self: ...
    def copy(self) -> Self: ...

4、练习

有如下值集合 [11,22,33,44,55,66,77,88,99,90...],
将所有大于 66 的值保存至字典的第一个key中,
将小于 66 的值保存至第二个key的值中。
即: {'k1': 大于66 , 'k2': 小于66}
  • 原生字典
# 原生字典
values = [11, 22, 33, 44, 55, 66, 77, 88, 99, 90]

my_dict = {}

for value in  values:
    if value > 66:
        if my_dict.has_key('k1'):
            my_dict['k1'].append(value)
        else:
            my_dict['k1'] = [value]
    else:
        if my_dict.has_key('k2'):
            my_dict['k2'].append(value)
        else:
            my_dict['k2'] = [value]
  • 默认字典
# 默认字典
from collections import defaultdict

values = [11, 22, 33, 44, 55, 66, 77, 88, 99, 90]

my_dict = defaultdict(list)

for value in  values:
    if value > 66:
        my_dict['k1'].append(value)
    else:
        my_dict['k2'].append(value)

defaultdict字典解决方法

四、可命名元组(namedtuple)

根据nametuple可以创建一个包含tuple所有功能以及其他功能的类

1、创建一个坐标类

import collections
# 创建类, defaultdict,坐标中会使用
MytupleClass = collections.namedtuple('MytupleClass',['x', 'y', 'z'])

obj = MytupleClass(11, 22, 33)

2、查询类中的x,y,z坐标

# 默认情况下元组只能使用索引进行访问,通过创建坐标类后,可以使用x,y,z进行访问
print(obj.x)
print(obj.y)
print(obj.z)

输出结果:
11
22
33

3、可命名元组的方法

# 获取MytupleClass的方法
print(help(MytupleClass))

五、双向队列(deque)

一个线程安全的双向队列

1、创建一个双向队列

import collections

d = collections.deque()

d.append('1')
d.appendleft('10')
d.appendleft('a')
d.appendleft('1')

2、查看双向队列

print(d)

输出结果:
deque(['1', 'a', '10', '1'])

3、查看双向队列的方法

>>> dir(d)
['append', 'appendleft', 'clear', 'count', 'extend', 'extendleft', 'maxlen', 'pop', 'popleft', 'remove', 'reverse', 'rotate']

4、双向队列常用的方法

  • append 右添加 appendleft 左添加
# append 右添加  appendleft 左添加 
import collections

d = collections.deque()
d.append('1')
d.appendleft('a')

print(d)

输出结果:
deque(['a', '1'])
  • count 统计队列中某元素出现的次数
# count 统计队列中某元素出现的次数
import collections

d = collections.deque()

d.append('1')
d.appendleft('a')
d.appendleft('1')

print(d)

r = d.count('1')

print(r)

输出结果:
deque(['1', 'a', '1'])
2
  • extend 右添加多个值 extendleft 左添加多个值
# extend 右添加多个值 extendleft 左添加多个值

import collections

d = collections.deque()

d.append('1')
d.appendleft('a')
d.appendleft('1')
print(d)

d.extend(['root', 'gm'])
print(d)

d.extendleft(['root', 'gm'])
print(d)

输出结果:
deque(['1', 'a', '1'])
deque(['1', 'a', '1', 'root', 'gm'])
deque(['gm', 'root', '1', 'a', '1', 'root', 'gm'])
  • rotate 把队列中最后的#个数放到最前面
# rotate 把队列中最后的#个数放到最前面
import collections

d = collections.deque()

d.append('1')
d.appendleft('10')
d.appendleft('a')
d.appendleft('b')

print(d)

d.rotate(2)

print(d)

输出结果:
deque(['b', 'a', '10', '1'])
deque(['10', '1', 'b', 'a'])

其他还有pop、remove、reverse<翻转>之类的方法

queue

六、单向队列(deque)

单向队列(先进先出 FIFO )

1、创建单向队列

import queue

q = queue.Queue()

q.put('123')
q.put('evescn')

2、查看单向队列

# 单向队列是先进先出,要查看单向队列,需使用get获取单向队列的值
print(q.get())
print(q.get())

输出结果:
123
evescn

3、单向队列和双向队列的区别

# 双向队列:
调用:import collections
collections.deque()

队列左边右边都可插入/获取数据

# 单向队列:
调用:import queue 
queue.Queue

只能右端插入数据,左端获取数据

4、查看单向队列的方法

>>> dir(q)
['all_tasks_done', 'empty', 'full', 'get', 'get_nowait', 'join', 'maxsize', 'mutex', 'not_empty', 'not_full', 'put', 'put_nowait', 'qsize', 'queue', 'task_done', 'unfinished_tasks']

5、单向队列常用的方法

  • put | get
# put 向队列中放入一个数 
# get 取出一个数
import queue

q = queue.Queue()

q.put('123')
q.put('evescn')

print(q.get())
print(q.get())

输出结果:     # 先进先出
123
evescn
  • qsize 统计当前队列长度
# qsize 统计当前队列长度
import queue

q = queue.Queue()

q.put('123')
q.put('evescn')

print(q.qsize()

# 输出:2
  • empty 队列是否为空
# empty 队列是否为空
import queue

q = queue.Queue()

print(q.empty()

# 输出:True
  • full 队列是否满了,返回True
# full 队列是否满了,返回True
import queue

q = queue.Queue(1)

print(q.empty())
print(q.full())

q.put('123')

print(q.full())

# 输出:
True
False
True