import collections
a = {1: 2, 2: 3}
b = {1: 3, 3: 4, 4: 5}
chains = collections.ChainMap(a, b)
# maps
# 注意maps是个属性，不是一个方法，其改变
print(chains.maps)  # [{1: 2, 2: 3}, {1: 3, 3: 4, 4: 5}]
# get
assert chains.get(1, -1) == 2
# parents
# 从第二个map开始找
assert chains.parents.get(1, -1) == 3
# popitem
assert chains.popitem() == (2, 3)
# pop
# 返回的是value
assert chains.pop(1) == 2
# new_child
assert chains.new_child()
print(chains.maps)  # [{}, {1: 3, 3: 4, 4: 5}]
chains[2] = 1
print(chains.maps)  # [{2: 1}, {1: 3, 3: 4, 4: 5}]
# setdedault
# 如果已经存在key，则不会添加
assert chains.setdefault(1, 10) == 3
# update
chains.update({2: 4, 3: 5})
print(chains.maps)  # [{1: 2, 2: 4, 3: 5}, {1: 3, 3: 4, 4: 5}]
# keys
print(chains.keys())  # KeysView(ChainMap({2: 4, 3: 5}, {1: 3, 3: 4, 4: 5}))
# KeysView 继承了mapping和set
print(2 in chains.keys())  # True
print(len(chains.keys()))  # 4（重复的不算）
# clear
chains.clear()
print(chains.maps)  # [{}, {1: 3, 3: 4, 4: 5}]

from collections import Counter
# init
# 可迭代
counter = Counter("accab")  # Counter({‘a‘: 2, ‘c‘: 2, ‘b‘: 1})
counter2 = Counter([1,2,3,4])  # Counter({1: 1, 2: 1, 3: 1, 4: 1})
counter5 = Counter([(‘a‘,3),(‘b‘, 2)])  # Counter({(‘a‘, 3): 1, (‘b‘, 2): 1})
# 字典
counter3 = Counter({‘a‘: 1, ‘b‘: 2, ‘a‘: 3})  # Counter({‘a‘: 3, ‘b‘: 2})
counter4 = Counter(a=1, b=2, c=1)  # Counter({‘b‘: 2, ‘a‘: 1, ‘c‘: 1})
# elements
# 键值以无序的方式返回，并且只返回值大于等于1的键值对
elememts = counter.elements()
print([x for x in elememts])  # [‘a‘, ‘a‘, ‘c‘, ‘c‘, ‘b‘]
# 为空是因为elements是generator
print(sorted(elememts))  # []
# most_common
# 键值以无序的方式返回
print(counter.most_common(1))  # [(‘a‘, 2)]
print(counter.most_common())  # [(‘a‘, 2), (‘c‘, 2), (‘b‘, 1)]
# update
# 单纯是增加的功能，而不是像dict.update()中的替换一样
counter.update("abb")
print(counter)  # Counter({‘a‘: 3, ‘b‘: 3, ‘c‘: 2})
# subtract
counter.subtract(Counter("accc"))
print(counter)  # Counter({‘b‘: 3, ‘a‘: 2, ‘c‘: -1})
print([x for x in counter.elements()])  # [‘a‘, ‘a‘, ‘b‘, ‘b‘, ‘b‘]
# get
# 键不存在则返回0，但是不会加入到counter键值对中
print(counter[‘d‘])
print(counter)  # Counter({‘b‘: 3, ‘a‘: 2, ‘c‘: -1})
del counter[‘d‘]
# 还可以使用数学运算
c = Counter(a=3, b=1)
d = Counter(a=1, b=2)
# add two counters together:  c[x] + d[x]
print(c + d)  # Counter({‘a‘: 4, ‘b‘: 3})
# subtract (keeping only positive counts)
print(c - d)  # Counter({‘a‘: 2})
# # intersection:  min(c[x], d[x])
print(c & d) # Counter({‘a‘: 1, ‘b‘: 1})
# union:  max(c[x], d[x])
print(c | d)  # Counter({‘a‘: 3, ‘b‘: 2})
# 一元加法和减法
c = Counter(a=3, b=-1)
# 只取正数
print(+c)  # Counter({‘a‘: 3})
print(-c)  # Counter({‘b‘: 1})

from collections import deque
# 从尾部进入，从头部弹出，保证长度为5
dq1 = deque(‘abcdefg‘, maxlen=5)
print(dq1)  # [‘c‘, ‘d‘, ‘e‘, ‘f‘, ‘g‘]
print(dq1.maxlen)  # 5
# 从左端入列
dq1.appendleft(‘q‘)
print(dq1)  # [‘q‘, ‘c‘, ‘d‘, ‘e‘, ‘f‘]
# 从左端入列
dq1.extendleft(‘abc‘)
print(dq1)  # [‘c‘, ‘b‘, ‘a‘, ‘q‘, ‘c‘]
# 从左端出列并且返回
dq1.popleft()  # c
print(dq1)  # [‘b‘, ‘a‘, ‘q‘, ‘c‘]
# 将队头n个元素进行右旋
dq1.rotate(2)
print(dq1)  # [‘q‘, ‘c‘, ‘b‘, ‘a‘]
# 将队尾两个元素进行左旋
dq1.rotate(-2)
print(dq1)  # [‘b‘, ‘a‘, ‘q‘, ‘c‘]
def tail(filename, n=10):
    ‘Return the last n lines of a file‘
    with open(filename) as f:
        return deque(f, n)
def delete_nth(d, n):
    """
    实现队列切片和删除，pop之后再放会原处
    :param d: deque
    :param n: int
    :return:
    """
    d.roatte(-n)
    d.popleft()
    d.rotate(n)

from collections import OrderedDict
items = {‘c‘: 3, ‘b‘: 2, ‘a‘: 1}
regular_dict = dict(items)
ordered_dict = OrderedDict(items)
print(regular_dict)  # {‘c‘: 3, ‘b‘: 2, ‘a‘: 1}
print(ordered_dict)  # [(‘c‘, 3), (‘b‘, 2), (‘a‘, 1)]
# 按照插入顺序进行排序而不是
ordered_dict[‘f‘] = 4
ordered_dict[‘e‘] = 5
print(ordered_dict)  # [(‘c‘, 3), (‘b‘, 2), (‘a‘, 1), (‘f‘, 4), (‘e‘, 5)]
# 把最近加入的删除
print(ordered_dict.popitem(last=True))  # (‘e‘, 5)
# 按照加入的顺序删除
print(ordered_dict.popitem(last=False))  # (‘c‘, 3)
print(ordered_dict)  # [(‘b‘, 2), (‘a‘, 1), (‘f‘, 4)]
# 移动到末尾
ordered_dict.move_to_end(‘b‘, last=True)
print(ordered_dict)  # [(‘a‘, 1), (‘f‘, 4), (‘b‘, 2)]
# 移动到开头
ordered_dict.move_to_end(‘b‘, last=False)
print(ordered_dict)  # [(‘b‘, 2), (‘a‘, 1), (‘f‘, 4)]
ordered_dict[‘a‘] = 3
# 说明更改值并不会影响加入顺序
print(ordered_dict.popitem(last=True))  # (‘f‘, 4)

from collections import namedtuple
Point = namedtuple(‘Point‘, [‘x‘, ‘y‘])
p = Point(10, y=20)
print(p)  # Point(x=10, y=20)
print(p.x + p.y)  # 30