好久没学习了,简单了解下
Ⅰ、namedtuple
1.1 简单回顾一下tuple
tuple是一个不可变的可迭代对象
①可迭代
In [1]: test_tuple = (1,2,3,4,5)
In [2]: for i in test_tuple:
...: print(i)
...:
1
2
3
4
5
②不可变
In [3]: test_tuple[0] = 5
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-3-8e5f1b05e43e> in <module>()
----> 1 test_tuple[0] = 5
TypeError: 'tuple' object does not support item assignment
不可变指的是其中的元素不可变
In [4]: test_tuple = (6,7,8)
其元素不可变也不是绝对的
In [5]: test_tuple = (1,[2,3])
In [6]: test_tuple[1].append(4)
In [7]: print(test_tuple)
(1, [2, 3, 4])
③拆包
In [8]: user_tuple = ("allen", 29, 175)
In [9]: name, age, height = user_tuple
In [10]: print(name, age, height)
allen 29 175
In [11]: name, *other = user_tuple
In [12]: print(name, other)
allen [29, 175]1.2 tuple对比list的优势
- 性能更好
- 线程安全
- 拆包特性
- 可以作为dict的key(可hash),如下：
In [14]: user_info_dict = {}
In [15]: user_info_dict[user_tuple] = "allen"
In [16]: user_info_dict
Out[16]: {('allen', 29, 175): 'allen'}
In [17]: user_info_dict[[1,2,3]] = "Kobe"
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-17-a32e731657ac> in <module>()
----> 1 user_info_dict[[1,2,3]] = "Kobe"
TypeError: unhashable type: 'list'对应C语言来说,tuple对应的是struct,而list对应的是array
1.3 nametuple创建一个User类
In [23]: from collections import namedtuple
In [24]: User = namedtuple("User", ["name", "age", "height", "edu"])
In [25]: user = User(name="Allen", age=29, height=175, edu="master")
In [26]: print(user.age, user.name, user.height, user.edu)
29 Allen 175 master
# 加一个字段
In [29]: user_tuple = ("Allen", 29, 175)
In [30]: user = User(*user_tuple, "master")
# user = User("Allen", 29, 175, "master")
In [31]: user
Out[31]: User(name='Allen', age=29, height=175, edu='master')
In [33]: user_dict = {"name":"allen","age":19,"height":175}
In [34]: user = User(**user_dict, edu="master")
In [35]: user
Out[35]: User(name='allen', age=19, height=175, edu='master')
1.4 看两个方法
- _make() 传入一个可迭代对象初始化namedtuple，灵活性不如前面直接传参
In [36]: user_tuple = ("Allen", 29, 175, "edu")
In [37]: user = User._make(user_tuple)
In [38]: user_list = ["Allen", 29, 175, "edu"]
In [39]: user = User._make(user_list)
In [42]: user_dict = {"name":"allen","age":19,"height":175, "edu":"master"}
In [43]: user = User._make(user_dict)- _asdict() 将tuple转为dict
- In [46]: user = User(name="Allen", age=29, height=175, edu="master")
In [47]: user_info_dict = user._asdict()
In [48]: user_info_dict
Out[48]:
OrderedDict([('name', 'Allen'),
('age', 29),
('height', 175),
('edu', 'master')])
tips:namedtuple也可以拆包,如下
In [49]: name, age, *other =user
In [50]: print(name, age, other)
Allen 29 [175, 'master']Ⅱ、defaultdict
需求：统计一个列表中每个元素的个数
- 老式方法
In [70]: a = [1,2,3,2,3,2,3,2,4,4,4,5]
In [71]: c_dict = {}
In [72]: for i in a:
...: if i not in c_dict:
...: c_dict[i] = 1
...: else:
...: c_dict[i] += 1
...:
In [73]: c_dict
Out[73]: {1: 1, 2: 4, 3: 3, 4: 3, 5: 1}- 另一个方法：
In [63]: a = [1,2,3,2,3,2,3,2,4,4,4,5]
In [64]: c_dict = {}
In [65]: for i in a:
...: c_dict.setdefault(i,0)
...: c_dict[i] += 1
...:
In [66]: c_dict
Out[66]: {1: 1, 2: 4, 3: 3, 4: 3, 5: 1}- 最好的方法
In [77]: from collections import defaultdict
In [78]: c_dict = defaultdict(int)
In [79]: a = [1,2,3,2,3,2,3,2,4,4,4,5]
In [80]: for i in a:
...: c_dict[i] += 1 #当key不存在时不会报错,绑一个0
...:
In [81]: c_dict
Out[81]: defaultdict(int, {1: 1, 2: 4, 3: 3, 4: 3, 5: 1})弄个自定义的带嵌套的玩一下
In [82]: def gen_default():
...: return {
...: "name":"",
...: "nums":0
...: }
...:
...:
In [83]: default_dict = defaultdict(gen_default)
In [85]: default_dict["test"]
Out[85]: {'name': '', 'nums': 0}Ⅲ、deque
deque GIL是线程安全的，list不是
deque叫做双端队列,那我们就得和list一起看,先看下list
In [86]: a_list = [1,2]
In [87]: t = a_list.pop()
In [88]: print(t, a_list)
2 [1]
可以看到最后一个元素被弹出来了3.1 弄最左边
- 插左边
In [94]: from collections import deque
In [95]: a_deque = deque([1, [2, 3], 4])
In [96]: a_deque.appendleft(5)
In [97]: a_deque
Out[97]: deque([5, 1, [2, 3], 4])- 弹左边
In [97]: a_deque
Out[97]: deque([5, 1, [2, 3], 4])
In [98]: a_deque.popleft()
Out[98]: 53.2 copy()
In [99]: a = deque([1,[2, 3], 4])
In [100]: b = a.copy()
In [101]: print(a, b)
deque([1, [2, 3], 4]) deque([1, [2, 3], 4])
In [102]: print(id(a), id(b))
139908896451392 139908896452328
In [103]: b[1].append(5)
In [104]: print(a, b)
deque([1, [2, 3, 5], 4]) deque([1, [2, 3, 5], 4])
由上可知,这种copy是浅拷贝- 深拷贝怎么弄？
deque本身没提供,需要用到copy这个包
In [106]: import copy
In [107]: a = deque([1,[2, 3], 4])
In [108]: b = copy.deepcopy(a)
In [109]: b[1].append(5)
In [110]: print(a, b)
deque([1, [2, 3], 4]) deque([1, [2, 3, 5], 4])3.3 extend()
In [111]: a = deque([1,2,3])
In [112]: b = deque([4,5,6])
In [113]: a.extend(b)
In [114]: a
Out[114]: deque([1, 2, 3, 4, 5, 6])
注意:这是直接在a上动态扩容,并不会返回一个新的列表,所以c = a.extend(b)是错的3.4 其他方法
In [114]: a
Out[114]: deque([1, 2, 3, 4, 5, 6])
In [115]: a.insert(1,888)
In [116]: a.reverse()
In [117]: print(a)
deque([6, 5, 4, 3, 2, 888, 1])
extendleft(),index(),count(),clear()等这里就不演示了Ⅳ、Counter()
In [122]: from collections import Counter
In [123]: tmp = "132531321059823"
In [124]: t_count = Counter(tmp)
In [125]: t_count
Out[125]: Counter({'1': 3, '3': 4, '2': 3, '5': 2, '0': 1, '9': 1, '8': 1})
# 找top
In [126]: t_count.most_common(2)
Out[126]: [('3', 4), ('1', 3)]两个串合起来统计
In [126]: t_count.most_common(2)
Out[126]: [('3', 4), ('1', 3)]
In [127]: t_count.update("3311")
In [128]: t_count.most_common(2)
Out[128]: [('3', 6), ('1', 5)]
再来一把
In [129]: tmp2 = "3311"
In [130]: t_count.update(tmp2)
In [131]: t_count.most_common(2)
Out[131]: [('3', 8), ('1', 7)]Ⅴ、OrderedDict
字典中的元素本来是无序的,下面这个用了OrderedDict,有序了(添加顺序),python2中和原来的Dict对比就能看出区别,3看不出来
In [141]: from collections import OrderedDict
In [142]: t_dict = OrderedDict()
In [143]: t_dict["b"] = "2b"
In [144]: t_dict["a"] = "1a"
In [145]: t_dict["c"] = "3c"
In [146]: t_dict
Out[146]: OrderedDict([('b', '2b'), ('a', '1a'), ('c', '3c')])5.1 一些方法
In [146]: t_dict
Out[146]: OrderedDict([('b', '2b'), ('a', '1a'), ('c', '3c')])
In [147]: print(t_dict.pop("a"))
1a
In [149]: print(t_dict.popitem())
('c', '3c')
pop()必须指定key,popitem()不用,默认弹出最后一组key:value把指定元素放到最后
In [155]: t_dict
Out[155]: OrderedDict([('a', '1a'), ('c', '3c'), ('b', '2b')])
In [156]: t_dict.move_to_end("a")
In [157]: t_dict
Out[157]: OrderedDict([('c', '3c'), ('b', '2b'), ('a', '1a')])Ⅵ、ChainMap
实现访问多个dict就像访问一个dict一样(在多个字典上套一个迭代器)
In [160]: from collections import ChainMap
In [161]: dt1 = {"a":"bb1","b":"bb2"}
In [162]: dt2 = {"a":"bb1","b":"bb2"}
In [163]: n_dt = ChainMap(dt1, dt2)
In [164]: n_dt
Out[164]: ChainMap({'a': 'bb1', 'b': 'bb2'}, {'a': 'bb1', 'b': 'bb2'})
# 加一个新的字典
In [165]: n_dt = n_dt.new_child({"e":"ee5"})
In [166]: n_dt
Out[166]: ChainMap({'e': 'ee5'}, {'a': 'bb1', 'b': 'bb2'}, {'a': 'bb1', 'b': 'bb2'})
# 将字典变成列表(maps指向原来的字典,并不是新生成的一个数据)
In [168]: n_dt.maps
Out[168]: [{'e': 'ee5'}, {'a': 'bb1', 'b': 'bb2'}, {'a': 'bb1', 'b': 'bb2'}]
In [171]: n_dt.maps[1]["a"] = "ahaha"
In [173]: for key, value in n_dt.items():
...: print(key, value)
...:
a ahaha
b bb2
e ee5