collections 共涉及到以下几个模块:
[‘deque’, ‘defaultdict’, ‘namedtuple’, ‘UserDict’, ‘UserList’,
‘UserString’, ‘Counter’, ‘OrderedDict’, ‘ChainMap’]
namedtuple详解
tuple
tuple 拆包特性
$ ipython
Python 3.7.3 (default, Mar 28 2019, 10:38:38) [MSC v.1915 32 bit (Intel)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: user_tuple = ('laoliu', 30, 175, 'beijing')
# python的拆包特性
In [2]: name, *other = user_tuple
In [3]: name
Out[3]: 'laoliu'
In [4]: other
Out[4]: [30, 175, 'beijing']
# 拆包
In [5]: name, age, hight, address = user_tuple
In [6]: print(name, age, hight, address)
laoliu 30 175 beijing
In [7]: d = {}
# 元组可以作为dict的键, 但list不可以,这是因为tuple是不可变对象
In [8]: d[user_tuple] = 'shanghai'
In [9]: d
Out[9]: {('laoliu', 30, 175, 'beijing'): 'shanghai'}
namedtuple的使用方法示例
Python 3.7.3 (default, Mar 28 2019, 10:38:38) [MSC v.1915 32 bit (Intel)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: from collections import namedtuple
In [2]: User = namedtuple("User",['name', 'age'])
In [3]: user = User(name='liu',age=18)
In [4]: print(user.name, user.age)
liu 18
In [5]: user_tuple = ('zhang', 19)
# 以 * 传参, 也是位置参数, 详解可参照函数的传参方式
In [6]: user2 = User(*user_tuple)
In [7]: print(user.name, user.age)
liu 18
In [8]: print(user2.name, user2.age)
zhang 19
In [9]: user_dict = {'name': 'wang', 'age':20}
# 以 ** 传参, 也就是关键字参数, 详解可参照函数的传参方式
In [10]: user3 = User(**user_dict)
In [11]: print(user3.name, user3.age)
wang 20
# 新增属性之后的解决办法
In [12]: User = namedtuple('User',['name', 'age', 'height'])
In [13]: user = User(name='liu',age=18)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
in
----> 1 user = User(name='liu',age=18)
TypeError: __new__() missing 1 required positional argument: 'height'
# 第一种解决办法就是添加关键字
In [14]: user = User(name='liu',age=18, height=170)
In [15]: print(user.name, user.age, user.height)
liu 18 170
In [16]: print(user2.name, user2.age, user2.height)
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
in
----> 1 print(user2.name, user2.age, user2.height)
AttributeError: 'User' object has no attribute 'height'
In [17]: user2
Out[17]: User(name='zhang', age=19)
# 这个就是函数传参中的位置参数与关键字参数混用的例子
In [18]: user2 = User(*user_tuple, height=171)
In [19]: user2
Out[19]: User(name='zhang', age=19, height=171)
In [20]: user3
Out[20]: User(name='wang', age=20)
# 错误传参示例
In [21]: use3 = User(**user_dict, 'height':172)
File "", line 1
use3 = User(**user_dict, 'height':172)
^
SyntaxError: invalid syntax
# 错误传参示例
In [22]: use3 = User(**user_dict, {'height': 172})
File "", line 1
use3 = User(**user_dict, {'height': 172})
^
SyntaxError: positional argument follows keyword argument unpacking
In [23]: user_dict
Out[23]: {'name': 'wang', 'age': 20}
In [24]: user_dict.update({'height':172})
In [25]: user_dict
Out[25]: {'name': 'wang', 'age': 20, 'height': 172}
In [26]: user3 = User(**user_dict)
In [27]: user3
Out[27]: User(name='wang', age=20, height=172)
defaultdict 使用详解
Python 3.7.3 (default, Mar 28 2019, 10:38:38) [MSC v.1915 32 bit (Intel)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: from collections import defaultdict
In [2]: users = ['liu1', 'liu2', 'liu3', 'liu1', 'liu1', 'liu2']
In [3]: # 统计users中每个名字出现的次数
In [4]: user_dict = {}
# 这是我们常用的方法来解决
In [5]: for user in users:
...: if user in user_dict:
...: user_dict[user] += 1
...: else:
...: user_dict[user] = 1
...:
In [6]: user_dict # 结果
Out[6]: {'liu1': 3, 'liu2': 2, 'liu3': 1}
# 第二种解决办法, 使用setdefault()
In [7]: user_dict2 = {}
In [8]: for user in users:
...: user_dict2.setdefault(user, 0)
...: user_dict2[user] += 1
...:
In [9]: user_dict2
Out[9]: {'liu1': 3, 'liu2': 2, 'liu3': 1}
# 第三种解决办法 使用defaultdict() (推荐使用)
In [10]: user_dict3 = defaultdict(int)
In [11]: for user in users:
...: user_dict3[user] += 1
...:
In [12]: user_dict3
Out[12]: defaultdict(int, {'liu1': 3, 'liu2': 2, 'liu3': 1})
# defaultdict() 扩展使用, 创建一些复杂的数据结构
# 求如下数据结构:
{
'group1':{
'name': '',
'nums': 0
}
}
In [13]: def gen_default():
...: return {'name': '', 'nums': 0}
...:
In [14]: default_dict = defaultdict(gen_default)
In [15]: default_dict['group1']
Out[15]: {'name': '', 'nums': 0}
deque 使用详解
deque 是线程安全的, list是非线程安全的,在多线程编程的情况下要多注意
queue (队列)其实是deque实现在的
deque是使用C语言编写的, 速度很快, 可以经常使用
Python 3.7.3 (default, Mar 28 2019, 10:38:38) [MSC v.1915 32 bit (Intel)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: from collections import deque
In [2]: a = ['b', 'c', 'd']
# 将list转为deque
In [3]: a_duque = deque(a)
In [4]: a_duque
Out[4]: deque(['b', 'c', 'd'])
# 将tuple转为deque
In [5]: b = (1,2, 3)
In [6]: b_duque = deque(b)
In [7]: b_duque
Out[7]: deque([1, 2, 3])
In [8]: c = {"a": 1, "b": 2, "c": 4}
# 将dice转为deque
In [9]: c_deque = deque(c)
In [10]: c_deque
Out[10]: deque(['a', 'b', 'c'])
# deque的append操作,同list
In [11]: c_deque.append('d')
In [12]: c_deque
Out[12]: deque(['a', 'b', 'c', 'd'])
# deque.appendleft() 将元素添加至左侧第0个位置
In [13]: c_deque.appendleft('e')
In [14]: c_deque
Out[14]: deque(['e', 'a', 'b', 'c', 'd'])
# 浅拷贝
In [15]: c_deque_copy = c_deque.copy()
In [16]: c_deque_copy.count()
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
in
----> 1 c_deque_copy.count()
TypeError: count() takes exactly one argument (0 given)
# 查找某个元素的出现的次数
In [17]: c_deque_copy.count('a')
Out[17]: 1
In [18]: c_deque_copy
Out[18]: deque(['e', 'a', 'b', 'c', 'd'])
In [19]: c_deque_copy[1] = 'a1'
In [20]: c_deque_copy
Out[20]: deque(['e', 'a1', 'b', 'c', 'd'])
In [21]: c_deque
Out[21]: deque(['e', 'a', 'b', 'c', 'd'])
# 合并两个deque, 将后者deque合并至前者的右侧,原址修改,返回None
In [22]: c_deque.extend(a_duque)
In [23]: c_deque
Out[23]: deque(['e', 'a', 'b', 'c', 'd', 'b', 'c', 'd'])
# 合并两个deque, 将全者的deque合并至前者的左侧, 原地修改, 返回None
In [24]: c_deque.extendleft(b_duque)
In [25]: c_deque
Out[25]: deque([3, 2, 1, 'e', 'a', 'b', 'c', 'd', 'b', 'c', 'd'])
# 返回元素的索引位置
In [26]: c_deque.index('b')
Out[26]: 5
# 在给出的位置插入元素
In [27]: c_deque.insert(2,'f')
In [28]: c_deque
Out[28]: deque([3, 2, 'f', 1, 'e', 'a', 'b', 'c', 'd', 'b', 'c', 'd'])
In [29]: c_deque.rotate('a')
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
in
----> 1 c_deque.rotate('a')
TypeError: 'str' object cannot be interpreted as an integer
# 将队尾的指定数量的元素放到队前, 默认为1
In [30]: c_deque.rotate()
In [31]: c_deque
Out[31]: deque(['d', 3, 2, 'f', 1, 'e', 'a', 'b', 'c', 'd', 'b', 'c'])
# 反转整个deque
In [32]: c_deque.reverse()
In [33]: c_deque
Out[33]: deque(['c', 'b', 'd', 'c', 'b', 'a', 'e', 1, 'f', 2, 3, 'd'])
In [34]:
Counter 使用详解
Python 3.7.3 (default, Mar 28 2019, 10:38:38) [MSC v.1915 32 bit (Intel)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: from collections import Counter
In [2]: users = ['liu1', 'liu2', 'liu3', 'liu1', 'liu1', 'liu2']
# 对列表中的数据进行统计
In [3]: users_counter = Counter(users)
In [4]: users_counter
Out[4]: Counter({'liu1': 3, 'liu2': 2, 'liu3': 1})
# 统计字符中,每个字符出现的次数
In [5]: test = Counter('abcddfdefadsfasdjfoaisdfjasdjfasdfasdfasdfgfhdf')
In [6]: test
Out[6]:
Counter({'a': 8,
'b': 1,
'c': 1,
'd': 11,
'f': 11,
'e': 1,
's': 7,
'j': 3,
'o': 1,
'i': 1,
'g': 1,
'h': 1})
# 统计两个字符串中的字符出现次数, 方法1
In [7]: test.update('aadfd')
In [8]: test
Out[8]:
Counter({'a': 10,
'b': 1,
'c': 1,
'd': 13,
'f': 12,
'e': 1,
's': 7,
'j': 3,
'o': 1,
'i': 1,
'g': 1,
'h': 1})无锡人流多少钱 http://www.bhnnk120.com/
# 统计两个字符串中的字符出现次数, 方法2
In [9]: test2 = Counter('abcde')
In [10]: test.update(test2)
In [11]: test
Out[11]:
Counter({'a': 11,
'b': 2,
'c': 2,
'd': 14,
'f': 12,
'e': 2,
's': 7,
'j': 3,
'o': 1,
'i': 1,
'g': 1,
'h': 1})
# TOP n 的统计方法
In [12]: test.most_common(3)
Out[12]: [('d', 14), ('f', 12), ('a', 11)]
OrderedDict 使用详解
Python 3.7.3 (default, Mar 28 2019, 10:38:38) [MSC v.1915 32 bit (Intel)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: from collections import OrderedDict
In [2]: user_dict = OrderedDict()
In [3]: user_dict['b'] = 'liu'
In [4]: user_dict['a'] = 'liu1'
In [5]: user_dict['c'] = 'liu2'
In [6]: user_dict
Out[6]: OrderedDict([('b', 'liu'), ('a', 'liu1'), ('c', 'liu2')])
# 弹出最后一个item.
In [7]: user_dict.popitem()
Out[7]: ('c', 'liu2')
In [8]: user_dict
Out[8]: OrderedDict([('b', 'liu'), ('a', 'liu1')])
# 移动元素至最后 使用场景优先级
In [9]: user_dict.move_to_end('b')
In [10]: user_dict
Out[10]: OrderedDict([('a', 'liu1'), ('b', 'liu')])
# 弹出最后一个key,返回key对应的值
In [11]: user_dict.pop('b')
Out[11]: 'liu'
In [12]: user_dict
Out[12]: OrderedDict([('a', 'liu1')])
ChainMap 使用详解
Python 3.7.3 (default, Mar 28 2019, 10:38:38) [MSC v.1915 32 bit (Intel)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: from collections import ChainMap
In [2]: dict1 = {'a': 'liu', 'b': "liu1"}
In [3]: dict2 = {"c": "liu3", 'd': 'liu4'}
# 最常使用的方法就是连接多个dict, 若键重复, 则只取第一个dict中的键值
In [4]: new_dict = ChainMap(dict1, dict2)
In [5]: for key, value in new_dict.items():
...: print(key, value)
...:
c liu3
d liu4
a liu
b liu1
# 修改dict2, 使其中一个键与dict1的键重复
In [6]: dict2 = {"b": "liu3", 'd': 'liu4'}
In [7]: new_dict = ChainMap(dict1, dict2)
In [8]: for key, value in new_dict.items():
...: print(key, value)
...:
b liu1
d liu4
a liu
In [9]: new_dict
Out[9]: ChainMap({'a': 'liu', 'b': 'liu1'}, {'b': 'liu3', 'd': 'liu4'})
# 返回dict的list, 但只是指向原来dict, 不是原dict的copy
In [10]: new_dict.maps
Out[10]: [{'a': 'liu', 'b': 'liu1'}, {'b': 'liu3', 'd': 'liu4'}]
In [11]: new_dict.maps[0]['a'] = 'liu333'
In [12]: new_dict
Out[12]: ChainMap({'a': 'liu333', 'b': 'liu1'}, {'b': 'liu3', 'd': 'liu4'})
In [13]: dict1
Out[13]: {'a': 'liu333', 'b': 'liu1'}