python高级编程改造_Python高级编程-collections模块（番外篇）
Python高级编程-collections模块(番外篇)
x.1 collections模块介绍
from collections import *
from collections.abc import *
"""
提供了更加高级的数据结构
"""
x.2 tuple的功能
不可变，iterable
拆包 # name, *others = ("coder", 20, 175)
tuple不可变性不是绝对的 # ("name", [20, 175])
tuple比list好的地方
immutable的重要性：
性能优化：指出元素全部为immutable的tuple会作为常量在编译时确定，因此产生了如此显著的速度差异
线程安全
可以作为dict的key
拆包特性
如果要拿C语言来类比，Tuple对应的是struct，而List对应的是array
x.3 nametuple 详解
from collections import nametuple
User = nametuple("User", ["name", "age", "height"])
user = User(name="booby", age=29, height=175)
print(user.age, user.name, user.height)
"""
为什么不用类封装，而用nametuple？
因为使用nametuple会比class少去很多默认属性，更加节约空间。
"""
user_tuple = ("bobby", 29, 175)
user = User(*user_tuple) # User._make(user_tuple)
print(user._asdict())
name, age, height = user
x.4 defaultdict 功能详解
from collections import defaultdict
# 统计数量
user_dict = {}
users = ["bobby1", "bobby2", "bobby3", "bobby1", "bobby2", "bobby2"]
for user in users:
if user not in user_dict:
user_dict[user] = 1
else:
user_dict[user] += 1
print(user_dict)
# 实际上dict中有一个方法setdefault
user_dict = {}
users = ["bobby1", "bobby2", "bobby3", "bobby1", "bobby2", "bobby2"]
for user in users:
user_dict.setdefault(user, 0) # 如果键不存在，就会设置默认值，性能更高，少了一次查
user_dict[user] += 1
print(user_dict)
# 代码可否再简洁，使用defaultdict
default_dict = defaultdict(int) # 传入一个可调用对象
users = ["bobby1", "bobby2", "bobby3", "bobby1", "bobby2", "bobby2"]
for user in users:
default_dict[user] += 1
print(default_dict)
"""
defaultdict如何实现？
使用__missing__魔法函数
"""
x.5 deque 功能详解
"""
双端队列
线程安全，list不是线程安全
"""
from collections import deque
# 只能队列尾部操作
user_list = ["bobby", "bobby1"]
user_name = user_list.pop()
print(user_name, user_list)
# 使用双端队列
user_list = deque(("bobby1", "bobb2"))
user_list_dict = deque(
{
"bobby1": 1,
"bobby2": 2
}
)
print(user_list)
print(user_list_dict)
user_list.appendleft("bobby8")
user_list_copy = user_list.copy() # 浅拷贝只是拷贝元素，可变元素会直接指向
print(id(user_list), id(user_list_copy))
class deque(MutableSequence[_T], Generic[_T]):
@property
def maxlen(self) -> Optional[int]: ...
def __init__(self, iterable: Iterable[_T] = ...,
maxlen: int = ...) -> None: ...
def append(self, x: _T) -> None: ...
def appendleft(self, x: _T) -> None: ...
def clear(self) -> None: ...
if sys.version_info >= (3, 5):
def copy(self) -> deque[_T]: ...
def count(self, x: _T) -> int: ...
def extend(self, iterable: Iterable[_T]) -> None: ...
def extendleft(self, iterable: Iterable[_T]) -> None: ...
def insert(self, i: int, x: _T) -> None: ...
def index(self, x: _T, start: int = ..., stop: int = ...) -> int: ...
def pop(self, i: int = ...) -> _T: ...
def popleft(self) -> _T: ...
def remove(self, value: _T) -> None: ...
def reverse(self) -> None: ...
def rotate(self, n: int) -> None: ...
x.6 Counter功能详解
"""
用来做统计
"""
from collections import Counter
users = ["bobby1", "bobby2", "bobby3", "bobby1", "bobby2", "bobby2"]
user_counter = Counter(users)
counter_str = Counter("dffdfdfd")
counter_str2 = Counter("dfdfdfdf")
user_counter.update("gwdesd")
user_counter.update(counter_str2) # 任意可迭代对象
print(counter_str)
print(user_counter)
# top n问题 headq 堆数据结构
print(user_counter.most_common(2)) # 返回前两名统计结果
x.7 OrderedDict 功能详解
"""
有序字典
"""
from collections import OrderedDict
user_dict = dict() # Python3中字典默认有序的，Python2是无需的
user_dict["b"] = "bobby2"
user_dict["a"] = "bobby1"
user_dict["c"] = "bobby3"
print(user_dict)
user_dict = OrderedDict()
user_dict["b"] = "bobby2"
user_dict["a"] = "bobby1"
user_dict["c"] = "bobby3"
print(user_dict)
print(user_dict.pop("a")) # pop方法必须传一个key
print(user_dict.popitem()) # 返回一个元组(key, val)
print(user_dict.move_to_end("b"))
x.8 ChainMap 功能
"""
用于连接dict
"""
from collections import ChainMap
user_dict1 = {"a":"bobby1", "b":"bobby2"}
user_dict2 = {"c":"bobby2", "d":"bobby3"}
for key,value in user_dict1.items():
print(key, value)
for key, value in user_dict2.items():
print(key, value)
new_dict = ChainMap(user_dict1, user_dict2)
new_dict.new_child({"aa":"aa", "bb":"bb"}) # 动态添加
print(new_dict["c"])
print(new_dict.maps)
new_dict.maps[0]["a"] = "bobby" # 只是指向对应字典，并没有完全复制
for key, value in new_dict.items():
print(key, value)
# 有重复怎么办？
user_dict1 = {"a":"bobby1", "b":"bobby2"}
user_dict2 = {"b":"bobby2", "d":"bobby3"}
new_dict = ChainMap(user_dict1, user_dict2) # 只取重复值第一个
for key, value in new_dict.items():
print(key, value)