1.多繼承
1).多繼承
python中的類支持多繼承(讓一個(gè)類同時(shí)繼承多個(gè)類);
多繼承的時(shí)候铃拇,子類只能繼承第一個(gè)父類所有的屬性和方法钞瀑,后面的父類中只有字段和方法可以被繼承
class Animal(object):
num = 100
def __init__(self):
self.age = 0
self.gender = '雌'
@classmethod
def func1(cls):
print('動(dòng)物類的類方法')
def func2(self):
print('動(dòng)物類中的對象方法')
class Fly(object):
name = '飛行器'
def __init__(self):
self.height = 100
self.time = 5
self.speed = 100
def func2(self):
print('飛行的對象方法')
class Bird(Animal, Fly):
pass
bird1 = Bird()
# 字段都能繼承
print(Bird.num, Bird.name)
Bird.func1()
bird1.func2()
# print(bird1.age, bird1.gender)
# print(bird1.speed, bird1.height, bird1.time)
2.運(yùn)算符重載
1).運(yùn)算符
python中所有的類型都是類,所以所有的數(shù)據(jù)都是對象;
python中使用任意的運(yùn)算符都是在調(diào)用相應(yīng)類中的相應(yīng)方法锚贱,每一個(gè)運(yùn)算符對應(yīng)什么方法是固定的仔戈,
某種數(shù)據(jù)是否支持某個(gè)運(yùn)算符操作就看這個(gè)數(shù)據(jù)類型中是否實(shí)現(xiàn)了對應(yīng)的方法
2).運(yùn)算符重載指的是在不同的類中實(shí)現(xiàn)同樣的運(yùn)算符對應(yīng)的函數(shù)
類的對象默認(rèn)情況下只支持: ==, !=
10 * 20
'abc'+'123'
[1, 2] + [2, 3, 4]
# {'a': 10} + {'b': 20}
class Student:
def __init__(self, name, age, score=0):
self.name = name
self.age = age
self.score = score
def __repr__(self):
return '<%s, id:%s>' % (str(self.__dict__)[1:-1], hex(id(self)))
# a+b -> a.__add__(b)
# self -> 當(dāng)前類的對象,也是+前面的那個(gè)數(shù)據(jù)
# other -> +后面的那個(gè)數(shù)據(jù), 類型根據(jù)運(yùn)算規(guī)則的設(shè)計(jì)可以是任何類型的數(shù)據(jù)
def __add__(self, other):
# return self.age + other.age
return self.score + other.score
# return Student(self.name+other.name, self.age+other.age, self.score + other.score)
# return self.score + other
# a*b -> a.__mul__(b)
def __mul__(self, other):
list = []
for _ in range(other):
list.append(copy.copy(self))
return list
# a<b -> a.__lt__(b)
# 注意: <和>符號只需要重載其中一個(gè)就可以
def __lt__(self, other):
return self.score < other.score
stu1 = Student('小明', 19, 90)
stu2 = Student('小花', 20, 78)
# stu1.__add__(stu2)
print(stu1 + stu2)
# stu1.__add__(100)
# print(stu1 + 100)
print(stu1 * 3)
students = [stu1, stu2, Student('小紅', 12, 100)]
students.sort()
print(students)
print(stu1 < stu2)
print(stu1 > stu2)
3.淺拷貝和深拷貝
from copy import copy, deepcopy
class Dog:
def __init__(self, name, color):
self.name = name
self.color = color
def __repr__(self):
return '<%s,id: %s>' % (str(self.__dict__)[1:-1], hex(id(self)))
class Person:
def __init__(self, name, age, dog=None):
self.name = name
self.age = age
self.dog = dog
def __repr__(self):
return '<%s,id: %s>' % (str(self.__dict__)[1:-1], hex(id(self)))
p1 = Person('小明', 18, Dog('大黃', '黃色'))
1).直接賦值
將變量中的地址直接賦給新的變量拧廊;賦值后兩個(gè)變量的地址相同
p2 = p1
print(id(p1), id(p2)) # 4537270848 4537270848
p1.name = '小花'
print(p1.name, p2.name) # 小花 小花
p2.dog.color = '綠色'
print(p1.dog.color, p2.dog.color) # 綠色 綠色
2).拷貝
不管是淺拷貝還是深拷貝都會(huì)對原數(shù)據(jù)進(jìn)行賦值產(chǎn)生新的地址
list1 = [1, 2, 3]
list2 = copy(list1)
list3 = deepcopy(list1)
print(id(list1), id(list2), id(list3))
list1.append(100)
print(list2, list3)
p3 = copy(p1)
p4 = deepcopy(p1)
print(id(p1), id(p3), id(p4))
p1.name = '小紅'
print(p3.name, p4.name)
3).淺拷貝
字符串监徘、列表和元祖的切片;對象.copy(); copy模塊中的copy方法都是淺拷貝
淺拷貝指拷貝當(dāng)前對象,不會(huì)對子對象進(jìn)行拷貝
print('==========淺拷貝=========')
p3 = copy(p1)
print(id(p1), id(p3)) # 4401914600 4401916336
print(id(p1.dog), id(p3.dog)) # 4567065152 4567065152
p1.name = 'Tom'
print(p1.name, p3.name) # Tom 小紅
p1.dog.color = '紅色'
print(p1.dog.color, p3.dog.color) # 紅色 紅色
4).深拷貝
copy模塊中的deepcopy方法是深拷貝
print('==========深拷貝=========')
p4 = deepcopy(p1)
print(id(p1), id(p4)) # 4486472592 4486474384
print(id(p1.dog), id(p4.dog)) # 4485063960 4486474552
p1.name = 'Bob'
print(p1.name, p4.name) # Bob Tom
p1.dog.color = '橙色'
print(p1.dog.color, p4.dog.color) # 橙色 紅色
# 練習(xí):
a = ['color', 'height', 'background']
b = [a, 'aaa', 'bbb']
c1 = b
c2 = copy(b)
c3 = deepcopy(b)
a[-1] = ['BG']
b.append('ccc')
# b = [['color', 'height', 'background'], 'aaa', 'bbb']
# 問題: print(c1), print(c2), print(c3)的結(jié)果分別是
# c1 = [['color', 'height', ['BG']], 'aaa', 'bbb', 'ccc']
# c2 = [['color', 'height', ['BG']], 'aaa', 'bbb']
# c3 = [['color', 'height', 'background'], 'aaa', 'bbb']
print(c1)
print(c2)
print(c3)
4.枚舉
枚舉值的特點(diǎn):
1).可以通過有意義的屬性名直接顯示數(shù)據(jù)
2).每個(gè)數(shù)據(jù)的值不能修改
3).可以做到不同的數(shù)據(jù)的值唯一
@unique
class PokerNum(Enum):
J = 11
Q = 12
K = 13
A = 1
class Color:
RED = (255, 0, 0)
print(PokerNum.J)
print(PokerNum.K.value, PokerNum.J.value > PokerNum.Q.value)
5.內(nèi)存管理
1).內(nèi)存的開辟
內(nèi)存區(qū)間分為棧區(qū)間和堆區(qū)間吧碾;棧區(qū)間的內(nèi)存自動(dòng)開辟自動(dòng)釋放凰盔,堆區(qū)間的內(nèi)存需要程序員手動(dòng)開辟,手動(dòng)釋放倦春;
但是python已經(jīng)將堆區(qū)間內(nèi)存的開辟和釋放自動(dòng)化
a.當(dāng)每次給變量賦值的時(shí)候户敬,系統(tǒng)會(huì)先在堆區(qū)間中開辟空間將數(shù)據(jù)存起來,然后再將數(shù)據(jù)在堆中的地址存到變量中睁本,變量存在棧區(qū)間;
b.數(shù)字和字符串?dāng)?shù)據(jù)在開辟空間的時(shí)候尿庐,會(huì)先檢測內(nèi)存中之前是否已經(jīng)有這個(gè)數(shù)據(jù),
如果有就直接使用之前的數(shù)據(jù)呢堰,沒有才開辟新的空間保存數(shù)據(jù)
a = [1, 2, 3, [1, 2]] # [1, 2, 0xccc]
b = [1, 2, 3, [1, 2]] # [1, 2, 0xccc]
print(id(a), id(b)) # 4484287176 4484285768
print(id(a[3]), id(b[3]))
print(id(a[0]), id(b[0]), id(a[3][0]), id(b[3][0])) # 4449482688 4449482688 4449482688 4449482688
a1 = 100
b1 = 100
print(id(a1), id(b1)) # 4480746528 4480746528
a2 = 'hello'
b2 = 'hello'
print(id(a2), id(b2)) # 4492420184 4492420184
a3 = {'a': 10}
b3 = {'a': 10}
print(a3 == b3) # True
print(a3 is b3) # False
a4 = 200
b4 = deepcopy(a4)
print(id(a4), id(b4)) # 4367007904 4367007904
a5 = '安徽省打發(fā)卡機(jī)阿哈薩克阿按時(shí)發(fā)阿薩德回復(fù)杰卡斯發(fā)是打飛機(jī)啊山東礦機(jī)返回撒嬌快遞費(fèi)哈薩克京東方撒地方'
b5 = '安徽省打發(fā)卡機(jī)阿哈薩克阿按時(shí)發(fā)阿薩德回復(fù)杰卡斯發(fā)是打飛機(jī)啊山東礦機(jī)返回撒嬌快遞費(fèi)哈薩克京東方撒地方'
print(id(a5), id(b5)) # 4338364064 4338364064
# 2.內(nèi)存的釋放
"""
棧區(qū)間:全局棧區(qū)間在程序結(jié)束后銷毀抄瑟,函數(shù)棧區(qū)間在函數(shù)調(diào)用結(jié)束后銷毀(自動(dòng))
堆區(qū)間: 看一個(gè)對象是否銷毀,就看這個(gè)對象的引用計(jì)數(shù)是否為0枉疼;如果一個(gè)對象的引用為0皮假,這個(gè)對象就會(huì)銷毀;
(垃圾回收機(jī)制)
注意: python中針對對象的循環(huán)引用已經(jīng)做了處理鞋拟,程序員不需要寫額外的代碼來解決循環(huán)引用問題
"""
# 1
a6 = {'name': '小明', 'age': 18}
b6 = a6
print(getrefcount(a6))
b6 = 100
print(getrefcount(a6))
del a6
print(getrefcount(b6))
class Person:
def __init__(self, name):
self.name = name
self.dog = None
class Dog:
def __init__(self, name):
self.name = name
self.owner = None
dog1 = Dog('大黃')
p1 = Person('小明')
p1.dog = dog1
dog1.owner = p1
del dog1
del p1