1.如果列表元素是对象，对这样的列表排序有哪些方法？
class MyClass:
def __init__(self):
self.value = 0
my1 = MyClass()
my1.value = 20
my2 = MyClass()
my2.value = 10
my3 = MyClass()
my3.value = 30
a = [my1, my2, my3]
print(a)
a.sort()
TypeError: '<' not supported between instances of 'MyClass' and 'MyClass'
[<__main__.MyClass object at 0x0000020B8C4F9CC8>, <__main__.MyClass object at 0x0000020B8C4FC248>, <__main__.MyClass object at 0x0000020B8C4FD748>]
方法一：利用magic函数：__lt__ 或 __gt__
class MyClass:
def __init__(self):
self.value = 0
def __lt__(self, other):
return self.value < other.value
def __gt__(self, other):
return self.value > other.value
my1 = MyClass()
my1.value = 20
my2 = MyClass()
my2.value = 10
my3 = MyClass()
my3.value = 30
a = [my1, my2, my3]
print(a)
a.sort()
print(a[0].value)
print(a[1].value)
print(a[2].value)
[<__main__.MyClass object at 0x0000016E82E16808>, <__main__.MyClass object at 0x0000016E82E16BC8>, <__main__.MyClass object at 0x0000016E82E16B08>]
10
20
30
方法二：利用operator sort的key
class MyClass:
def __init__(self):
self.value = 0
my1 = MyClass()
my1.value = 20
my2 = MyClass()
my2.value = 10
my3 = MyClass()
my3.value = 30
a = [my1, my2, my3]
print(a)
import operator
a.sort(key = operator.attrgetter('value'))
print(a[0].value)
print(a[1].value)
print(a[2].value)
方法三：利用operator sorted 的key
class MyClass:
def __init__(self):
self.value = 0
my1 = MyClass()
my1.value = 20
my2 = MyClass()
my2.value = 10
my3 = MyClass()
my3.value = 30
a = [my1, my2, my3]
print(a)
import operator
b = sorted(a, key=operator.attrgetter('value'))
print(b)
print(b[0].value)
print(b[1].value)
print(b[2].value)
2. 如果列表元素是对象，进行倒序排列的方法有哪些？
最简单方法
import operator
a.sort(key = operator.attrgetter('value'), reverse=True)
或者更改__lt__或__gt__
def __lt__(self, other):
return self.value > other.value
a.sort()
3. 总结
为类添加__lt__或__gt__方法，可以让该类的实例支持sort方法和sorted函数
通过改变__lt__或__gt__方法的返回值，或者设置sort方法和sorted函数的reverse参数，可以让列表倒序排列
