具有对象列表的Python最小函数

Question

如何使用min函数的key参数来比较对象的1属性列表？

例

class SpecialNumber: 
    def __init__(self, i): 
     self.number = i 

li = [SpecialNumber(1), SpecialNumber(3), SpecialNumber(2)] 

Answer 1

http://docs.python.org/library/operator.html#operator.attrgetter

from operator import attrgetter 
min_num = min(li,key=attrgetter('number')) 

样品交互式会话：

>>> li = [SpecialNumber(1), SpecialNumber(3), SpecialNumber(2)] 
>>> [i.number for i in li] 
[1, 3, 2] 
>>> min_num = min(li,key=attrgetter('number')) 
>>> print min_num.number 
1 

Answer 2

它是：

min(li, key=lambda x: x.number) 

你需要接受一个SpecialNumber并返回其元素的功能。

Answer 3

我想通过覆盖__cmp__

class SpecialNumber: 
    def __init__(self, i): 
     self.number = i 

    def __repr__(self): 
     return '<SpecialNumber(%d)>' % self.number 

    def __cmp__(self, other): 
     return cmp(self.number, other.number) 

li = [SpecialNumber(1), SpecialNumber(3), SpecialNumber(2)] 
print min(li) # <SpecialNumber(1)> 

Answer 4

的GETATTR版本做比较快

import random 
from operator import attrgetter 

class Test: 
    def __init__(self): 
     self.a = random.random() 

t = [Test() for i in range(10000)] 

%timeit min(t, key=lambda x: x.a) 
1000 loops, best of 3: 790 µs per loop 

%timeit min(t,key=attrgetter('a')) 
1000 loops, best of 3: 582 µs per loop