pythonソートsorted最初の条件が同じ場合は2番目の条件でソート

      list        
 
1.               
2.                2        
a = [[2,3],[4,1],(2,8),(2,1),(3,4)]
b = sorted(a,key=lambda x: (x[0], -x[1]))
print b
 
  
a = [[2,{'a':8}],[4,{'a':1}],(2,{'a':8}),(2,{'a':1}),(3,{'a':4})]
array = [7,8]
def fun(x):
    return (-x[0],x[1].get('a'))
b = sorted(a,key=fun)
print b
  
             
import time
 
n = xrange(1000000)
x = zip(n, n)
 
start = time.time()
x.sort(key = lambda x: (x[0], -x[1]))
end = time.time()
print 'key', end-start
 
x = zip(n, n)
start = time.time()
x.sort(cmp=lambda x, y: x[0] - y[0] or y[1] - x[1])
end = time.time()
print 'cmp', end-start
  ：
 
key 2.34500002861 cmp 0.269000053406
 
key        ，       ，      ，     key  。
           '-'      ，             ：         
class Reversinator(object):
    def __init__(self, obj):
        self.obj = obj
    def __lt__(self, other):
        return other.obj < self.obj
 
a = [(2,'3'),(4,'1'),(2,'8'),(2,'1'),(3,'4')]
 
print sorted(a, key=lambda x: (x[0], Reversinator(x[1])))
  ：
Python         ，         ， key                
    list.sort         key           
   key  ，               ，          key
  
#python list   
 
    
 
def my_key1(x):
 
    return x % 10
 
aList = [4, 5, 1, 2, 12, 34, 56, 9 ,80]
 
aList.sort() #       
 
print(aList)
 
aList.sort(reverse = True) #     
 
print(aList)
 
aList.sort(key = my_key1) #  key  ，           
 
print(aList)
 
     
 
def my_key2(x):
 
    return x[1]
 
aList = [(4,'ab'), (56,'c'), (1,'bb'), (102, 'a')]
 
aList.sort(key = my_key2) #        2  ，      
 
print(aList)