python sort排序（当list列表元素为dict字典时）

# 简单的dict
lst = [('d', 2), ('a', 4), ('b', 3), ('c', 2)]
 
# 按照value排序
lst.sort(key=lambda k: k[1])
print(lst)
 
# 按照key排序
lst.sort(key=lambda k: k[0])
print(lst)
 
# 先按value排序再按key排序
lst.sort(key=lambda k: (k[1], k[0]))
print(lst)
 
# 输出---->>>
[('d', 2), ('c', 2), ('b', 3), ('a', 4)]
[('a', 4), ('b', 3), ('c', 2), ('d', 2)]
[('c', 2), ('d', 2), ('b', 3), ('a', 4)]
 
 
# 复杂的dict，按照dict对象中某一个属性进行排序
lst = [{'level': 19, 'star': 36, 'time': 1},
       {'level': 20, 'star': 40, 'time': 2},
       {'level': 20, 'star': 40, 'time': 3},
       {'level': 20, 'star': 40, 'time': 4},
       {'level': 20, 'star': 40, 'time': 5},
       {'level': 18, 'star': 40, 'time': 1}]
 
# 需求:
# level越大越靠前;
# level相同, star越大越靠前;
# level和star相同, time越小越靠前;
 
# 先按time排序
lst.sort(key=lambda k: (k.get('time', 0)))
 
# 再按照level和star顺序
# reverse=True表示反序排列，默认正序排列
lst.sort(key=lambda k: (k.get('level', 0), k.get('star', 0)), reverse=True)
 
 
 
for idx, r in enumerate(lst):
    print('idx[%d]\tlevel: %d\t star: %d\t time: %d\t' % (idx, r['level'], r['star'],r['time']))
 
# 输出---->>>
idx[0]   level: 20       star: 40        time: 2        
idx[1]   level: 20       star: 40        time: 3        
idx[2]   level: 20       star: 40        time: 4        
idx[3]   level: 20       star: 40        time: 5        
idx[4]   level: 19       star: 36        time: 1        
idx[5]   level: 18       star: 40        time: 1   

转载地址：https://blog.csdn.net/t8116189520/article/details/84395319