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