py 排序.md

dict排序

# 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])

# 按照key排序
lst.sort(key=lambda k: k[0])

# 先按value排序再按key排序
lst.sort(key=lambda k: (k[1], k[0]))

# 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

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
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