这篇文章主要介绍Python怎么实现字典排序、按照list中字典的某个key排序，文中介绍的非常详细，具有一定的参考价值，感兴趣的小伙伴们一定要看完！
1.给字典按照value按照从大到小排序
排序
dict = {'a':21, 'b':5, 'c':3, 'd':54, 'e':74, 'f':0}
new_dict = sorted(dict.iteritems(), key=lambda d:d[1], reverse = True)
print new_dict输出：
[('e', 74), ('d', 54), ('a', 21), ('b', 5), ('c', 3), ('f', 0)]
2. python按照list中的字典的某key排序:
例子:
s=[
{"no":28,"score":90},
{"no":25,"score":90},
{"no":1,"score":100},
{"no":2,"score":20},
]
print "original s: ",s
# 单级排序，仅按照score排序
new_s = sorted(s,key = lambda e:e.__getitem__('score'))
print "new s: ", new_s
# 多级排序,先按照score，再按照no排序
new_s_2 = sorted(new_s,key = lambda e:(e.__getitem__('score'),e.__getitem__('no')))
print "new_s_2: ", new_s_2输出:
original s: [{'score': 90, 'no': 28}, {'score': 90, 'no': 25}, {'score': 100, 'no': 1}, {'score': 20, 'no': 2}]
new s: [{'score': 20, 'no': 2}, {'score': 90, 'no': 28}, {'score': 90, 'no': 25}, {'score': 100, 'no': 1}]
new_s_2: [{'score': 20, 'no': 2}, {'score': 90, 'no': 25}, {'score': 90, 'no': 28}, {'score': 100, 'no': 1}]
说明
1.new_s和new_s2的区别在于当score均为90的时候,重新按照no排序
2.顺序为从小到大,若在sorted函数的参数加上reverse = True则为从大到小
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