python list排序,list嵌套字典根据key排序

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 
 

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:

说明

1.new_s和new_s2的区别在于当score均为90的时候,从新按照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