python list排序,list嵌套字典根據key排序

1.給字典按照value按照從大到小排序

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

 

輸出 
 

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
# 單級排序，僅按照score排序
new_s = sorted(s,key = lambda e:e.__getitem__('score'))
print "new s: ", new_s
# 多級排序,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

輸出:

說明

1.new_s和new_s2的區別在於當score均爲90的時候,從新按照no排序

2.順序爲從小到大,若在sorted函數的參數加上reverse = True則爲從大到小