用lambda()實現對列表,字典的排序


1、對list排序foo = [-5,8,0,4,9,-4,-20,-2,8,2,-4],使用lambda函數從小到大排序
foo = [-5,8,0,4,9,-4,-20,-2,8,2,-4]
k=sorted(foo,key=lambda x:x)

2、使用lambda函數對list排序foo = [-5,8,0,4,9,-4,-20,-2,8,2,-4],輸出結果為
[0,2,4,8,8,9,-2,-4,-4,-5,-20],正數從小到大,負數從大到小

foo = [-5,8,0,4,9,-4,-20,-2,8,2,-4]
k=sorted(foo,key=lambda x:(x<0,abs(x)))

3、列表嵌套字典的排序,分別根據年齡和姓名排序
foo = [{"name":"zs","age":19},
{"name":"ll","age":54},
{"name":"wa","age":17},
{"name":"df","age":23}]
k=sorted(foo,key=lambda x:x["name"])
k=sorted(foo,key=lambda x:x["age"])
4、列表嵌套元組,分別按字母和數字排序
foo = [("zs",19),
("ll",54),
("wa",17),
("df",23)]
k=sorted(foo,key=lambda x:x[0])
k=sorted(foo,key=lambda x:x[1])

5、列表嵌套列表排序,年齡數字相同怎么辦?
foo = [["zs",19],["ll",54],["wa",17],["sd",17],["df",73],["df",45]]
k=sorted(foo,key=lambda x:(x[1],x[0]))

6、根據鍵對字典排序(方法一,zip函數)
foo={"name":"zs","age":19,'city':"shanghai","sex":"F"}
k=zip(foo.keys(),foo.values())
kk=[x for x in k]
ss=sorted(kk,key=lambda x:x[0])
new_dict={}
for i in ss:
new_dict.setdefault(i[0],i[1])
print(new_dict)

7、根據鍵對字典排序(方法二,不用zip)
foo={"name":"zs","age":19,'city':"shanghai","sex":"F"}
k=sorted(foo.keys())
new_dict={}
for i in k:
new_dict.setdefault(i,foo[i])


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