将表格信息封装成字典,每个学生的信息为一个键值对,键为学号,其它信息组织成列表,当作键值对的值。
按键排序,显示如图所示,请问字典建立有什么问题吗?
两个问题,你使用名字那些要用字符串处理,还有就是你应该想按学号排序,如果你这么写的话只能得到key,要得到键值对要用items
students = {"512020": ["张三","1班"],\
"512022": ["王五","1班"],\
"512025": ["lisi","2班"],\
"512026": ["陈荣","1班"],\
"512010": ["赵荣","1班"]}
students = sorted(list(students.items()),key=lambda x:x[0])
for i in students:
print(i)
你定义的是字典,不是函数,为什么要加()
张三,1班
这些都要用''包裹
dic_students = {"5120200000": ["张三", "1班", 175, 70],
"5120200002": ["王五", "1班", 185, 85],
"5120200005": ["李四", "2班", 165, 67],
"5120200006": ["陈荣", "1班", 170, 67],
"5120200010": ["赵荣", "2班", 179, 70],
}
dic_result = sorted(list(dic_students), key=lambda x:x[0])
print(dic_result)