Python：列表字典-输入的平均回报

Question

我需要考虑一个学生列表，创建一个字典，然后从一个输入，我必须返回平均从学生。

你们已经帮过我了：

我需要返回我的平均计算值，但它不起作用

StudentGrades = {
"Alcott": [5, 9, 7],
"Jerry": [3, 5, 2],
"David": [8, 9, 8, 9]
}

## I fixed this, thank you
print("Students:")
for x, y in StudentGrades.items():
    print(x, y)

while True:
    print("Select a student: ")
    name = str(input())

    if name in StudentGrades.keys():
        print("You have chosen: " + name)
        print (res = sum(StudentGrades[name]/len(StudentGrades[name]))
    else:
        print('You chosen wrong!')
       break

## Also, I didn't wrap right John's value, thank you for showing me
#use of update function

print("Original Dictionary:")
print(StudentGrades)

StudentGrades.update(student)
print("Dictionary after update:")
print (StudentGrades)

#John's Average (not working? int object is not iterable)
res = sum(StudentGrades["John"]) / len(StudentGrades["John"])
print("John's average is:[10]/1 =" + str(res))

但我也在想我是否可以退回这样的东西：

#Alcott's average
resAlcott = sum(StudentGrades["Alcott"]) / len(StudentGrades["Alcott"])


print("Students:")
for x, y in StudentGrades.items():
    print(x, y)

while True:
    print("Select a student: ")
    name = str(input())

    if name in StudentGrades.keys() == "Alcott":
        print("You have chosen: " + name + "'s average is", StudentGrades["Alcott"], resAlcott)
        
        
    else:
        print('You chosen wrong!')
        break

谢谢你，我再次感到抱歉

Answer 1

你需要这样的东西：

StudentGrades = {
"Alcott": [5, 9, 7], 
"Jerry": [3, 5, 2], 
"David": [8, 9, 8, 9]
}

print("Students:")
for x, y in StudentGrades.items():
    print(x, y)
 
while True:
    print("Select a student: ")
    name = str(input())
 
    if name in StudentGrades.keys():
        print("You have chosen: " + name)
        print( sum(StudentGrades[name]) / len(StudentGrades[name]) ). # This is the part of your code that was wrong.
    else:
        print('You chosen wrong!')
        break

此外，要添加新学生，不应使用student = {'Name': 10}，而应使用student = {'Name': [10]}。与所有其他值一样，该值必须是一个列表