我試圖去適應一個鍛鍊,將吐出一個學生的成績,而不是把名字中的代碼如Santising用戶輸入引用字典?
print get_letter_grade (get_average(alice))
讓我想用用戶輸入。 我個人傾向於使用大寫字母(即愛麗絲)自動輸入姓名,所以我需要使用.lower()
來打造我的輸入,只要我不使用大寫字母,代碼就不需要.lower
。
的錯誤似乎是所有小寫輸入AttributeError: 'dict' object has no attribute 'lower'
而對於大寫(即愛麗絲)輸入:NameError: name 'Tyler' is not defined:
pupil = input("Which student?").lower()
print get_letter_grade (get_average(
pupil))
嘗試:
pupil = input("Which student?").lower
student = pupil
print get_letter_grade (get_average(
student))
也試過:
pupil = input("Which student?")
student = pupil.lower()
print get_letter_grade (get_average(
student))
和:
pupil = input("Which student?")
student = pupil
print get_letter_grade (get_average(input("Which student?").lower()))
下面是完整的代碼:
lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
def average(numbers):
total=sum(numbers)
total=float(total)
total =total/len(numbers)
return total
def get_average(student):
homework = average(student["homework"])
quizzes = average(student["quizzes"])
tests = average(student["tests"])
return float(0.1*homework + 0.3*quizzes + 0.6*tests)
def get_letter_grade(score):
if score>=90:
return "A"
elif score>=80:
return "B"
elif score>=70:
return "C"
elif score>=60:
return "D"
else:
return "F"
pupil = input("Which student?").lower()
print get_letter_grade (get_average(
pupil))
我覺得我誤解有關用戶輸入的是如何使用/處理,或者說與格式字符串,字典做的事情?
如果輸入的是''alice''你得到的''student''實際上只是字符串'' 「愛麗絲」'' - 它絕不會突然變成對全局alice字典的引用。提示:從小處開始。根據用戶的輸入,開始識別和引用正確的字典,而不是編寫完整的程序。這樣可以減少潛在的錯誤。 –