我必須編寫一個接受 5 個分數并確定每個分數等效字母等級的程序。結果將以表格格式打印。我了解問題的要求以及我沒有轉述的所有其他步驟，因為它們與我的問題無關。我無法弄清楚如何將我的代碼壓縮成某種循環。我覺得我擁有的東西很多余，應該有一種方法可以濃縮它。另外，我不允許使用列表來解決這個問題。# Letter grade constantsA_SCORE = 90B_SCORE = 80C_SCORE = 70D_SCORE = 60# Gather scores from usernum1 = int(input("Enter score 1: "))if num1 >= A_SCORE: letter1 = "A"elif num1 >= B_SCORE: letter1 = "B"elif num1 >= C_SCORE: letter1 = "C"elif num1 >= D_SCORE: letter1 = "D"else: letter1 = "F"num2 = int(input("Enter score 2: "))if num2 >= A_SCORE: letter2 = "A"elif num2 >= B_SCORE: letter2 = "B"elif num2 >= C_SCORE: letter2 = "C"elif num2 >= D_SCORE: letter2 = "D"else: letter2 = "F"num3 = int(input("Enter score 3: "))if num3 >= A_SCORE: letter3 = "A"elif num3 >= B_SCORE: letter3 = "B"elif num3 >= C_SCORE: letter3 = "C"elif num3 >= D_SCORE: letter3 = "D"else: letter3 = "F"num4 = int(input("Enter score 4: "))if num4 >= A_SCORE: letter4 = "A"elif num4 >= B_SCORE: letter4 = "B"elif num4 >= C_SCORE: letter4 = "C"elif num4 >= D_SCORE: letter4 = "D"else: letter4 = "F"num5 = int(input("Enter score 5: "))if num5 >= A_SCORE: letter5 = "A"elif num5 >= B_SCORE: letter5 = "B"elif num5 >= C_SCORE: letter5 = "C"elif num5 >= D_SCORE: letter5 = "D"else: letter5 = "F"# Print resultsprint("score 1: \t\t" + str(num1) + "\t\t" + letter1)print("score 2: \t\t" + str(num2) + "\t\t" + letter2)print("score 3: \t\t" + str(num3) + "\t\t" + letter3)print("score 4: \t\t" + str(num4) + "\t\t" + letter4)print("score 5: \t\t" + str(num5) + "\t\t" + letter5)這是顯示內容的示例。它的格式正確，我剛剛排除了表格標題。Enter score 1: 75Enter score 2: 88Enter score 3: 91Enter score 4: 80Enter score 5: 85score 1: 75 Cscore 2: 88 Bscore 3: 91 Ascore 4: 80 Bscore 5: 85 B
3 回答
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TA貢獻1836條經驗 獲得超13個贊
是否沒有家庭作業限制以完全阻止列表并以艱苦的方式進行工作。
我能想象的最好的縮短代碼將是一個函數：
# Letter grade constants
A_SCORE = 90
B_SCORE = 80
C_SCORE = 70
D_SCORE = 60
# Gather scores from user
def getscore(num):
if num >= A_SCORE:
return "A"
elif num >= B_SCORE:
return "B"
elif num >= C_SCORE:
return "C"
elif num >= D_SCORE:
return "D"
else:
return "F"
num1 = int(input("Enter score 1: "))
num2 = int(input("Enter score 2: "))
num3 = int(input("Enter score 3: "))
num4 = int(input("Enter score 4: "))
num5 = int(input("Enter score 5: "))
letter1 = getscore(num1)
letter2 = getscore(num2)
letter3 = getscore(num3)
letter4 = getscore(num4)
letter5 = getscore(num5)
# Print results
print("score 1: \t\t" + num1 + "\t\t" + letter1)
print("score 2: \t\t" + num2 + "\t\t" + letter2)
print("score 3: \t\t" + num3 + "\t\t" + letter3)
print("score 4: \t\t" + num4 + "\t\t" + letter4)
print("score 5: \t\t" + num5 + "\t\t" + letter5)
另請注意，無需將整數轉換為字符串。打印函數隱式地執行此操作
ABOUTYOU
TA貢獻1812條經驗 獲得超5個贊
您可以使用bisect模塊來獲取成績。在這里，get_score將采用數字等級并返回字母等級。
from bisect import bisect
def get_score(A):
i = bisect([60, 70, 80, 90], A)
return 'FDCBA'[i]
mm = []
for i in range(5):
mm += [input('Enter score {}: '.format(i))]
for m in mm:
print('score {}: {} {}'.format(i, m, get_score(int(m))))
#Outputs
Enter score 0: 75
Enter score 1: 88
Enter score 2: 91
Enter score 3: 80
Enter score 4: 85
score 4: 75 C
score 4: 88 B
score 4: 91 A
score 4: 80 B
score 4: 85 B
叮當貓咪
TA貢獻1776條經驗 獲得超12個贊
您的代碼是使用循環的教科書案例。每當您發現自己編寫了充滿命令的重復代碼，每次只更改次要細節時，看看是否可以使用循環：
A_SCORE = 90
B_SCORE = 80
C_SCORE = 70
D_SCORE = 60
scores = []
grades = []
for n in range(1, 6):
grade = int(input("Enter score {}: ".format(n)))
grades.append(grade)
if grade >= A_SCORE:
scores.append('A')
elif grade >= B_SCORE:
scores.append('B')
elif grade >= C_SCORE:
scores.append('C')
elif grade >= D_SCORE:
scores.append('D')
else:
scores.append('F')
for n, (score, grade) in enumerate(zip(scores, grades), 1):
print("score {}: \t\t".format(n) + str(grade) + "\t\t" + score)
輸出：
Enter score 1: 75
Enter score 2: 88
Enter score 3: 91
Enter score 4: 80
Enter score 5: 85
score 1: 75 C
score 2: 88 B
score 3: 91 A
score 4: 80 B
score 5: 85 B
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