if语句\\如何循环出现问题

Question

Okay so when i run this code and put in 50 for example it will list all the grades under that. How do i fix this? Also how can i make this more efficient? this is quite a lot of code for quite a simple task put i'm not sure how to achieve it. thanks for all replies.

print("-------WELCOME-------")
testscore = int(input("What mark did you get?"))
if testscore>= 80:
    print("You got an A!!")
else:
    if testscore>= 70:
        print("You got an B!")
    if testscore>= 60:
        print("You got an C.")
    if testscore>= 50:
        print("You got an D.")
    if testscore>= 40:
        print("You got an F.")
    if testscore>= 30:
        print("You got a G. Wow.")
    if testscore<= 20:
        print("There is nothing this low. Maybe you should have studied more? re-think your life please.")

second = input("do you want to get another test score?")
if second == "yes":
    testscore = int(input("What mark did you get?"))

if testscore>= 80:
    print("You got an A!!")
else:
    if testscore<= 70:
        print("You got an B!")
    if testscore<= 60:
        print("You got an C.")
    if testscore<= 50:
        print("You got an D.")
    if testscore<= 40:
        print("You got an F.")
    if testscore<= 30:
        print("You got a G. Wow.")
    if testscore<= 20:
        print("There is nothing this low. Maybe you should have studied more? re-think your life please.")
if second == "no":
    print("Thanks for using MARK TO GRADE CONVERTER. See you soon!")

Answer 1

The problem is that all those conditions in the else part are tested independently.

if testscore>= 80:
    print("You got an A!!")
else:
    if testscore>= 70:
        print("You got an B!")
    if testscore>= 60:
        print("You got an C.")
    if ...

If testscore is 75 , then the first condition is true, so print("You got an B!") is executed. It then tests the second condition, which is also true, so it executes print("You got an C!") , and so on. For the first two conditions (A and B) you used else: if ... , which went in the right direction, but using else: if for all would result in a huge cascade of nested blocks. Instead, you can use elif for all but the first condition. This way, the next condition is only tested if the previous one evaluated to false:

if testscore>= 80:
    print("You got an A!!")
elif testscore>= 70:
    print("You got an B!")
elif testscore>= 60:
    print("You got an C.")
elif ...

And similar for the second block of if statements further down. (The comparison is reversed in that block, but actually I'd guess that those blocks should be the same; in this case, you should make that a function and call the function twice instead of replicating that code.)

---

Alternatively, you could eg create a list of grades and scores and then find the next score that satisfies the condition in a single line:

>>> grades = (('A', 80), ('B', 70), ('C', 60), ('D', 50), ('F', 0))
>>> score = 56
>>> next((g for g, n in grades if score >= n))
'D'

Answer 2

If you want to reduce the code and make it more efficient you can use a dictionary to store the grades and use integer divison on the grade to get the key, as the following:

    grades = {7:'B', 6:'C', 5:'D', 4:'F', 3:'G'}               
    testscore = int(input("What mark did you get?"))
    if testscore>= 80:
        print("You got an A!!")
    elif testscore < 30:
        print("There is nothing this low. Maybe you should have studied more? re-think your life please.")
    else:
        print("You got an " + grades[testscore // 10] + "!")

and if you want to loop you can use:

    loop = 'yes'
    while loop == 'yes':
        #code here
        loop = raw_input('Do you want to get another test score?')