How to keep the correct inputs in the range?

Question

For 10 inputs I want to have passed and failed. When an input other than 1 or 2 loops continues but 1 output is missing how can I fix it?

num_pass = 0
num_fail = 0
loopcount = 11
r = range(1, loopcount)

for student in r:
    grade = int(input('1 for pass 2 for fail'))
    if grade == 1:
        num_pass = num_pass + 1
    elif grade == 2:
        num_fail = num_fail + 1
    else:
        loopcount += 1
        continue



print('Passes:', num_pass)
print('Failed:', num_fail)

Please don't post code as an image. Post the code instead. It'll make it possible to copy/paste it into our IDE — Thomas Weller, Dec 02 '21 at 19:23
Show code as properly formatted text in the question, not as image or external link. — Michael Butscher, Dec 02 '21 at 19:23
A. You can't increment the counter in a for loop from inside the loop. — swaggg, Dec 02 '21 at 19:25
B. When you save the range to a variable, it's defined at that point and changing the variable that was used to define it doesn't do anything. But the same actually happens if you don't save it to a variable. — swaggg, Dec 02 '21 at 19:27

score 0 · Answer 1 · answered Dec 02 '21 at 19:31

One option is to have an inner infinite loop (i.e. while True) that you break from when you get a valid answer. For example:

num_pass = 0
num_fail = 0
loopcount = 11
r = range(1, loopcount)

for student in r:
    while True:
        grade = int(input('1 for pass 2 for fail '))
        if grade == 1:
            num_pass = num_pass + 1
            break
        elif grade == 2:
            num_fail = num_fail + 1
            break
        else:
            print('please try again')
        
print('Passes:', num_pass)
print('Failed:', num_fail)

The else block here is optional, but it is confusing without it, because there is no indication whether the next occurrence of the question is still asking about the same student or the next one.

You might also want to guard against non-numeric input that might otherwise cause your program to abort with an error when it tries to convert it to an int. For this, use exception handling, replace the line:

        grade = int(input('1 for pass 2 for fail '))

with:

        try:
            grade = int(input('1 for pass 2 for fail '))
        except ValueError:
            print('that was not a number')
            continue

How to keep the correct inputs in the range?

1 Answers1