How to loop after 'Except ERROR' line in Python

Question

as part of the Chapter 3 exercise for Automate the Boring Stuff, I need to write a short program that mimics the Collatz sequence, where:

If number inputted is even number, divide it by 2, repeat until it equals 1;
If number inputted is odd number, multiply it by 3 then plus 1, repeat until it equals 1;
Create a clean exit for Ctrl+C.
Detect whether the user typed in a noninteger string.

Below is my code so far, which seems to work but I would appreciate any advice/best practice for improving it.

My main question is, after the program prints 'Enter integers only', is there any short and simple way to loop back to the 'Enter any number: ' line? I can't think of anything atm besides complicated if loops. Thanks.

def collatz(number):
    if number % 2 == 0 :
        results = number // 2
        print(results)
        if results != 1:
            collatz(results)
    elif number % 2 == 1 :
        results = 3 * number + 1
        print(results)
        if results != 1:
            collatz(results)

try: 
    entry = input('Enter any number : ')
    number = int(entry)
    print(number)
    collatz(number)
except ValueError:
    print('Enter integers only.')
except KeyboardInterrupt:
    sys.end()

Warning: DO NOT use a recursive function when calculating Collatz numbers: given the right number, you'd get millions or recursion. Just try with 27 :) — gimix
– gimix, Commented Jun 13, 2022 at 10:39
Which part of it is a recursive function? I tried 27 with the code and it seems to work fine and not too many lines. Sorry I'm a beginner so can't really tell. — yudontfly
– yudontfly, Commented Jun 13, 2022 at 11:42
I suggested 27 because it takes more than one hundred steps to be solved, which can be surprising for such a small number. But there are larger numbers which literally require millions of steps, so you will get a "recursion depth exceeded" error. Your function is recursive because it calls itself - that's what recursion means — gimix
– gimix, Commented Jun 13, 2022 at 12:36

Mr. Developer · Accepted Answer · 2022-06-13 10:37:18Z

1

You can try using a while loop. The below code will be helpful for you.

def collatz(number):
    if number % 2 == 0 :
        results = number // 2
        print(results)
        if results != 1:
            collatz(results)
    elif number % 2 == 1 :
        results = 3 * number + 1
        print(results)
        if results != 1:
            collatz(results)

while True:
    try: 
        entry = input('Enter any number : ')
        number = int(entry)
        print(number)
        collatz(number)
    except ValueError:
        print('Enter integers only.')
    except KeyboardInterrupt:
        break # Stop the while loop.
        sys.end()

answered Jun 13, 2022 at 10:37

Mr. Developer

12711 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Vvamp · Accepted Answer · 2022-06-13 10:42:45Z

0

As you want to keep prompting the same, until the program exits. I'd suggest just encapsulating the program code in a while(true) loop:

while True:
    try: 
        entry = input('Enter any number : ')
        number = int(entry)
        print(number)
        collatz(number)
    except ValueError:
        print('Enter integers only.')
    except KeyboardInterrupt:
        sys.end()

edited Jun 13, 2022 at 10:42

answered Jun 13, 2022 at 10:35

Vvamp

4144 silver badges13 bronze badges

2 Comments

Anony Mous Over a year ago

just quick sidenote, true shouldn't be in brackets and should have a capitalised "t".

Vvamp Over a year ago

@AnonyMous woops, thanks for the heads-up. It's still early here :)

M. Yasin · Accepted Answer · 2022-06-13 10:44:48Z

0

You can define an iteration to make sure the input is integer type.

while True: # executes the loop body indefinitely
    try: 
        entry = input('Enter any number : ')
        number = int(entry)
        print(number)
        collatz(number)
        break  # immediately terminates the loop entirely
    except ValueError:
        print('Enter integers only.')
    except KeyboardInterrupt:
        sys.end()

answered Jun 13, 2022 at 10:44

M. Yasin

666 bronze badges

Comments

Rawres · Accepted Answer · 2022-06-13 12:45:10Z

0

Put it in a "while" loop.

while True:
    try:
        ...
    except:
        ...

edited Jun 13, 2022 at 12:45

answered Jun 13, 2022 at 10:37

Rawres

351 silver badge9 bronze badges

Collectives™ on Stack Overflow

How to loop after 'Except ERROR' line in Python

4 Answers 4

Comments

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related