How to repeat the input until a special condition is met in Python?

Question

I need to take integer inputs from the user and add them to a set. The number of integers is unknown. The input process will end when the user input is "Done".

Here is my code:

s = set()
print('Please type the number, when you are done please type "Done":')
while True:
    try:
        a = int(input())
        s.add(a)
    except:
        if a == "Done":
            break
        else:
            print('Integer only, please re-type:')
            continue
print(s)

But it didn’t work as intended.

My idea is simple. If the user types an integer then it’s fine, just add to set s. If they type a string (which int(input() is wrong), then it will go down to except, if the string is "Done", then break the while True loop, if it isn’t then ask the user to re-type and continue the loop.

Here is the error if I type a string first and if I type normally:

I tried to use a simple if-else statement at the start but it causes traceback right at int(input()).

Asked By: Quang Hoàng Minh

||

Source

Answer 1

Use the two-argument form of iter to read the input until Done is entered. You can iterate over the resulting iterator, handling integer conversions and updating the set in the body of a for loop.

s = set()
def get_input():
    return input('''Please type the number, when you're done please type "Done": ''')

for x in iter(get_input, "Done"):
    # x is guaranteed to *not* be Done here, or the loop
    # would have already exited.
    try:
        a = int(x)
    except ValueError:
        print("Integer only, please re-type")
        continue
    s.add(a)

If you weren’t concerned about error handling, you could reduce this to a single set comprehension:

s = {int(x) for x in iter(get_input, "Done")}

Answered By: chepner

Answer 2

When the exception happens, a remains unset. Try this instead.

s = set()
# Notice also correct quoting
print("Please type the number, when you're done please type 'Done':")
while True:
    a = input()
    try:
        n = int(a)
        s.add(n)
    # Avoid blanket except
    except ValueError:
        if a == "Done":
            break
        else:
            print('Integer only, please re-type:')
            continue
print(s)

Answered By: tripleee

Answer 3

Take the input as a string first and then convert it to integer before adding to the set. This way you can check if the value entered by the user is "Done" and exit the loop if it is.

Something like the code below. Please also check some syntax errors that you have in your code.

s = {}
print('Please type the number, when you're done please type "Done":')
while True:
    a = input()
    if a == 'Done':
        break
    else:
        try:
            s.add(int(a))
        except:
            print('Integer only, please re-type:')
print(s)

Answered By: LearnerMantis

Answer 4

s = set()
b = True

print("Please type the number, when you're done please type 'Done'.")

while b:
    try:
        a = input("Input: ")
        if a == "Done":
            b = False
            break
        a = int(a)
        s.add(a)
    except ValueError:
        print("Whoops, please type an integer!")
print(s)

Answered By: Fireye

Answer 5

You can try the code below.

s = set()
print('Please type the number, when you're done please type "Done":')
while True:
    try:
        a = input()
        if a == "Done":
            break
        s.add(int(a))
    except:
        if a == "Done":
            break
        else:
            print('Integer only, please re-type:')
            continue
print(s)

Answered By: E M

How to repeat the input until a special condition is met in Python?

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