How can you make a program prompt again if the input is invalid in Python?

Question

I need the choices to be asked again until the user chooses a valid option.

choice = input('Please choose 1,2,3\n')

if choice == 1:
print('You have chosen the first option')

elif choice == 2:
print('You have chosen the second option')

elif choice == 3:
print('You have chosen the third option')

else:
print('This is not an option, please try again.')

Have you learned about loops? You should learn about loops. Go do that. They are fundamental in almost any programming language. — TigerhawkT3
– TigerhawkT3, Commented Aug 3, 2015 at 22:37

Tom · Accepted Answer · 2015-08-03 23:02:55Z

2

Perhaps I am mistaken, as I am just a hack, but I believe a more "Pythonic" answer would be:

choices = {1:'first', 2:'second', 3:'third'}

while True:
    choice = input('Please choose 1, 2, or 3.\n')
    try:
        print 'You have chosen the {} option.'.format(choices[choice])
        break
    except KeyError:
        print 'This is not an option; please try again.'

Or at least:

while True:
    choice = input('Please choose 1, 2, or 3.\n')

    if choice == 1:
        print 'You have chosen the first option'
        break
    elif choice == 2:
        print 'You have chosen the second option'
        break
    elif choice == 3:
        print 'You have chosen the third option'
        break
    else:
        print 'This is not an option; please try again.'

Both of these avoid creating an unnecessary test variable, and the first one reduces overall code needed.

For Python 3, I believe adding parentheses around the printed statements should be the only change. The question wasn't tagged with a version.

edited Aug 3, 2015 at 23:02

answered Aug 3, 2015 at 22:59

Tom

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5 Comments

a p Over a year ago

Like this, but I'd suggest use of if int(choice) in choices: print 'You have chosen the {} option.'.format(choices[choice]) -- '1', '2', and '3' can be strings in the choices dict too, saving some trouble with converting to int.

cphlewis Over a year ago

I'd include an option 'q' for 'quit'. Pleasanter for users who aren't sophisticated enough to Ctrl-C out, and if you hook the logic into a GUI, you already have something for the Cancel button.

Tom Over a year ago

@ap, I used to always do it that way too, but I seem to remember coming across an explanation that for large dictionaries, it was less efficient. I realize that, in this case, it's trivial, but I got in the habit of catching the KeyError instead. Upon further inspection, it looks like it depends, with your approach just edging mine out based on clarity.

a p Over a year ago

@Tom it's possible, but I can't imagine how either way would be less efficient. Checking membership in a hash (dict) is O(1) -- very fast, even on large data sets. And in order to throw a KeyError, the work of looking something up needs to be done anyway.

Dalen Over a year ago

@Tom I'd argue about Pythonic. Using dictionary as choice-chooser is cool but it's also considered a replacement for switch-case. I'd also use dict.

Woksin · Accepted Answer · 2015-08-03 22:58:04Z

0

Try this:

valid = False
while(valid == False):

   choice = input("Please choose 1,2,3\n")

   if choice == 1:
      valid = True
      print('You have chosen the first option')

   elif choice == 2:
     valid = True
     print('You have chosen the second option')

   elif choice == 3:
     valid = True
     print('You have chosen the third option')

   else:
     valid = False
     print('This is not an option, please try again.')

edited Aug 3, 2015 at 22:58

answered Aug 3, 2015 at 22:41

Woksin

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6 Comments

Woksin Over a year ago

Used to C/C++ structuring so I had to think twice for python

Francis Over a year ago

It says that 'false' is not defined

Woksin Over a year ago

Try capitalizing False and True

TigerhawkT3 Over a year ago

You keep entering code that is either wrong or entirely invalid Python syntax.

TigerhawkT3 Over a year ago

As far as I'm aware, testing your code to make sure it compiles and runs as expected is just as important in C/C++ as it is in Python.

|

Martin Evans · Accepted Answer · 2015-08-04 15:47:52Z

0

You could make it into a function, which takes the required prompt, valid selections and only returns when a valid input is made.

def get_input(prompt, choices):
    while True:
        choice = input("%s %s or %s: " % (prompt, ", ".join(choices[:-1]), choices[-1]))
        if str(choice) in choices:
            return choice

choice = get_input("Please choose", ["1", "2", "3"])
print("You have chosen {}".format(choice))

Which would give the following type of output:

Please choose 1, 2 or 3: 1
You have chosen 1

answered Aug 4, 2015 at 15:47

Martin Evans

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