Validating an input in Python

Question

I have a piece of code that asks for an input. The input must not contain any numbers:

def main():
 invalid = []
 foo = ""
 foo = str(input("Enter a word: ") )
 invalid = ["1","2","3","4","5","6","7","8","9","0"]
 for eachInvalid in invalid:
 if eachInvalid in foo:
 print("Not a real word!")
 main()
 else:
 pass
main()

So far, it works but I am not proud of it. What would be the better way of doing this be? For example, a word does not contain things like ~@: so I would have to add those to that list. How would I avoid this but keep the code rather clean and readable?

Also, another error, if I run the code and type something in like hello, that is valid but if I type something invalid it takes me back to the start but when I type a valid word it still says invalid? For example:

Enter a word: hello
>>> ================================ RESTART ================================
>>> 
Enter a word: 123
Not a real word!
Enter a word: hello
Not a real word!

How would I fix this error and what would be the best way of looking for invalid characters in an input?

edit: nevermind, regular expression is fine.

• user554546: "I would like to avoid regular expression if possible" Why?
• Michał Szydłowski: Regexes are actually the best and simplest way to achieve it.
• Zak: I want to keep my code simple and not import too many modules however, if it is easier to be done in regular expression then that's ok.
• jonrsharpe: if any(char.isdigit() for char in foo):? Also, have a look at stackoverflow.com/q/23294658/3001761
• jonrsharpe: Note too that you don't have to "initialise variables" in Python - the first assignments to invalid and foo are totally redundant.

Answer 1

While a more complex validation is an appropriate use case for a regular expression, in this simple case there is a built in function isalpha() which checks whether a string only contains alphabetic characters.

foo.isalpha()

Returns True or False.

Note that in Python 3 this will deal with all unicode characters defined as "letters".

• Zak: Thanks. Exactly what I was looking for, even works for words with an umlaut, etc.

Answer 2

You could use a simpler method to test for a digit in the whole string, rather than testing each letter individually.

if any(x.isdigit() for x in foo):
 print("Invalid Word")
 else:
 print("valid word")

Answer 3

You could reverse the way you are doing, instead of having a list of invalids, it is better to have a list of valids. Then check every character of your string :

valids = ['a', 'b', 'c', 'd']
for letter in foo:
 if (not letter in valids):
 print("Not a real word!")

It is even easier with regex as it is easier to list all valids options :

import re
if (not re.match("^[a-zA-Z_ ]*$", foo)):
 print("Not a real word!")

The regex ^[a-zA-Z_ ]*$ meaning a word that contain only symbols in [a-zA-Z_ ]

If you want to stay with a list of invalids, use negated regular expression :

if (re.match("[^0-9@]", foo)):
 print("Not a real word!")

where [^0-9@] means anything but the characters defined between the brackets

• user: ^[a-zA-Z_ ]*$ might be more clear if it includes \s instead of whitespace (though i think it should be excluded since it would mean more than 1 word is given). Also, [^0-9@] wouldn't work since it matches characters like ,.!?- etc.