Python list-comprehension for words that do not consist solely of digits

Question

At a high level, what I'm trying to accomplish is:

given a list of words, return all the words that do not consist solely of digits

My first thought of how to do that is:

import string
result = []
for word in words:
    for each_char in word:
        if each_char not in string.digit:
            result.append(word)
            break
return result

This works fine. To be more Pythonic, I figured - list comprehensions, right? So:

return [word for word in words for char in word if not char in string.digits]

Unfortunately, this adds a copy of word to the result for each character that isn't a digit. So for f(['foo']), I end up with ['foo', 'foo', 'foo'].

Is there a clever way to do what I'm trying to do? My current solution is to just write an is_all_digits function, and say [word for word in words if not is_all_digits(word)]. My general understanding is that list comprehensions allow this sort of operation to be declarative, and the helper function is plenty declarative for me; just curious as to whether there's some clever way to make it one compound statement.

Thanks!

Answer 1

Why not just check the whole string for isdigit():

>>> words = ['foo', '123', 'foo123']
>>> [word for word in words if not word.isdigit()]
['foo', 'foo123']

Or, turn the logic other way and use any():

>>> [word for word in words if any(not char.isdigit() for char in word)]
['foo', 'foo123']

any() would stop on the first non-digit character for a word and return True.