Checking for a digit in a string - without loops and without libraries

Question

I want to have a string, for example I want to type in 'I have 2 dogs' and I want the shell to return True (A function will do) if a number is shown in the string. I found it so hard to code without using any loops and without any 're' or 'string' library.

Any help please?

Example:

input: "I own 23 dogs"
output: True

input: "I own one dog"
output: False

In my learnings they didn't teach us this yet so I don't want to do anything too ahead — Lidor Cohen, Mar 09 '19 at 13:25
Folks, they haven't learnt about loops yet so why are you suggesting generators/list comphrensions? That said, the question is bad and likely a dupe — Chris_Rands, Mar 09 '19 at 13:39

Mojtaba Kamyabi · Answer 1 · 2019-03-09T14:01:24.607

1

>>> a = "I own 23 dogs"
>>> print(bool([i for i in a if i.isdigit()]))
True

>>> b = "I own some dogs"
>>> print(bool([i for i in b if i.isdigit()]))
False

However the better solution is to use any and use generators instead of lists for better performance (just like @Alderven solution):

>>> a = "I own 23 dogs"
>>> any(i.isdigit() for i in a)
True

edited Mar 09 '19 at 14:01

answered Mar 09 '19 at 13:27

Mojtaba Kamyabi

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this builds the whole list, not a good idea when you can short circuit with `any()` – Chris_Rands Mar 09 '19 at 13:41
@Chris_Rands you are right. I update my answer – Mojtaba Kamyabi Mar 09 '19 at 14:04

score 1 · Answer 2 · answered Mar 09 '19 at 13:28

1

List comprehension helps you:

def is_digit(string):
    return any(c.isdigit() for c in string)

print(is_digit("I own 23 dogs"))
print(is_digit("I own one dog"))

Output:

True
False

answered Mar 09 '19 at 13:28

Alderven

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Checking for a digit in a string - without loops and without libraries

2 Answers2