Using a boolean to check if the number inside the string is a natural

Question

So i'm trying to make a function to check if the numbers inside a string is natural.

So my code is like this

def natural(x):
     while True:
          return x.isdigit() and 0 <= int(x) <= 9

I want my outputs to be like this

natural('05')
True

natural('asasassaas')
False

natural('243,432,355')
False

My question is how would i account for exponentially large numbers?

Answer 1

我认为您只是在寻找isdigit函数-它适用于您在此处列出的所有示例，并且用c编写，几乎可以肯定比纯python解决方案要快。

Answer 2

There is no need for you to do the additional check here:

and 0 <= int(x) <= 9

The isdigit method does this for you already. You pretty much want to stick with isdigit

Furthermore, a simplification can be made with the code where you can just stick to using isdigit like so:

def natural(x):
    return x.isdigit()

Thanks to @SteveJessop for the discussion on this.

Answer 3

You can loop over all characters in the provided string:

def natural(x):
    for c in x:
        if not c.isdigit():
            return False
    return True

This will stop once it encounters the first character that is not a digit.