How to verify if a string is an integer in Python without try/except?

General Discussions
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1

How can I check if a string represents an integer in Python, without using try/except?

For example, I want to verify if a string like ‘3’ or ‘-17’ is an integer, but strings like ‘3.14’ or ‘asfasfas’ should return False. Here’s an example of the expected behavior:

is_int('3.14') == False
is_int('-7') == True
2

I’ve been working with Python for years, and when you need to verify if a string is an integer without using try/except, one simple method is to use the str.isdigit() function. This works perfectly for non-negative integers and is straightforward to implement:

string = "12345"
if string.isdigit():
    print("It's an integer!")
else:
    print("Not an integer.")

However, this doesn’t handle cases like negative integers or whitespace, so it’s more suitable for clean, positive numeric strings.

3

Building on Netra’s point, while str.isdigit() is great for basic checks, you might encounter scenarios where you need to handle negative integers too. A more flexible approach is to use str.lstrip('-').isdigit(). This method strips the leading - and then checks if the rest is numeric:

string = "-12345"
if string.lstrip('-').isdigit():
    print("It's an integer!")
else:
    print("Not an integer.")

This method ensures that even negative numbers are caught accurately, but it might not cover cases like numbers with leading or trailing spaces.

4

Let’s refine this further. If you want a robust way to handle all cases—like strings with spaces, multiple - signs, or empty strings—you can combine .strip() and .isdigit() intelligently. Alternatively, regular expressions provide a more comprehensive solution:

import re

def is_integer(string):
    return bool(re.fullmatch(r"-?\d+", string.strip()))

# Examples
print(is_integer("  -12345  "))  # True
print(is_integer("123abc"))     # False

This method handles leading/trailing whitespace, single negative signs, and strictly numeric strings, making it versatile and reliable.