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

Question

Is there any way to tell whether a string represents an integer (e.g., \'3\', \'-17\' but not \'3.14\' or \'asf

Answer 1

with positive integers you could use .isdigit:

>>> '16'.isdigit()
True

it doesn't work with negative integers though. suppose you could try the following:

>>> s = '-17'
>>> s.startswith('-') and s[1:].isdigit()
True

it won't work with '16.0' format, which is similar to int casting in this sense.

edit:

def check_int(s):
    if s[0] in ('-', '+'):
        return s[1:].isdigit()
    return s.isdigit()

Answer 2

I suggest the following:

import ast

def is_int(s):
    return isinstance(ast.literal_eval(s), int)

From the docs:

Safely evaluate an expression node or a string containing a Python literal or container display. The string or node provided may only consist of the following Python literal structures: strings, bytes, numbers, tuples, lists, dicts, sets, booleans, and None.

I should note that this will raise a ValueError exception when called against anything that does not constitute a Python literal. Since the question asked for a solution without try/except, I have a Kobayashi-Maru type solution for that:

from ast import literal_eval
from contextlib import suppress

def is_int(s):
    with suppress(ValueError):
        return isinstance(literal_eval(s), int)
    return False

¯\_(ツ)_/¯

Answer 3

str.isdigit() should do the trick.

Examples:

str.isdigit("23") ## True
str.isdigit("abc") ## False
str.isdigit("23.4") ## False

EDIT: As @BuzzMoschetti pointed out, this way will fail for minus number (e.g, "-23"). In case your input_num can be less than 0, use re.sub(regex_search,regex_replace,contents) before applying str.isdigit(). For example:

import re
input_num = "-23"
input_num = re.sub("^-", "", input_num) ## "^" indicates to remove the first "-" only
str.isdigit(input_num) ## True

Answer 4

Uh.. Try this:

def int_check(a):
    if int(a) == a:
        return True
    else:
        return False

This works if you don't put a string that's not a number.

And also (I forgot to put the number check part. ), there is a function checking if the string is a number or not. It is str.isdigit(). Here's an example:

a = 2
a.isdigit()

If you call a.isdigit(), it will return True.

Answer 5

def is_int_or_float(x):
    try:    
        if int(float(x)):
            if float(x).is_integer():  return 'int'
            else:                      return 'float'
    except ValueError:  return False


for  i  in ['*', '-1', '01', '1.23', '+0011.002', ]:
    print(i, is_int_or_float(i))

# * False
# -1 int
# 01 int
# 1.23 float
# +0011.002 float

Answer 6

If you're really just annoyed at using try/excepts all over the place, please just write a helper function:

def RepresentsInt(s):
    try: 
        int(s)
        return True
    except ValueError:
        return False

>>> print RepresentsInt("+123")
True
>>> print RepresentsInt("10.0")
False

It's going to be WAY more code to exactly cover all the strings that Python considers integers. I say just be pythonic on this one.