Checking whether the user input is a number

Question

I'm trying to compare a input that the user receives that checks if the input is actually a number. Here's what I have so far:

numstring = input("Choose a number between 1 and 10")

and then to compare I have it in a method like this:

def check(a):
   if int(a) == int:
       print("It's a number!")
   else:
       print("It's not a number!")

It always prints out that it's not a number, even though it is.

Goal of this program: I'm making a guessing game:

import random

numstring = input("Choose a number between 1 and 10")

def check(a):
    if int(a) == int:
        print("It's a number!")
    else:
        print("It's not a number!")

    if abs(int(a)) < 1:
        print("You chose a number less than one!")
    elif abs(int(a)) > 10:
        print("You chose a number more than ten!")
    else:
        randomnum = random.randint(1, 10)
        randomnumstring = str(abs(randomnum))
        if randomnum == abs(int(a)):
            print("You won! You chose " + a + "and the computer chose " + randomnumstring)
        else:
            print("You lost! You chose " + a + " and the computer chose " + randomnumstring)


check(numstring)

Thanks for any help! The actual code works, but it just fails at checking it.

@AshwiniChaudhary. He is using `print()` function. So, I guess 3. — Rohit Jain, Aug 18 '13 at 19:26
@RohitJain `print()` works in py2 as well, but prints tuples. — Ashwini Chaudhary, Aug 18 '13 at 19:27
@AshwiniChaudhary. Yeah. Thanks for the info. I know that :) — Rohit Jain, Aug 18 '13 at 19:29

Rohit Jain · Answer 1 · 2013-08-18T19:33:52.910

10

You can just use str.isdigit() method:

def check(a):
    if a.isdigit(): 
        print("It's a number!")
    else:
        print("It's not a number!")

The above approach would fail for negative number input. '-5'.isdigit() gives False. So, an alternative solution is to use try-except:

try:
    val = int(a)
    print("It's a number!")
except ValueError:
    print("It's not a number!")

edited Aug 18 '13 at 19:33

answered Aug 18 '13 at 19:17

Rohit Jain

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Ashwini Chaudhary · Answer 2 · 2013-08-18T19:52:59.850

int(a) and int are not equal because int() returns an integer and just int with no () is a type in python2.(or class in python3)

>>> a = '10'
>>> int(a)
10
>>> int
<type 'int'>

If you're only dealing with integers then use try-except with int as others have shown. To deal with any type of number you can use ast.literal_eval with try-except:

>>> from ast import literal_eval
>>> from numbers import Number
def check(a):
    try:
        return isinstance(literal_eval(a), Number)
    except ValueError:
        return False

>>> check('foo')
False
>>> check('-100')
True
>>> check('1.001')
True
>>> check('1+1j')
True
>>> check('1e100')
True

`isinstance(int("a"), int)` will throw exception. And `'-1'.isdigit()` will return `False`. — Rohit Jain, Aug 18 '13 at 19:37

score 4 · Answer 3 · edited May 23 '17 at 12:09

4

Try to cast an input to int in try/except block:

numstring = input("Choose a number between 1 and 10")

try:
    a = int(numstring)
    print("It's a number!")
except ValueError:
    print("It's not a number!")

If you are using python 2, consider switching to raw_input() instead of input(). input() accepts any python expression and this is bad and unsafe, like eval(). raw_input() just reads the input as a string:

numstring = raw_input("Choose a number between 1 and 10")

try:
    a = int(numstring)
    print("It's a number!")
except ValueError:
    print("It's not a number!")

Note that raw_input() has been renamed to input() in python 3.

Also see:

Hope that helps.

edited May 23 '17 at 12:09

Community

1
1

answered Aug 18 '13 at 19:17

alecxe

462,703
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1,195

May be OP is using py3. – Ashwini Chaudhary Aug 18 '13 at 19:26
@AshwiniChaudhary yeah, thanks, I've added a link to the relevant thread. – alecxe Aug 18 '13 at 19:27
Your first approach will fail. `isinstance("5", int)` will return `False`. And `isinstance(int("a"), int)` will throw exception. – Rohit Jain Aug 18 '13 at 19:38
@RohitJain sure, it's fragile. But if the OP is using python 2 and the input is 5 - it'll work. – alecxe Aug 18 '13 at 19:40
@RohitJain "5" is not of `int` type - I understand. But, I mean that `input()` will return 5 (int), not "5" - it evaluates the input, reads it as an expression in python 2. – alecxe Aug 18 '13 at 19:44
@RohitJain but, anyway, thank you - I've removed `isinstance` option in order to avoid confusion. – alecxe Aug 18 '13 at 19:47

Checking whether the user input is a number

3 Answers3