How can I check if string input is a number?

Question

How do I check if a user's string input is a number (e.g., -1, 0, 1, etc.)?

user_input = input("Enter something:")

if type(user_input) == int:
    print("Is a number")
else:
    print("Not a number")

The above won't work since input always returns a string.

Answer 1

Simply try converting it to an int and then bailing out if it doesn't work.

try:
    val = int(userInput)
except ValueError:
    print("That's not an int!")

See Handling Exceptions in the official tutorial.

Answer 2

Apparently this will not work for negative values, but it will for positive numbers.

Use isdigit()

if userinput.isdigit():
    #do stuff

Answer 3

The method isnumeric() will do the job:

>>>a = '123'
>>>a.isnumeric()
True

But remember:

>>>a = '-1'
>>>a.isnumeric()
False

isnumeric() returns True if all characters in the string are numeric characters, and there is at least one character.

So negative numbers are not accepted.

Answer 4

For Python 3 the following will work.

userInput = 0
while True:
  try:
     userInput = int(input("Enter something: "))       
  except ValueError:
     print("Not an integer!")
     continue
  else:
     print("Yes an integer!")
     break 

Answer 5

EDITED: You could also use this below code to find out if its a number or also a negative

import re
num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$")
isnumber = re.match(num_format,givennumber)
if isnumber:
    print "given string is number"

you could also change your format to your specific requirement. I am seeing this post a little too late.but hope this helps other persons who are looking for answers :) . let me know if anythings wrong in the given code.

Answer 6

If you specifically need an int or float, you could try "is not int" or "is not float":

user_input = ''
while user_input is not int:
    try:
        user_input = int(input('Enter a number: '))
        break
    except ValueError:
        print('Please enter a valid number: ')

print('You entered {}'.format(user_input))

If you only need to work with ints, then the most elegant solution I've seen is the ".isdigit()" method:

a = ''
while a.isdigit() == False:
    a = input('Enter a number: ')

print('You entered {}'.format(a))

Answer 7

Works fine for check if an input is a positive Integer AND in a specific range

def checkIntValue():
    '''Works fine for check if an **input** is
   a positive Integer AND in a specific range'''
    maxValue = 20
    while True:
        try:
            intTarget = int(input('Your number ?'))
        except ValueError:
            continue
        else:
            if intTarget < 1 or intTarget > maxValue:
                continue
            else:
                return (intTarget)

Answer 8

natural: [0, 1, 2 ... ∞]

Python 2

it_is = unicode(user_input).isnumeric()

Python 3

it_is = str(user_input).isnumeric()

integer: [-∞, .., -2, -1, 0, 1, 2, ∞]

try:
    int(user_input)
    it_is = True
except ValueError:
    it_is = False
 

float: [-∞, .., -2, -1.0...1, -1, -0.0...1, 0, 0.0...1, ..., 1, 1.0...1, ..., ∞]

try:
    float(user_input)
    it_is = True
except ValueError:
    it_is = False

Answer 9

I would recommend this, @karthik27, for negative numbers

import re
num_format = re.compile(r'^\-?[1-9][0-9]*\.?[0-9]*')

Then do whatever you want with that regular expression, match(), findall() etc

Answer 10

The most elegant solutions would be the already proposed,

a = 123
bool_a = a.isnumeric()

Unfortunately, it doesn't work neither for negative integers nor for general float values of a. If your point is to check if 'a' is a generic number beyond integers, I'd suggest the following one, which works for every kind of float and integer :). Here is the test:

def isanumber(a):

    try:
        float(repr(a))
        bool_a = True
    except:
        bool_a = False

    return bool_a


a = 1 # Integer
isanumber(a)
>>> True

a = -2.5982347892 # General float
isanumber(a)
>>> True

a = '1' # Actually a string
isanumber(a)
>>> False

Answer 11

This solution will accept only integers and nothing but integers.

def is_number(s):
    while s.isdigit() == False:
        s = raw_input("Enter only numbers: ")
    return int(s)


# Your program starts here    
user_input = is_number(raw_input("Enter a number: "))

Answer 12

This works with any number, including a fraction:

import fractions

def isnumber(s):
   try:
     float(s)
     return True
   except ValueError:
     try: 
       Fraction(s)
       return True
     except ValueError: 
       return False

Answer 13

You can use the isdigit() method for strings. In this case, as you said the input is always a string:

    user_input = input("Enter something:")
    if user_input.isdigit():
        print("Is a number")
    else:
        print("Not a number")

Answer 14

I've been using a different approach I thought I'd share. Start with creating a valid range:

valid = [str(i) for i in range(-10,11)] #  ["-10","-9...."10"] 

Now ask for a number and if not in list continue asking:

p = input("Enter a number: ")

while p not in valid:
    p = input("Not valid. Try to enter a number again: ")

Lastly convert to int (which will work because list only contains integers as strings:

p = int(p)

Answer 15

Why not divide the input by a number? This way works with everything. Negatives, floats, and negative floats. Also Blank spaces and zero.

numList = [499, -486, 0.1255468, -0.21554, 'a', "this", "long string here", "455 street area", 0, ""]

for item in numList:

    try:
        print (item / 2) #You can divide by any number really, except zero
    except:
        print "Not A Number: " + item

Result:

249
-243
0.0627734
-0.10777
Not A Number: a
Not A Number: this
Not A Number: long string here
Not A Number: 455 street area
0
Not A Number: 

Answer 16

I know this is pretty late but its to help anyone else that had to spend 6 hours trying to figure this out. (thats what I did):

This works flawlessly: (checks if any letter is in the input/checks if input is either integer or float)

a=(raw_input("Amount:"))

try:
    int(a)
except ValueError:
    try:
        float(a)
    except ValueError:
        print "This is not a number"
        a=0


if a==0:
    a=0
else:
    print a
    #Do stuff

Answer 17

Here is a simple function that checks input for INT and RANGE. Here, returns 'True' if input is integer between 1-100, 'False' otherwise

def validate(userInput):

    try:
        val = int(userInput)
        if val > 0 and val < 101:
            valid = True
        else:
            valid = False

    except Exception:
        valid = False

    return valid

Answer 18

If you wanted to evaluate floats, and you wanted to accept NaNs as input but not other strings like 'abc', you could do the following:

def isnumber(x):
    import numpy
    try:
        return type(numpy.float(x)) == float
    except ValueError:
        return False

Answer 19

while True:
    b1=input('Type a number:')
    try:
        a1=float(b1)
    except ValueError:
        print('"%(a1)s" is not a number. Try again.' %{'a1':b1})       
    else:
        print ('You typed "{}".'.format(a1))
        break

This makes a loop to check whether input is an integer or not, result would look like below:

>>> %Run 1.1.py
Type a number:d
"d" is not a number. Try again.
Type a number:
>>> %Run 1.1.py
Type a number:4
You typed 4.
>>> 

Answer 20

I also ran into problems this morning with users being able to enter non-integer responses to my specific request for an integer.

This was the solution that ended up working well for me to force an answer I wanted:

player_number = 0
while player_number != 1 and player_number !=2:
    player_number = raw_input("Are you Player 1 or 2? ")
    try:
        player_number = int(player_number)
    except ValueError:
        print "Please enter '1' or '2'..."

I would get exceptions before even reaching the try: statement when I used

player_number = int(raw_input("Are you Player 1 or 2? ") 

and the user entered "J" or any other non-integer character. It worked out best to take it as raw input, check to see if that raw input could be converted to an integer, and then convert it afterward.

Answer 21

This will work:

print(user_input.isnumeric())

This checks if the string has only numbers in it and has at least a length of 1. However, if you try isnumeric with a string with a negative number in it, isnumeric will return False.

Now this is a solution that works for both negative and positive numbers

try:
    user_input = int(user_input)
except ValueError:
    process_non_numeric_user_input()  # user_input is not a numeric string!
else:
    process_user_input()

Answer 22

Looks like there's so far only two answers that handle negatives and decimals (the try... except answer and the regex one?). Found a third answer somewhere a while back somewhere (tried searching for it, but no success) that uses explicit direct checking of each character rather than a full regex.

Looks like it is still quite a lot slower than the try/exceptions method, but if you don't want to mess with those, some use cases may be better compared to regex when doing heavy usage, particularly if some numbers are short/non-negative:

>>> from timeit import timeit

On Python 3.10 on Windows shows representative results for me:

Explicitly check each character:

>>> print(timeit('text="1234"; z=text[0]; (z.isdigit() or z == "-" or z == ".") and all(character.isdigit() or character == "." for character in text[1:])'))
0.5673831000458449
>>> print(timeit('text="-4089175.25"; z=text[0]; (z.isdigit() or z == "-" or z == ".") and all(character.isdigit() or character == "." for character in text[1:])'))
1.0832774000009522
>>> print(timeit('text="-97271851234.28975232364"; z=text[0]; (z.isdigit() or z == "-" or z == ".") and all(character.isdigit() or character == "." for character in text[1:])'))
1.9836419000057504

A lot slower than the try/except:

>>> def exception_try(string):
...   try:
...     return type(float(string)) == int
...   except:
...     return false

>>> print(timeit('text="1234"; exception_try(text)', "from __main__ import exception_try"))
0.22721579996868968
>>> print(timeit('text="-4089175.25"; exception_try(text)', "from __main__ import exception_try"))
0.2409859000472352
>>> print(timeit('text="-97271851234.28975232364"; exception_try(text)', "from __main__ import exception_try"))
0.45190039998851717

But a fair bit quicker than regex, unless you have an extremely long string?

>>> print(timeit('import re'))
0.08660140004940331

(In case you're using it already)... and then:

>>> print(timeit('text="1234"; import re; num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$"); re.match(num_format,text)'))
1.3882658999646083
>>> print(timeit('text="-4089175.25"; import re; num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$"); re.match(num_format,text)'))
1.4007637000177056
>>> print(timeit('text="-97271851234.28975232364"; import re; num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$"); re.match(num_format,text)'))
1.4191589000402018

None are close to the simplest isdecimal, but that of course won't catch the negatives...

>>> print(timeit('text="1234"; text.isdecimal()'))
0.04747540003154427

Always good to have options depending on needs?

Answer 23

I have found that some Python libraries use assertions to make sure that the value supplied by the programmer-user is a number.

Sometimes it's good to see an example 'from the wild'. Using assert/isinstance:

def check_port(port):
    assert isinstance(port, int), 'PORT is not a number'
    assert port >= 0, 'PORT < 0 ({0})'.format(port)

Answer 24

I think not doing a simple thing in one line is not Pythonic.

A version without try..except, using a regex match:

Code:

import re

if re.match('[-+]?\d+$', the_str):
  # Is integer

Test:

>>> import re
>>> def test(s): return bool(re.match('[-+]?\d+$', s))

>>> test('0')
True
>>> test('1')
True
>>> test('-1')
True

>>> test('-0')
True
>>> test('+0')
True
>>> test('+1')
True


>>> test('-1-1')
False
>>> test('+1+1')
False

Answer 25

Try this! It worked for me even if I input negative numbers.

def length(s):
    return len(s)

s = input("Enter the string: ")
try:
    if (type(int(s))) == int:
        print("You input an integer")

except ValueError:
    print("it is a string with length " + str(length(s)))

Answer 26

We can use regex to validate the given input string is integer:

import re

def isInt(input_str: str):
    """
        Validate the given input is int type
        Arguments:
            input_str: Input string
    """
    if isinstance(input_str, int) or re.search("^[+-]?\d+$", input_str):
        print(f"{input_str} is integer type")
    else:
        print(f"{input_str} is not integer type")

Output:

In [41]: isInt("5435a")
5435a is not integer type

In [42]: isInt("5a435")
5a435 is not integer type

In [43]: isInt("+5435")
+5435 is integer type

In [44]: isInt("-5435")
-5435 is integer type

In [45]: isInt(5435)
5435 is integer type

In [46]: isInt(-5435)
-5435 is integer type

In [47]: isInt(+5435)
5435 is integer type

Answer 27

Here is the simplest solution:

a= input("Choose the option\n")

if(int(a)):
    print (a);
else:
    print("Try Again")

Answer 28

Checking for Decimal type:

import decimal
isinstance(x, decimal.Decimal)

Answer 29

You can type:

user_input = input("Enter something: ")

if type(user_input) == int:
    print(user_input, "Is a number")
else:
    print("Not a number")
  
try:
    val = int(user_input)
except ValueError:
    print("That's not an int!")

Answer 30

This is based on inspiration from an answer. I defined a function as below. It looks like it’s working fine.

def isanumber(inp):
    try:
        val = int(inp)
        return True
    except ValueError:
        try:
            val = float(inp)
            return True
        except ValueError:
            return False