How can I check if string input is a number?

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

Use isdigit()

if userinput.isdigit():
    #do stuff

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 

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

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)

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.

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

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.

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

The method isnumeric() will do the job (Documentation for python3.x):

>>>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.

:):):)

a=10

isinstance(a,int)  #True

b='abc'

isinstance(b,int)  #False

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

Here is the simplest solution:

a= input("Choose the optionn")

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

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

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

while True:
    b1=input('Type a number:')
    try:
        a1=int(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.
>>> 

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

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")

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))

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)

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: 

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

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: "))

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)))

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()

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!")

Checking for Decimal type:

import decimal
isinstance(x, decimal.Decimal)

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

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?

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)