How to check if string input is a number?

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.

Answers:

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Method 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.

Method 2

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

Use isdigit()

if userinput.isdigit():
    #do stuff

Method 3

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.

:):):)

Method 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

Method 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.

Method 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))

Method 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)

Method 8

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

Method 9

the most elegant solutions would be the already proposed,

a=123
bool_a = a.isnumeric()

Unfortunatelly it doesn’t work both for negative integers and 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

I hope you find it useful 🙂

Method 10

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

Method 11

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")

Method 12

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: "))

Method 13

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

Method 14

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:

Method 15

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

Method 16

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

Method 17

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)

Method 18

Here is the simplest solution:

a= input("Choose the optionn")

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

Method 19

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.
>>>

Method 20

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

Method 21

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.

Method 22

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)))

Method 23

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!")

Method 24

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()

Method 25

Checking for Decimal type:

import decimal
isinstance(x, decimal.Decimal)

Method 26

I think not done simple thing in one line is not pythonnic.

A version without try..except, use 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

Method 27

Based on inspiration from answer. I defined a function as below. Looks like its working fine. Please let me know if you find any issue

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

Method 28

a=10

isinstance(a,int)  #True

b='abc'

isinstance(b,int)  #False

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0