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.
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.
val = int(str(userInput)) if you want only integer values, not floats
int(True) == 1 and int(False) == 0Apparently this will not work for negative values, but it will for positive numbers.
Use isdigit()
if userinput.isdigit(): #do stuff 
isdecimal not isdigit because isdigit is an unsafe test that recognises characters like unicode power-of-2, ² , as a digit that can not be converted to integers.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.
"123456".isnumeric()) yields AttributeError: 'str' object has no attribute 'isnumeric', while U"12345".numeric() = True
a = '\U0001f10a'. a.isnumeric() is True, but int(a) raises a ValueError.
'3.3'.isnumeric() is FalseFor 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 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.
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)) 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) 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 
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
num_format = re.compile(r'^\-?[1-9]+\.?[0-9]*')
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 
float, rather than a bool, if the conversion doesn't fail.
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: ")) 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 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") 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: 
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 
0, one of the examples specifically mentioned by the OP.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 
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 
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. >>> 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.
ValueError.
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() 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) 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 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))) 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 Here is the simplest solution:
a= input("Choose the option\n") if(int(a)): print (a); else: print("Try Again") 
SyntaxError: 'return' outside function. Also, a is int will never evaluate to True.
Checking for Decimal type:
import decimal isinstance(x, decimal.Decimal) 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!") 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 
if type(eval(user_input)) == int:this might work.