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Implement IsNumber() function in Python
  • Last Updated : 06 May, 2021

In this article we will see how to implement isNumber() method using Python. This method takes in a string as input and returns True or False according to whether the string is a number or not. 

Input :  "12345"
Output :  True

Input : "-12345" 
Output : True 

Input : "abc"
Output : False

Approach : 
Here we will take advantage of the int() built-in function available in Python. Also with this approach we will see how Exception Handling comes to our aid. Using try-catch construct we try to convert the string to integer. In case the string cannot be converted the handler catches the exception that is thrown.
Below is the Implementation: 


# Implementation of isNumber() function
def isNumber(s):
    # handle for negative values
    negative = False
    if(s[0] =='-'):
        negative = True
    if negative == True:
        s = s[1:]
    # try to convert the string to int
        n = int(s)
        return True
    # catch exception if cannot be converted
    except ValueError:
        return False
s1 = "9748513"
s2 = "-9748513"
s3 = "GeeksforGeeks"
print("For input '{}' isNumber() returned : {}".format(s1, isNumber(s1)))
print("For input '{}' isNumber() returned : {}".format(s2, isNumber(s2)))
print("For input '{}' isNumber() returned : {}".format(s3, isNumber(s3)))

Output : 

For input '9748513' isNumber() returned : True
For input '-9748513' isNumber() returned : True
For input 'GeeksforGeeks' isNumber() returned : False

Note : This is not the only way to implement the isNumber() function but it is arguably the fastest way of doing so. Try/Catch doesn’t introduce much overhead because the most common exception is caught without an extensive search of stack frames.

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