Python - Input Validation

This should do the job:

def valid_user_input(x):
    try:
        return int(x) > 2
    except ValueError:
        return False

maximum_number_input = input("Maximum Number: ")

while valid_user_input(maximum_number_input):
    maximum_number_input = input("Maximum Number: ")
    print("You have successfully entered a valid number")

Or even shorter:

def valid_user_input():
    try:
        return int(input("Maximum Number: ")) > 2
    except ValueError:
        return False

while valid_user_input():
    print('You have successfully entered a valid number')

My take:

from itertools import dropwhile
from numbers import Integral
from functools import partial
from ast import literal_eval

def value_if_type(obj, of_type=(Integral,)):
    try:
        value = literal_eval(obj)
        if isinstance(value, of_type):
            return value
    except ValueError:
        return None

inputs = map(partial(value_if_type), iter(lambda: input('Input int > 2'), object()))

gt2 = next(dropwhile(lambda L: L <= 2, inputs))

def take_user_in():
    try:
        return int(raw_input("Enter a value greater than 2 -> "))  # Taking user input and converting to string
    except ValueError as e:  # Catching the exception, that possibly, a inconvertible string could be given
        print "Please enter a number as" + str(e) + " as a number"
        return None


if __name__ == '__main__':  # Somethign akin to having a main function in Python

    # Structure like a do-whole loop
    # func()
    # while()
    #     func()
    var = take_user_in()  # Taking user data
    while not isinstance(var, int) or var < 2:  # Making sure that data is an int and more than 2
        var = take_user_in()  # Taking user input again for invalid input

    print "Thank you"  # Success

def check_value(some_value):
    try:
       y = int(some_value)
    except ValueError:
       return False
    return y > 2 

Hope this helps

import str
def validate(s):       
    return str.isdigit(s) and int(s) > 2

• str.isdidig() will eliminate all strings containing non-integers, floats ('.') and negatives ('-') (which are less than 2)
• int(user_input) confirms that it's an integer greater than 2
• returns True if both are True

This verifies that the input is an integer, but does reject values that look like integers (like 3.0):

def is_valid(x):
    return isinstance(x,int) and x > 2

x = 0
while not is_valid(x):
    # In Python 2.x, use raw_input() instead of input()
    x = input("Please enter an integer greater than 2: ")
    try:
        x = int(x)
    except ValueError:
        continue

The problem with using the int() built-in shown in other answers is that it will convert float and booleans to integers, so it's not really a check that your argument was an integer.

It's tempting to use the built-in isinstance(value, int) method on its own, but unfortunately, it will return True if passed a boolean. So here's my short and sweet Python 3.7 solution if you want strict type checking:

def is_integer(value):
    if isinstance(value, bool):
        return False
    else:
        return isinstance(value, int)

Results:

is_integer(True)  --> False
is_integer(False) --> False
is_integer(0.0)   --> False
is_integer(0)     --> True
is_integer((12))  --> True
is_integer((12,)) --> False
is_integer([0])   --> False

etc...