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...

• python
• validation
• input
• python-3.x
• python-3.3