Checking whether a variable is an integer or not [duplicate]

Answer 1

Why not try something like:

if x%1 == 0: 

Answer 2

Here is a simple example how you can determine an integer

def is_int(x):
    print round(x),
    if x == round(x):
        print 'True',
    else:
        print 'False'

is_int(7.0)   # True
is_int(7.5)   # False
is_int(-1)    # True    

Answer 3

If you are reading from a file and you have an array or dictionary with values of multiple datatypes, the following will be useful. Just check whether the variable can be type casted to int(or any other datatype you want to enforce) or not.

try :
    int(a);
    #Variable a is int
except ValueError : 
    # Variable a is not an int

Answer 4

If you need to do this, do

isinstance(<var>, int)

unless you are in Python 2.x in which case you want

isinstance(<var>, (int, long))

Do not use type. It is almost never the right answer in Python, since it blocks all the flexibility of polymorphism. For instance, if you subclass int, your new class should register as an int, which type will not do:

class Spam(int): pass
x = Spam(0)
type(x) == int # False
isinstance(x, int) # True

This adheres to Python's strong polymorphism: you should allow any object that behaves like an int, instead of mandating that it be one.

BUT

The classical Python mentality, though, is that it's easier to ask forgiveness than permission. In other words, don't check whether x is an integer; assume that it is and catch the exception results if it isn't:

try:
    x += 1
except TypeError:
    ...

This mentality is slowly being overtaken by the use of abstract base classes, which let you register exactly what properties your object should have (adding? multiplying? doubling?) by making it inherit from a specially-constructed class. That would be the best solution, since it will permit exactly those objects with the necessary and sufficient attributes, but you will have to read the docs on how to use it.

Answer 5

If the variable is entered like a string (e.g. '2010'):

if variable and variable.isdigit():
    return variable #or whatever you want to do with it. 
else: 
    return "Error" #or whatever you want to do with it.

Before using this I worked it out with try/except and checking for (int(variable)), but it was longer code. I wonder if there's any difference in use of resources or speed.

Answer 6

#!/usr/bin/env python

import re

def is_int(x):

    if(isinstance(x,(int,long))):

        return True
    matchObj = re.match(r'^-?\d+\.(\d+)',str(x))

        if matchObj:

        x = matchObj.group(1)

        if int(x)-0==0:

            return True

     return False

print is_int(6)

print is_int(1.0)

print is_int(1.1)

print is_int(0.1)

print is_int(-956.0)