Here’s how I did it:

inNumber = somenumber
inNumberint = int(inNumber)
if inNumber == inNumberint:
    print "this number is an int"
else:
    print "this number is a float"

Something like that.
Are there any nicer looking ways to do this?

Use isinstance.

>>> x = 12
>>> isinstance(x, int)
True
>>> y = 12.0
>>> isinstance(y, float)
True

So:

>>> if isinstance(x, int):
        print 'x is a int!'

x is a int!

_EDIT:_

As pointed out, in case of long integers, the above won’t work. So you need to do:

>>> x = 12L
>>> import numbers
>>> isinstance(x, numbers.Integral)
True
>>> isinstance(x, int)
False

I like @ninjagecko’s answer the most.

This would also work:

for Python 2.x

isinstance(n, (int, long, float)) 

Python 3.x doesn’t have long

isinstance(n, (int, float))

there is also type complex for complex numbers

One-liner:

isinstance(yourNumber, numbers.Real)

This avoids some problems:

>>> isinstance(99**10,int)
False

Demo:

>>> import numbers

>>> someInt = 10
>>> someLongInt = 100000L
>>> someFloat = 0.5

>>> isinstance(someInt, numbers.Real)
True
>>> isinstance(someLongInt, numbers.Real)
True
>>> isinstance(someFloat, numbers.Real)
True

It’s easier to ask forgiveness than ask permission. Simply perform the operation. If it works, the object was of an acceptable, suitable, proper type. If the operation doesn’t work, the object was not of a suitable type. Knowing the type rarely helps.

Simply attempt the operation and see if it works.

inNumber = somenumber
try:
    inNumberint = int(inNumber)
    print "this number is an int"
except ValueError:
    pass
try:
    inNumberfloat = float(inNumber)
    print "this number is a float"
except ValueError:
    pass

What you can do too is usingtype() Example:

if type(inNumber) == int : print "This number is an int"
elif type(inNumber) == float : print "This number is a float"

Here’s a piece of code that checks whether a number is an integer or not, it works for both Python 2 and Python 3.

import sys

if sys.version < '3':
    integer_types = (int, long,)
else:
    integer_types = (int,)

isinstance(yourNumber, integer_types)  # returns True if it's an integer
isinstance(yourNumber, float)  # returns True if it's a float

Notice that Python 2 has both types int and long, while Python 3 has only type int. Source.

If you want to check whether your number is a float that represents an int, do this

(isinstance(yourNumber, float) and (yourNumber).is_integer())  # True for 3.0

If you don’t need to distinguish between int and float, and are ok with either, then ninjagecko’s answer is the way to go

import numbers

isinstance(yourNumber, numbers.Real)

how about this solution?

if type(x) in (float, int):
    # do whatever
else:
    # do whatever

I know it’s an old thread but this is something that I’m using and I thought it might help.

It works in python 2.7 and python 3< .

def is_float(num):
    """
    Checks whether a number is float or integer

    Args:
        num(float or int): The number to check

    Returns:
        True if the number is float
    """
    return not (float(num)).is_integer()


class TestIsFloat(unittest.TestCase):
    def test_float(self):
        self.assertTrue(is_float(2.2))

    def test_int(self):
        self.assertFalse(is_float(2))

You can use modulo to determine if x is an integer numerically. The isinstance(x, int) method only determines if x is an integer by type:

def isInt(x):
    if x%1 == 0:
        print "X is an integer"
    else:
        print "X is not an integer"

Tried in Python version 3.6.3 Shell

>>> x = 12
>>> import numbers
>>> isinstance(x, numbers.Integral)
True
>>> isinstance(x,int)
True

Couldn’t figure out anything to work for.

pls check this: import numbers

import math

a = 1.1 - 0.1
print a 

print isinstance(a, numbers.Integral)
print math.floor( a )
if (math.floor( a ) == a):
    print "It is an integer number"
else:
    print False

Although X is float but the value is integer, so if you want to check the value is integer you cannot use isinstance and you need to compare values not types.

absolute = abs(x)
rounded = round(absolute)
if absolute - rounded == 0:
  print 'Integer number'
else:
  print 'notInteger number'

Update: Try this

inNumber = [32, 12.5, 'e', 82, 52, 92, '1224.5', '12,53',
            10000.000, '10,000459', 
           'This is a sentance, with comma number 1 and dot.', '121.124']

try:

    def find_float(num):
        num = num.split('.')
        if num[-1] is not None and num[-1].isdigit():
            return True
        else:
            return False

    for i in inNumber:
        i = str(i).replace(',', '.')
        if '.' in i and find_float(i):
            print('This is float', i)
        elif i.isnumeric():
            print('This is an integer', i)
        else:
            print('This is not a number ?', i)

except Exception as err:
    print(err)

You can do it with simple if statement

To check for float

if type(a)==type(1.1)

To check for integer type

if type(a)==type(1)

def is_int(x):
  absolute = abs(x)
  rounded = round(absolute)
  if absolute - rounded == 0:
    print str(x) + " is an integer"
  else:
    print str(x) +" is not an integer"


is_int(7.0) # will print 7.0 is an integer

Try this…

def is_int(x):
  absolute = abs(x)
  rounded = round(absolute)
  return absolute - rounded == 0

variable.isnumeric checks if a value is an integer:

 if myVariable.isnumeric:
    print('this varibale is numeric')
 else:
    print('not numeric')