Check if a number is int or float [duplicate]

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! 

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

(note: this will return True for type bool, at least in cpython, which may not be what you want. Thank you commenters.)

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 

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" 

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" 

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.

I am not sure why this hasn't been proposed before, but how about using the built-in Python method on a float called is_integer()? Basically you could give it some number cast as a float, and ask whether it is an integer or not. For instance:

>>> (-13.0).is_integer() True >>> (3.14).is_integer() False 

For completeness, however, consider:

isinstance(i, int) or i.is_integer() 

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 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)

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) 

Use the most basic of type inference that python has:

>>> # Float Check >>> myNumber = 2.56 >>> print(type(myNumber) == int) False >>> print(type(myNumber) == float) True >>> print(type(myNumber) == bool) False >>> >>> # Integer Check >>> myNumber = 2 >>> print(type(myNumber) == int) True >>> print(type(myNumber) == float) False >>> print(type(myNumber) == bool) False >>> >>> # Boolean Check >>> myNumber = False >>> print(type(myNumber) == int) False >>> print(type(myNumber) == float) False >>> print(type(myNumber) == bool) True >>> 

Easiest and Most Resilient Approach in my Opinion

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