Check if a number is int or float

Question

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"

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"