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

To check if a float value is a whole number, use the float.is_integer() method:

>>> (1.0).is_integer()
True
>>> (1.555).is_integer()
False

The method was added to the float type in Python 2.6.

Take into account that in Python 2, 1/3 is 0 (floor division for integer operands!), and that floating point arithmetic can be imprecise (a float is an approximation using binary fractions, not a precise real number). But adjusting your loop a little this gives:

>>> for n in range(12000, -1, -1):
...     if (n ** (1.0/3)).is_integer():
...         print n
... 
27
8
1
0

which means that anything over 3 cubed, (including 10648) was missed out due to the aforementioned imprecision:

>>> (4**3) ** (1.0/3)
3.9999999999999996
>>> 10648 ** (1.0/3)
21.999999999999996

You'd have to check for numbers close to the whole number instead, or not use float() to find your number. Like rounding down the cube root of 12000:

>>> int(12000 ** (1.0/3))
22
>>> 22 ** 3
10648

If you are using Python 3.5 or newer, you can use the math.isclose() function to see if a floating point value is within a configurable margin:

>>> from math import isclose
>>> isclose((4**3) ** (1.0/3), 4)
True
>>> isclose(10648 ** (1.0/3), 22)
True

For older versions, the naive implementation of that function (skipping error checking and ignoring infinity and NaN) as mentioned in PEP485:

def isclose(a, b, rel_tol=1e-9, abs_tol=0.0):
    return abs(a - b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)

The reason you get this error is because you divide the integer 10 by 3 using integer division, getting the integral number 3 in the form of an int instance as a result. You then try to call the method is_integer() on that result but that method is in the float class and not in the int class, just as the error message says.

A quick fix would be to change your code and divide by 3.0 instead of 3 which would result in floating point division and give you a float instance on which you can call the is_integer() method like you are trying to. Do this:

n = 10
((n/3.0)).is_integer()