How to check if a float value is a whole number

I am trying to find the largest cube root that is a whole number, that is less than 12,000.

    processing = True
    n = 12000
    while processing:
        n -= 1
        if n ** (1/3) == #checks to see if this has decimals or not

I am not sure how to check if it is a whole number or not though! I could convert it to a string then use indexing to check the end values and see whether they are zero or not, that seems rather cumbersome though. Is there a simpler way?

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)

From: stackoverflow.com/q/21583758

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