[python] 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?

You can use the round function to compute the value.

Yes in python as many have pointed when we compute the value of a cube root, it will give you an output with a little bit of error. To check if the value is a whole number you can use the following function:

def cube_integer(n):
    if round(n**(1.0/3.0))**3 == n:
        return True
    return False

But remember that int(n) is equivalent to math.floor and because of this if you find the int(41063625**(1.0/3.0)) you will get 344 instead of 345.

So please be careful when using int withe cube roots.

All the answers are good but a sure fire method would be

def whole (n):
     return (n*10)%10==0

The function returns True if it's a whole number else False....I know I'm a bit late but here's one of the interesting methods which I made...

Edit: as stated by the comment below, a cheaper equivalent test would be:

def whole(n):
    return n%1==0

Just a side info, is_integer is doing internally:

import math
isInteger = (math.floor(x) == x)

Not exactly in python, but the cpython implementation is implemented as mentioned above.

The above answers work for many cases but they miss some. Consider the following:

fl = sum([0.1]*10)  # this is 0.9999999999999999, but we want to say it IS an int

Using this as a benchmark, some of the other suggestions don't get the behavior we might want:

fl.is_integer() # False

fl % 1 == 0     # False

Instead try:

def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):
    return abs(a-b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)

def is_integer(fl):
    return isclose(fl, round(fl))

now we get:

is_integer(fl)   # True

isclose comes with Python 3.5+, and for other Python's you can use this mostly equivalent definition (as mentioned in the corresponding PEP)

You don't need to loop or to check anything. Just take a cube root of 12,000 and round it down:

r = int(12000**(1/3.0))
print r*r*r # 10648

You could use this:

if k == int(k):
    print(str(k) + " is a whole number!")

You can use a modulo operation for that.

if (n ** (1.0/3)) % 1 != 0:
    print("We have a decimal number here!")

>>> def is_near_integer(n, precision=8, get_integer=False):
...     if get_integer:
...         return int(round(n, precision))
...     else:
...         return round(n) == round(n, precision)
...
>>> print(is_near_integer(10648 ** (1.0/3)))
True
>>> print(is_near_integer(10648 ** (1.0/3), get_integer=True))
22
>>> for i in [4.9, 5.1, 4.99, 5.01, 4.999, 5.001, 4.9999, 5.0001, 4.99999, 5.000
01, 4.999999, 5.000001]:
...     print(i, is_near_integer(i, 4))
...
4.9 False
5.1 False
4.99 False
5.01 False
4.999 False
5.001 False
4.9999 False
5.0001 False
4.99999 True
5.00001 True
4.999999 True
5.000001 True
>>>

Wouldn't it be easier to test the cube roots? Start with 20 (20**3 = 8000) and go up to 30 (30**3 = 27000). Then you have to test fewer than 10 integers.

for i in range(20, 30):
    print("Trying {0}".format(i))
    if i ** 3 > 12000:
        print("Maximum integral cube root less than 12000: {0}".format(i - 1))
        break

How about

if x%1==0:
    print "is integer"

We can use the modulo (%) operator. This tells us how many remainders we have when we divide x by y - expresses as x % y. Every whole number must divide by 1, so if there is a remainder, it must not be a whole number.

This function will return a boolean, True or False, depending on whether n is a whole number.

def is_whole(n):
    return n % 1 == 0

Try using:

int(val) == val

It will give lot more precision than any other methods.