How to check if a float value is a whole number

Question

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?

Answer 1

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)

Answer 2

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

Answer 3

You could use this:

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

Answer 4

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

Answer 5

You can use a modulo operation for that.

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

Answer 6

How about

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

Answer 7

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

Answer 8

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)

Answer 9

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.

Answer 10

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

Answer 11

You can use something like:

num = 1.9899
bool(int(num)-num)
#returns True

if it is True, It means it holds some value, hence not a whole number. Else

num = 1.0
bool(int(num)-num)
# returns False

Answer 12

>>> 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
>>>

Answer 13

This problem has been solved, but I would like to propose an additional mathematical-based solution for funcies.

The benefit of this approach is that it calculates the whole number part of your number, which may be beneficial depending on your general task.

Algorithm:

• Decompose whole number part of your number its a sum of its decimals (e.g., 327=3*100+2*10+7*1)
• take difference between calculated whole number and number itself
• decide whether difference is close enough to be considered an integer.

from math import ceil, log, isclose

def is_whole(x: float) -> bool:
    n_digits = ceil(log(x,10)) # number of digits of decimals at or above ones
    digits = [(n//(10**i))%10 for i in range(n_digits)] # parse digits of `x` at or above ones decimal
    whole = 0 # will equal the whole number part of `x`
    for i in range(n_digits):
        decimal = 10**i
        digit = digits[i]
        whole += digit*decimal
    
    diff = whole - x
    return isclose(diff, 0.0)

NOTE: the idea of parsing digits of a number was realized from here

Answer 14

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.

Answer 15

Try using:

int(val) == val

It will give lot more precision than any other methods.