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?

14 Answers

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) 

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 

You could use this:

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

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 can use a modulo operation for that.

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

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" 

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)

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.

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 

>>> 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 >>> 

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 

Try using:

int(val) == val 

It will give lot more precision than any other methods.

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.