How to check if a float value is a whole number
To check if a float value is a whole number, use the
The method was added to the
float type in Python 2.6.
Take into account that in Python 2,
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 n27810
which means that anything over 3 cubed, (including 10648) was missed out due to the aforementioned imprecision:
4**3) ** (1.0/3)3.999999999999999610648 ** (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
int(12000 ** (1.0/3))2222 ** 310648
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)Trueisclose(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,
False, depending on whether
n is a whole number.
def is_whole(n): return n % 1 == 0