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
>>> (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,
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
>>> 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)
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