chopper draw lion4 - 1 year ago 111

Python 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

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
```

Source (Stackoverflow)