my_var = (int)my_var;
As simple as that. Basically you don't need it if the variable is int.
Answer from Zach P on Stack OverflowVideos
All integers that can be represented by floating point numbers have an exact representation. So you can safely use int on the result. Inexact representations occur only if you are trying to represent a rational number with a denominator that is not a power of two.
That this works is not trivial at all! It's a property of the IEEE floating point representation that int∘floor = ⌊⋅⌋ if the magnitude of the numbers in question is small enough, but different representations are possible where int(floor(2.3)) might be 1.
To quote from Wikipedia,
Any integer with absolute value less than or equal to 224 can be exactly represented in the single precision format, and any integer with absolute value less than or equal to 253 can be exactly represented in the double precision format.
Use int(your non integer number) will nail it.
print int(2.3) # "2"
print int(math.sqrt(5)) # "2"
Using Math.round() will round the float to the nearest integer.
Actually, there are different ways to downcast float to int, depending on the result you want to achieve:
(for int i, float f)
round (the closest integer to given float)
i = Math.round(f); f = 2.0 -> i = 2 ; f = 2.22 -> i = 2 ; f = 2.68 -> i = 3 f = -2.0 -> i = -2 ; f = -2.22 -> i = -2 ; f = -2.68 -> i = -3note: this is, by contract, equal to
(int) Math.floor(f + 0.5f)truncate (i.e. drop everything after the decimal dot)
i = (int) f; f = 2.0 -> i = 2 ; f = 2.22 -> i = 2 ; f = 2.68 -> i = 2 f = -2.0 -> i = -2 ; f = -2.22 -> i = -2 ; f = -2.68 -> i = -2ceil/floor (an integer always bigger/smaller than a given value if it has any fractional part)
i = (int) Math.ceil(f); f = 2.0 -> i = 2 ; f = 2.22 -> i = 3 ; f = 2.68 -> i = 3 f = -2.0 -> i = -2 ; f = -2.22 -> i = -2 ; f = -2.68 -> i = -2 i = (int) Math.floor(f); f = 2.0 -> i = 2 ; f = 2.22 -> i = 2 ; f = 2.68 -> i = 2 f = -2.0 -> i = -2 ; f = -2.22 -> i = -3 ; f = -2.68 -> i = -3
For rounding positive values, you can also just use (int)(f + 0.5), which works exactly as Math.Round in those cases (as per doc).
You can also use Math.rint(f) to do the rounding to the nearest even integer; it's arguably useful if you expect to deal with a lot of floats with fractional part strictly equal to .5 (note the possible IEEE rounding issues), and want to keep the average of the set in place; you'll introduce another bias, where even numbers will be more common than odd, though.
See
http://mindprod.com/jgloss/round.html
http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html
for more information and some examples.
To convert an integer to a float in Python you can use the following:
float_version = float(int_version)
The reason you are getting 0 is that Python 2 returns an integer if the mathematical operation (here a division) is between two integers. So while the division of 144 by 314 is 0.45~~~, Python converts this to integer and returns just the 0 by eliminating all numbers after the decimal point.
Alternatively you can convert one of the numbers in any operation to a float since an operation between a float and an integer would return a float. In your case you could write float(144)/314 or 144/float(314). Another, less generic code, is to say 144.0/314. Here 144.0 is a float so it’s the same thing.
Other than John's answer, you could also make one of the variable float, and the result will yield float.
>>> 144 / 314.0
0.4585987261146497