How to round a float with a 0.25 step

[Alessandro Camel]

I need to round a float in this way:

2.0----> 2.0
2.125-> 2.0
2.25--> 2.25
2.375-> 2.25
2.5----> 2.5
2.625-> 2.5
2.75--> 2.75
2.875-> 2.75
3.0----> 3.0

Can someone help me please ?
I am writing a lot of code, but I am sure there is a "smart" way to do that!! ;)

Thanks!

[Hebert Coelho]

Maybe this will help:
http://www.java-forums.org/advanced-java/4130-rounding-double-two-decimal-places.html
http://www.java-tips.org/java-se-tips/java.math/how-to-round-a-double.html

[Campbell Ritchie]

Why are you using floats rather than doubles? Try multiplying by 4, adding 0.125, using the floor(double) method, and dividing by 4 again. You may, however, introduce imprecision into your results like that. Note you will end up with a double by that technique.

[Alessandro Camel]

I end up with something like this:

public class testMath {

	public static void main(String[] args) {
		
		Float step = 0.25f;
		Float[] myFloat = {2.0f, 2.125f, 2.25f , 2.375f, 2.5f , 2.625f , 2.75f , 2.875f , 3.0f };
	
		for (int i=0; i< myFloat.length; i++) {
			System.out.print(myFloat[i] + "\t");
			if ((myFloat[i] % step) >= (0.13)) {
				System.out.println( myFloat[i] + (step - (myFloat[i] % step)));
			} else
			{
				System.out.println( myFloat[i] - ((myFloat[i] % step)));
			}
		}
	}
}

output:

2.0	2.0
2.125	2.0
2.25	2.25
2.375	2.25
2.5	2.5
2.625	2.5
2.75	2.75
2.875	2.75
3.0	3.0

I am using float, because these are really simple numbers: the larger number used in the program could be 15! ;)

[Rob Spoor]

Multiply by 4, floor (not round), divide by 4 again. You'll get (original, *4, rounded, /4):

2.0     8.0   8.0  2.0
2.125   8.5   8.0  2.0
2.25    9.0   9.0  2.25
2.375   9.5   9.0  2.25
2.5    10.0  10.0  2.5
2.625  10.5  10.0  2.5
2.75   11.0  11.0  2.75
2.875  11.5  11.0  2.75
3.0    12.0  12.0  3.0

As for the imprecision that Campbell was talking about, I don't think that will happen since you're only working with multiples of 2. These can be represented 100% accurately in doubles (as long as neither the integer part becomes too large or the decimal part becomes too small).

[Campbell Ritchie]

You should still not use floats, even for small numbers. Where are you getting 0.13 from? That will not give the right answer at all.
You will get imprecision for numbers like 2.3, but presumably, as Rob suggests, the “floor”ing will get rid of that.
I think I should have said add 0.5, not 0.125. Sorry for that mistake.

[Rob Spoor]

I never said that flooring would get rid of precision problems (although it probably will). What I said is that because the decimal parts are all of a form of a*1/2 + b*1/4 + c*1/8, these can be represented in float / double with accurate precision, as long as the integer part doesn't cause it to be cut off.

As for float, because these numbers are so small, there will also be no problem with precision loss.

[Campbell Ritchie]

I think we actually agree, Rob. Only I am not expressing myself clearly.

[Jesper de Jong]

Here is another approach, using a NavigableSet:

NavigableSet<Float> set = new TreeSet<>();

set.addAll(Arrays.asList(2.0f, 2.25f, 2.5f, 2.75f, 3.0f));

System.out.printf("%f -> %f%n", 2.0f, set.floor(2.0f));
System.out.printf("%f -> %f%n", 2.125f, set.floor(2.125f));
System.out.printf("%f -> %f%n", 2.25f, set.floor(2.25f));
System.out.printf("%f -> %f%n", 2.375f, set.floor(2.375f));
System.out.printf("%f -> %f%n", 2.5f, set.floor(2.5f));
System.out.printf("%f -> %f%n", 2.625f, set.floor(2.625f));
System.out.printf("%f -> %f%n", 2.75f, set.floor(2.75f));
System.out.printf("%f -> %f%n", 2.875f, set.floor(2.875f));
System.out.printf("%f -> %f%n", 3.0f, set.floor(3.0f));