Round Double to 0.5

round double to 0.5

x = 13000 / 9000.0;

denominator = 2;
a.text = String(round(x*denominator )/denominator );

First convert 1.444 to 2.888, then round to 3.0 and then divide by 2 to get 1.5. In this case, the denominator of 0.5 is 2 (i.e. 1/2). If you want to round to nearest quarter (0.25,0.5, 0.75, 0.00), then denominator=4

I Should point out that this works perfectly if the denominator is a power of 2. If it is not, say denominator=3, then you can get weird answers like 1.99999999 instead of 2 for particular values.

How do I round to the nearest 0.5?

Multiply your rating by 2, then round using Math.Round(rating, MidpointRounding.AwayFromZero), then divide that value by 2.

Math.Round(value * 2, MidpointRounding.AwayFromZero) / 2

Java round to nearest .5

@SamiKorhonen said this in a comment:

Multiply by two, round and finally divide by two

So this is that code:

public static double roundToHalf(double d) {
    return Math.round(d * 2) / 2.0;
}

public static void main(String[] args) {
    double d1 = roundToHalf(1.1);
    double d2 = roundToHalf(1.3);
    double d3 = roundToHalf(2.5);
    double d4 = roundToHalf(3.223920);
    double d5 = roundToHalf(3);

    System.out.println(d1);
    System.out.println(d2);
    System.out.println(d3);
    System.out.println(d4);
    System.out.println(d5);
}

Output:

1.0
1.5
2.5
3.0
3.0

Round double to number in interval, defining by step

Java has instruments which can round numbers to n decimal places, see How to round a number to n decimal places in Java. For rounding to any interval you specified, you may have to manually use Math.round.

Formula:

Given an interval r and a double value x to round, a simple formula is:

• x_rounded = Math.round(x/r)*r;

Examples:

double x = 0.59999;
double r = 0.25; // Quarters
x = Math.round(x/r)*r;
System.out.println(x); // Result is 0.5

double x = 0.59999;
double r = 0.1; // Tenths
x = Math.round(x/r)*r;
System.out.println(x); // Result is approximately 0.6

double x = 0.31421;
double r = 0.125; // Eighths
x = Math.round(x/r)*r;
System.out.println(x); // Result is exactly 0.375

Proof:

• The interval r can be thought as the value of a fractional unit.
  
  
  • When r = 0.25, the fractional unit is a quarter.
• The value x/r represents the number of fractional units that make up x.
  
  
  • When x = 0.75, r = 0.25, x/r == 3, because x contains three fractional unit, which is the quarter. x/r represents the number of quarters.
• Math.round(x) rounds x to the nearest integral value. Similarly, Math.round(x/r) rounds x/r to the nearest integral multiple of that fraction.
  
  
  • For x = 0.7, r = 0.25, we have x/r = 2.8, representing 2.8 quarters. Math.round(x/r) therefore rounds the value to the nearest quarter, 3 quarters.
• Math.round(x/r)*r therefore rounds x to the nearest fractional interval r. The multiplier is needed because r is the value of each fractional unit.
  
  
  • For x = 0.7, r = 0.25, Math.round(x/r) represents 3 quarters. It has to be multiplied by r=0.25 to get the rounded value of x.

Rounding down a float to nearest 0.5 in Objective C / Cocoa Touch

floor() will round down floating point numbers to the next integer. If you first multiply by 2, round down, then divide by two, you will get the desired result.

float rounded = value < 0.5f ? 0.5f : floorf(value * 2) / 2;

How to round up and down double to the nearest interval in dart/flutter

• Multiply your number by two before any calculations.
• Use the desired round feature. (Floor or Ceiling).
• Divide the result by two again:

  double a = 134.78;
  print((2*a).floorToDouble()/2); // prints 134.5
  print((2*a).ceilToDouble()/2);// prints 135

Which is the desired result.

You can read more here.

Rounding of nearest 0.5

If it is required for 13.1, round to 13.5 and 13.9, round to 14.0, then:

double a = 13.1;
double rounded = Math.Ceil(a * 2) / 2;

How can I round to next floor 0.5 in Kotlin?

Multiply by 2, take the floor, then divide by 2 again.

There doesn't seem to be a built-in way, but you can always write it yourself:

fun Double.roundDownToMultipleOf(base: Double): Double = base * floor(this / base)

Use as 12.56.roundDownToMultipleOf(0.5).

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