Double value to round up in Java

Question

I have a double value = 1.068879335 i want to round it up with only two decimal values like 1.07.

I tried like this

DecimalFormat df=new         


        

Answer 1

You could try defining a new DecimalFormat and using it as a Double result to a new double variable.

Example given to make you understand what I just said.

double decimalnumber = 100.2397;
DecimalFormat dnf = new DecimalFormat( "#,###,###,##0.00" );
double roundednumber = new Double(dnf.format(decimalnumber)).doubleValue();

Answer 2

Note the comma in your string: "1,07". DecimalFormat uses a locale-specific separator string, while Double.parseDouble() does not. As you happen to live in a country where the decimal separator is ",", you can't parse your number back.

However, you can use the same DecimalFormat to parse it back:

DecimalFormat df=new DecimalFormat("0.00");
String formate = df.format(value); 
double finalValue = (Double)df.parse(formate) ;

But you really should do this instead:

double finalValue = Math.round( value * 100.0 ) / 100.0;

Note: As has been pointed out, you should only use floating point if you don't need a precise control over accuracy. (Financial calculations being the main example of when not to use them.)

Answer 3

Live @Sergey's solution but with integer division.

double value = 23.8764367843;
double rounded = (double) Math.round(value * 100) / 100;
System.out.println(value +" rounded is "+ rounded);

prints

23.8764367843 rounded is 23.88

EDIT: As Sergey points out, there should be no difference between multipling double*int and double*double and dividing double/int and double/double. I can't find an example where the result is different. However on x86/x64 and other systems there is a specific machine code instruction for mixed double,int values which I believe the JVM uses.

for (int j = 0; j < 11; j++) {
    long start = System.nanoTime();
    for (double i = 1; i < 1e6; i *= 1.0000001) {
        double rounded = (double) Math.round(i * 100) / 100;
    }
    long time = System.nanoTime() - start;
    System.out.printf("double,int operations %,d%n", time);
}
for (int j = 0; j < 11; j++) {
    long start = System.nanoTime();
    for (double i = 1; i < 1e6; i *= 1.0000001) {
        double rounded = (double) Math.round(i * 100.0) / 100.0;
    }
    long time = System.nanoTime() - start;
    System.out.printf("double,double operations %,d%n", time);
}

Prints

double,int operations 613,552,212
double,int operations 661,823,569
double,int operations 659,398,960
double,int operations 659,343,506
double,int operations 653,851,816
double,int operations 645,317,212
double,int operations 647,765,219
double,int operations 655,101,137
double,int operations 657,407,715
double,int operations 654,858,858
double,int operations 648,702,279
double,double operations 1,178,561,102
double,double operations 1,187,694,386
double,double operations 1,184,338,024
double,double operations 1,178,556,353
double,double operations 1,176,622,937
double,double operations 1,169,324,313
double,double operations 1,173,162,162
double,double operations 1,169,027,348
double,double operations 1,175,080,353
double,double operations 1,182,830,988
double,double operations 1,185,028,544

Answer 4

Try this: org.apache.commons.math3.util.Precision.round(double x, int scale)

See: http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/util/Precision.html

Apache Commons Mathematics Library homepage is: http://commons.apache.org/proper/commons-math/index.html

The internal implemetation of this method is:

public static double round(double x, int scale) {
    return round(x, scale, BigDecimal.ROUND_HALF_UP);
}

public static double round(double x, int scale, int roundingMethod) {
    try {
        return (new BigDecimal
               (Double.toString(x))
               .setScale(scale, roundingMethod))
               .doubleValue();
    } catch (NumberFormatException ex) {
        if (Double.isInfinite(x)) {
            return x;
        } else {
            return Double.NaN;
        }
    }
}