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

this is giving me this following exception

java.lang.NumberFormatException: For input string: "1,07"
     at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:1224)
     at java.lang.Double.parseDouble(Double.java:510)

can some one tell me what is wrong with my code.

finaly i need the finalValue = 1.07;

Answer 1

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 2

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 3

The problem is that you use a localizing formatter that generates locale-specific decimal point, which is "," in your case. But Double.parseDouble() expects non-localized double literal. You could solve your problem by using a locale-specific parsing method or by changing locale of your formatter to something that uses "." as the decimal point. Or even better, avoid unnecessary formatting by using something like this:

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

Answer 4

There is something fundamentally wrong with what you're trying to do. Binary floating-points values do not have decimal places. You cannot meaningfully round one to a given number of decimal places, because most "round" decimal values simply cannot be represented as a binary fraction. Which is why one should never use float or double to represent money.

So if you want decimal places in your result, that result must either be a String (which you already got with the DecimalFormat), or a BigDecimal (which has a setScale() method that does exactly what you want). Otherwise, the result cannot be what you want it to be.

Read The Floating-Point Guide for more information.

Answer 5

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;
        }
    }
}

Answer 6

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 7

This is not possible in the requested way because there are numbers with two decimal places which can not be expressed exactly using IEEE floating point numbers (for example 1/10 = 0.1 can not be expressed as a Double or Float). The formatting should always happen as the last step before presenting the result to the user.

I guess you are asking because you want to deal with monetary values. There is no way to do this reliably with floating-point numbers, you shoud consider switching to fixed-point arithmetics. This probably means doing all calculations in "cents" instead of "dollars".

Answer 8

double TotalPrice=90.98989898898;

  DecimalFormat format_2Places = new DecimalFormat("0.00");

    TotalPrice = Double.valueOf(format_2Places.format(TotalPrice));

Answer 9

You can use format like here,

  public static double getDoubleValue(String value,int digit){
    if(value==null){
        value="0";
     }
    double i=0;
     try {
         DecimalFormat digitformat = new DecimalFormat("#.##");
         digitformat.setMaximumFractionDigits(digit);
        return Double.valueOf(digitformat.format(Double.parseDouble(value)));

    } catch (NumberFormatException numberFormatExp) {
        return i;   
    }
}

Answer 10

If you do not want to use DecimalFormat (e.g. due to its efficiency) and you want a general solution, you could also try this method that uses scaled rounding:

public static double roundToDigits(double value, int digitCount) {
    if (digitCount < 0)
        throw new IllegalArgumentException("Digit count must be positive for rounding!");

    double factor = Math.pow(10, digitCount);
    return (double)(Math.round(value * factor)) / factor;
}