How can I truncate a double to only two decimal pl

DecimalFormat df = new DecimalFormat(fmt);
df.setRoundingMode(RoundingMode.DOWN);
s = df.format(d);

Check available RoundingMode and DecimalFormat.

回答3:

Bit Old Forum, None of the above answer worked for both positive and negative values ( I mean for the calculation and just to do truncate without Rounding). From the How to round a number to n decimal places in Java link

private static BigDecimal truncateDecimal(double x,int numberofDecimals)
{
    if ( x > 0) {
        return new BigDecimal(String.valueOf(x)).setScale(numberofDecimals, BigDecimal.ROUND_FLOOR);
    } else {
        return new BigDecimal(String.valueOf(x)).setScale(numberofDecimals, BigDecimal.ROUND_CEILING);
    }
}

This method worked fine for me .

System.out.println(truncateDecimal(0, 2));
    System.out.println(truncateDecimal(9.62, 2));
    System.out.println(truncateDecimal(9.621, 2));
    System.out.println(truncateDecimal(9.629, 2));
    System.out.println(truncateDecimal(9.625, 2));
    System.out.println(truncateDecimal(9.999, 2));
    System.out.println(truncateDecimal(-9.999, 2));
    System.out.println(truncateDecimal(-9.0, 2));

Results :

0.00
9.62
9.62
9.62
9.62
9.99
-9.99
-9.00

回答4:

Note first that a double is a binary fraction and does not really have decimal places.

If you need decimal places, use a BigDecimal, which has a setScale() method for truncation, or use DecimalFormat to get a String.

回答5:

If, for whatever reason, you don't want to use a BigDecimal you can cast your double to an int to truncate it.

If you want to truncate to the Ones place:

simply cast to int

To the Tenths place:

multiply by ten
cast to int
cast back to double
and divide by ten.

Hundreths place

multiply and divide by 100 etc.

Example:

static double truncateTo( double unroundedNumber, int decimalPlaces ){
    int truncatedNumberInt = (int)( unroundedNumber * Math.pow( 10, decimalPlaces ) );
    double truncatedNumber = (double)( truncatedNumberInt / Math.pow( 10, decimalPlaces ) );
    return truncatedNumber;
}

In this example, decimalPlaces would be the number of places PAST the ones place you wish to go, so 1 would round to the tenths place, 2 to the hundredths, and so on (0 rounds to the ones place, and negative one to the tens, etc.)

回答6:

Maybe Math.floor(value * 100) / 100? Beware that the values like 3.54 may be not exactly represented with a double.

回答7:

You can use NumberFormat Class object to accomplish the task.

// Creating number format object to set 2 places after decimal point
NumberFormat nf = NumberFormat.getInstance();
nf.setMaximumFractionDigits(2);            
nf.setGroupingUsed(false);

System.out.println(nf.format(precision));// Assuming precision is a double type variable

回答8:

3.545555555 to get 3.54. Try Following for this:

    DecimalFormat df = new DecimalFormat("#.##");

    df.setRoundingMode(RoundingMode.FLOOR);

    double result = new Double(df.format(3.545555555);

This will give= 3.54!

回答9:

Here is the method I use:

double a=3.545555555; // just assigning your decimal to a variable
a=a*100;              // this sets a to 354.555555
a=Math.floor(a);      // this sets a to 354
a=a/100;              // this sets a to 3.54 and thus removing all your 5's

This can also be done:

a=Math.floor(a*100) / 100;

回答10:

Formating as a string and converting back to double i think will give you the result you want.

The double value will not be round(), floor() or ceil().

A quick fix for it could be:

 String sValue = (String) String.format("%.2f", oldValue);
 Double newValue = Double.parseDouble(sValue);

You can use the sValue for display purposes or the newValue for calculation.

回答11:

Maybe following :

double roundTwoDecimals(double d) { 
      DecimalFormat twoDForm = new DecimalFormat("#.##"); 
      return Double.valueOf(twoDForm.format(d));
}

回答12:

A quick check is to use the Math.floor method. I created a method to check a double for two or less decimal places below:

public boolean checkTwoDecimalPlaces(double valueToCheck) {

    // Get two decimal value of input valueToCheck 
    double twoDecimalValue = Math.floor(valueToCheck * 100) / 100;

    // Return true if the twoDecimalValue is the same as valueToCheck else return false
    return twoDecimalValue == valueToCheck;
}

回答13:

I have a slightly modified version of Mani's.

private static BigDecimal truncateDecimal(final double x, final int numberofDecimals) {
    return new BigDecimal(String.valueOf(x)).setScale(numberofDecimals, BigDecimal.ROUND_DOWN);
}

public static void main(String[] args) {
    System.out.println(truncateDecimal(0, 2));
    System.out.println(truncateDecimal(9.62, 2));
    System.out.println(truncateDecimal(9.621, 2));
    System.out.println(truncateDecimal(9.629, 2));
    System.out.println(truncateDecimal(9.625, 2));
    System.out.println(truncateDecimal(9.999, 2));
    System.out.println(truncateDecimal(3.545555555, 2));

    System.out.println(truncateDecimal(9.0, 2));
    System.out.println(truncateDecimal(-9.62, 2));
    System.out.println(truncateDecimal(-9.621, 2));
    System.out.println(truncateDecimal(-9.629, 2));
    System.out.println(truncateDecimal(-9.625, 2));
    System.out.println(truncateDecimal(-9.999, 2));
    System.out.println(truncateDecimal(-9.0, 2));
    System.out.println(truncateDecimal(-3.545555555, 2));

}

Output:

0.00
9.62
9.62
9.62
9.62
9.99
9.00
3.54
-9.62
-9.62
-9.62
-9.62
-9.99
-9.00
-3.54

回答14:

This worked for me:

double input = 104.8695412  //For example

long roundedInt = Math.round(input * 100);
double result = (double) roundedInt/100;

//result == 104.87

I personally like this version because it actually performs the rounding numerically, rather than by converting it to a String (or similar) and then formatting it.

回答15:

//if double_v is 3.545555555
String string_v= String.valueOf(double_v);
int pointer_pos = average.indexOf('.');//we find the position of '.'
string_v.substring(0, pointer_pos+2));// in this way we get the double with only 2 decimal in string form
double_v = Double.valueOf(string_v);//well this is the final result

well this might be a little awkward, but i think it can solve your problem :)