How to round a number to n decimal places in Java

问题:

What I would like is a method to convert a double to a string which rounds using the half-up method - i.e. if the decimal to be rounded is 5, it always rounds up to the next number. This is the standard method of rounding most people expect in most situations.

I also would like only significant digits to be displayed - i.e. there should not be any trailing zeroes.

I know one method of doing this is to use the String.format method:

String.format("%.5g%n", 0.912385);

returns:

0.91239

which is great, however it always displays numbers with 5 decimal places even if they are not significant:

String.format("%.5g%n", 0.912300);

returns:

0.91230

Another method is to use the DecimalFormatter:

DecimalFormat df = new DecimalFormat("#.#####");
df.format(0.912385);

returns:

0.91238

However as you can see this uses half-even rounding. That is it will round down if the previous digit is even. What I'd like is this:

0.912385 -> 0.91239
0.912300 -> 0.9123

What is the best way to achieve this in Java?

回答1:

Use setRoundingMode, set the RoundingMode explicitly to handle your issue with the half-even round, then use the format pattern for your required output.

Example:

DecimalFormat df = new DecimalFormat("#.####");
df.setRoundingMode(RoundingMode.CEILING);
for (Number n : Arrays.asList(12, 123.12345, 0.23, 0.1, 2341234.212431324)) {
    Double d = n.doubleValue();
    System.out.println(df.format(d));
}

gives the output:

12
123.1235
0.23
0.1
2341234.2125

---

EDIT: The original answer does not address the accuracy of the double values. That is fine if you don't care much whether it rounds up or down. But if you want accurate rounding, then you need to take the expected accuracy of the values into account. Floating point values have a binary representation internally. That means that a value like 2.7735 does not actually have that exact value internally. It can be slightly larger or slightly smaller. If the internal value is slightly smaller, then it will not round up to 2.7740. To remedy that situation, you need to be aware of the accuracy of the values that you are working with, and add or subtract that value before rounding. For example, when you know that your values are accurate up to 6 digits, then to round half-way values up, add that accuracy to the value:

Double d = n.doubleValue() + 1e-6;

To round down, subtract the accuracy.

回答2:

Assuming value is a double, you can do:

(double)Math.round(value * 100000d) / 100000d

That's for 5 digits precision. The number of zeros indicate the number of decimals.

回答3:

new BigDecimal(String.valueOf(double)).setScale(yourScale, BigDecimal.ROUND_HALF_UP);

will get you a BigDecimal. To get the string out of it, just call that BigDecimal's toString method, or the toPlainString method for Java 5+ for a plain format string.

Sample program:

package trials;
import java.math.BigDecimal;

public class Trials {

    public static void main(String[] args) {
        int yourScale = 10;
        System.out.println(BigDecimal.valueOf(0.42344534534553453453-0.42324534524553453453).setScale(yourScale, BigDecimal.ROUND_HALF_UP));
    }

回答4:

You can also use the

DecimalFormat df = new DecimalFormat("#.00000");
df.format(0.912385);

to make sure you have the trailing 0's.

回答5:

As some others have noted, the correct answer is to use either DecimalFormat or BigDecimal. Floating-point doesn't have decimal places so you cannot possibly round/truncate to a specific number of them in the first place. You have to work in a decimal radix, and that is what those two classes do.

I am posting the following code as a counter-example to all the answers in this thread and indeed all over StackOverflow (and elsewhere) that recommend multiplication followed by truncation followed by division. It is incumbent on advocates of this technique to explain why the following code produces the wrong output in over 92% of cases.

public class RoundingCounterExample
{

    static float roundOff(float x, int position)
    {
        float a = x;
        double temp = Math.pow(10.0, position);
        a *= temp;
        a = Math.round(a);
        return (a / (float)temp);
    }

    public static void main(String[] args)
    {
        float a = roundOff(0.0009434f,3);
        System.out.println("a="+a+" (a % .001)="+(a % 0.001));
        int count = 0, errors = 0;
        for (double x = 0.0; x < 1; x += 0.0001)
        {
            count++;
            double d = x;
            int scale = 2;
            double factor = Math.pow(10, scale);
            d = Math.round(d * factor) / factor;
            if ((d % 0.01) != 0.0)
            {
                System.out.println(d + " " + (d % 0.01));
                errors++;
            }
        }
        System.out.println(count + " trials " + errors + " errors");
    }
}

Output of this program:

10001 trials 9251 errors

EDIT: To address some comments below I redid the modulus part of the test loop using BigDecimal and new MathContext(16) for the modulus operation as follows:

public static void main(String[] args)
{
    int count = 0, errors = 0;
    int scale = 2;
    double factor = Math.pow(10, scale);
    MathContext mc = new MathContext(16, RoundingMode.DOWN);
    for (double x = 0.0; x < 1; x += 0.0001)
    {
        count++;
        double d = x;
        d = Math.round(d * factor) / factor;
        BigDecimal bd = new BigDecimal(d, mc);
        bd = bd.remainder(new BigDecimal("0.01"), mc);
        if (bd.multiply(BigDecimal.valueOf(100)).remainder(BigDecimal.ONE, mc).compareTo(BigDecimal.ZERO) != 0)
        {
            System.out.println(d + " " + bd);
            errors++;
        }
    }
    System.out.println(count + " trials " + errors + " errors");
}

Result:

10001 trials 4401 errors

回答6:

Suppose you have

double d = 9232.129394d;

you can use BigDecimal

BigDecimal bd = new BigDecimal(d).setScale(2, RoundingMode.HALF_EVEN);
d = bd.doubleValue();

or without BigDecimal

d = Math.round(d*100)/100.0d;

with both solutions d == 9232.13

回答7:

You can use the DecimalFormat class.

double d = 3.76628729;

DecimalFormat newFormat = new DecimalFormat("#.##");
double twoDecimal =  Double.valueOf(newFormat.format(d));

回答8:

Real's Java How-to posts this solution, which is also compatible for versions before Java 1.6.

BigDecimal bd = new BigDecimal(Double.toString(d));
bd = bd.setScale(decimalPlace, BigDecimal.ROUND_HALF_UP);
return bd.doubleValue();

回答9:

double myNum = .912385;
int precision = 10000; //keep 4 digits
myNum= Math.floor(myNum * precision +.5)/precision;

回答10:

@Milhous: the decimal format for rounding is excellent:

You can also use the

DecimalFormat df = new DecimalFormat("#.00000");
df.format(0.912385);

to make sure you have the trailing 0's.

I would add that this method is very good at providing an actual numeric, rounding mechanism - not only visually, but also when processing.

Hypothetical: you have to implement a rounding mechanism into a GUI program. To alter the accuracy / precision of a result output simply change the caret format (i.e. within the brackets). So that:

DecimalFormat df = new DecimalFormat("#0.######");
df.format(0.912385);

would return as output: 0.912385

DecimalFormat df = new DecimalFormat("#0.#####");
df.format(0.912385);

would return as output: 0.91239

DecimalFormat df = new DecimalFormat("#0.####");
df.format(0.912385);

would return as output: 0.9124

[EDIT: also if the caret format is like so ("#0.############") and you enter a decimal, e.g. 3.1415926, for argument's sake, DecimalFormat does not produce any garbage (e.g. trailing zeroes) and will return: 3.1415926 .. if you're that way inclined. Granted, it's a little verbose for the liking of some dev's - but hey, it's got a low memory footprint during processing and is very easy to implement.]

So essentially, the beauty of DecimalFormat is that it simultaneously handles the string appearance - as well as the level of rounding precision set. Ergo: you get two benefits for the price of one code implementation. ;)

回答11:

You could use the following utility method-

public static double round(double valueToRound, int numberOfDecimalPlaces)
{
    double multipicationFactor = Math.pow(10, numberOfDecimalPlaces);
    double interestedInZeroDPs = valueToRound * multipicationFactor;
    return Math.round(interestedInZeroDPs) / multipicationFactor;
}

回答12:

Here is a summary of what you can use if you want the result as String:

1. DecimalFormat#setRoundingMode():

   DecimalFormat df = new DecimalFormat("#.#####");
   df.setRoundingMode(RoundingMode.HALF_UP);
   String str1 = df.format(0.912385)); // 0.91239
   

2. BigDecimal#setScale()

   String str2 = new BigDecimal(0.912385)
       .setScale(5, BigDecimal.ROUND_HALF_UP)
       .toString();
   

Here is a suggestion of what libraries you can use if you want double as a result. I wouldn't recommend it for string conversion, though, as double may not be able to represent what you want exactly (see e.g. here):

1. Precision from Apache Commons Math

   double rounded = Precision.round(0.912385, 5, BigDecimal.ROUND_HALF_UP);
   

2. Functions from Colt

   double rounded = Functions.round(0.00001).apply(0.912385)
   

3. Utils from Weka

   double rounded = Utils.roundDouble(0.912385, 5)
   

回答13:

A succinct solution:

   public static double round(double value, int precision) {
      int scale = (int) Math.pow(10, precision);
      return (double) Math.round(value * scale) / scale;
  }

See also, https://stackoverflow.com/a/22186845/212950 Thanks to jpdymond for offering this.

回答14:

You can use BigDecimal

BigDecimal value = new BigDecimal("2.3");
value = value.setScale(0, RoundingMode.UP);
BigDecimal value1 = new BigDecimal("-2.3");
value1 = value1.setScale(0, RoundingMode.UP);
System.out.println(value + "n" + value1);

Refer: http://www.javabeat.net/precise-rounding-of-decimals-using-rounding-mode-enumeration/

回答15:

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;
        }
    }
}

回答16:

Since I found no complete answer on this theme I've put together a class that should handle this properly, with support for:

• Formatting: Easily format a double to string with a certain number of decimal places
• Parsing: Parse the formatted value back to double
• Locale: Format and parse using the default locale
• Exponential notation: Start using exponential notation after a certain threshold

Usage is pretty simple:

(For the sake of this example I am using a custom locale)

public static final int DECIMAL_PLACES = 2;

NumberFormatter formatter = new NumberFormatter(DECIMAL_PLACES);

String value = formatter.format(9.319); // "9,32"
String value2 = formatter.format(0.0000005); // "5,00E-7"
String value3 = formatter.format(1324134123); // "1,32E9"

double parsedValue1 = formatter.parse("0,4E-2", 0); // 0.004
double parsedValue2 = formatter.parse("0,002", 0); // 0.002
double parsedValue3 = formatter.parse("3423,12345", 0); // 3423.12345

Here is the class:

import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.text.DecimalFormatSymbols;
import java.text.ParseException;
import java.util.Locale;

public class NumberFormatter {

    private static final String SYMBOL_INFINITE           = "\u221e";
    private static final char   SYMBOL_MINUS              = '-';
    private static final char   SYMBOL_ZERO               = '0';
    private static final int    DECIMAL_LEADING_GROUPS    = 10;
    private static final int    EXPONENTIAL_INT_THRESHOLD = 1000000000; // After this value switch to exponential notation
    private static final double EXPONENTIAL_DEC_THRESHOLD = 0.0001; // Below this value switch to exponential notation

    private DecimalFormat decimalFormat;
    private DecimalFormat decimalFormatLong;
    private DecimalFormat exponentialFormat;

    private char groupSeparator;

    public NumberFormatter(int decimalPlaces) {
        configureDecimalPlaces(decimalPlaces);
    }

    public void configureDecimalPlaces(int decimalPlaces) {
        if (decimalPlaces <= 0) {
            throw new IllegalArgumentException("Invalid decimal places");
        }

        DecimalFormatSymbols separators = new DecimalFormatSymbols(Locale.getDefault());
        separators.setMinusSign(SYMBOL_MINUS);
        separators.setZeroDigit(SYMBOL_ZERO);

        groupSeparator = separators.getGroupingSeparator();

        StringBuilder decimal = new StringBuilder();
        StringBuilder exponential = new StringBuilder("0.");

        for (int i = 0; i < DECIMAL_LEADING_GROUPS; i++) {
            decimal.append("###").append(i == DECIMAL_LEADING_GROUPS - 1 ? "." : ",");
        }

        for (int i = 0; i < decimalPlaces; i++) {
            decimal.append("#");
            exponential.append("0");
        }

        exponential.append("E0");

        decimalFormat = new DecimalFormat(decimal.toString(), separators);
        decimalFormatLong = new DecimalFormat(decimal.append("####").toString(), separators);
        exponentialFormat = new DecimalFormat(exponential.toString(), separators);

        decimalFormat.setRoundingMode(RoundingMode.HALF_UP);
        decimalFormatLong.setRoundingMode(RoundingMode.HALF_UP);
        exponentialFormat.setRoundingMode(RoundingMode.HALF_UP);
    }

    public String format(double value) {
        String result;
        if (Double.isNaN(value)) {
            result = "";
        } else if (Double.isInfinite(value)) {
            result = String.valueOf(SYMBOL_INFINITE);
        } else {
            double absValue = Math.abs(value);
            if (absValue >= 1) {
                if (absValue >= EXPONENTIAL_INT_THRESHOLD) {
                    value = Math.floor(value);
                    result = exponentialFormat.format(value);
                } else {
                    result = decimalFormat.format(value);
                }
            } else if (absValue < 1 && absValue > 0) {
                if (absValue >= EXPONENTIAL_DEC_THRESHOLD) {
                    result = decimalFormat.format(value);
                    if (result.equalsIgnoreCase("0")) {
                        result = decimalFormatLong.format(value);
                    }
                } else {
                    result = exponentialFormat.format(value);
                }
            } else {
                result = "0";
            }
        }
        return result;
    }

    public String formatWithoutGroupSeparators(double value) {
        return removeGroupSeparators(format(value));
    }

    public double parse(String value, double defValue) {
        try {
            return decimalFormat.parse(value).doubleValue();
        } catch (ParseException e) {
            e.printStackTrace();
        }
        return defValue;
    }

    private String removeGroupSeparators(String number) {
        return number.replace(String.valueOf(groupSeparator), "");
    }

}

回答17:

If you really want decimal numbers for calculation (and not only for output), do not use a binary-based floating point format like double.

Use BigDecimal or any other decimal-based format.

I do use BigDecimal for calculations, but bear in mind it is dependent on the size of numbers you're dealing with. In most of my implementations, I find parsing from double or integer to Long is sufficient enough for very large number calculations.

In fact, I've recently used parsed-to-Long to get accurate representations (as opposed to hex results) in a GUI for numbers as big as ################################# characters (as an example).

回答18:

To achieve this we can use this formatter:

 DecimalFormat df = new DecimalFormat("#.00");
 String resultado = df.format(valor)

or:

DecimalFormat df = new DecimalFormat("0.00"); :

Use this method to get always two decimals:

   private static String getTwoDecimals(double value){
      DecimalFormat df = new DecimalFormat("0.00"); 
      return df.format(value);
    }

---

Defining this values:

91.32
5.22
11.5
1.2
2.6

Using the method we can get this results:

91.32
5.22
11.50
1.20
2.60

demo online.

回答19:

Just in case someone still needs help with this. This solution works perfectly for me.

private String withNoTrailingZeros(final double value, final int nrOfDecimals) {
return new BigDecimal(String.valueOf(value)).setScale(nrOfDecimals,  BigDecimal.ROUND_HALF_UP).stripTrailingZeros().toPlainString();

}

returns a String with the desired output.

回答20:

I agree with the chosen answer to use DecimalFormat --- or alternatively BigDecimal.

Please read Update below first!

However if you do want to round the double value and get a double value result, you can use org.apache.commons.math3.util.Precision.round(..) as mentioned above. The implementation uses BigDecimal, is slow and creates garbage.

A similar but fast and garbage-free method is provided by the DoubleRounder utility in the decimal4j library:

 double a = DoubleRounder.round(2.0/3.0, 3);
 double b = DoubleRounder.round(2.0/3.0, 3, RoundingMode.DOWN);
 double c = DoubleRounder.round(1000.0d, 17);
 double d = DoubleRounder.round(90080070060.1d, 9);
 System.out.println(a);
 System.out.println(b);
 System.out.println(c);
 System.out.println(d);

Will output

 0.667
 0.666
 1000.0
 9.00800700601E10

See https://github.com/tools4j/decimal4j/wiki/DoubleRounder-Utility

Disclaimer: I am involved in the decimal4j project.

Update: As @iaforek pointed out DoubleRounder sometimes returns counterintuitive results. The reason is that it performs mathematically correct rounding. For instance DoubleRounder.round(256.025d, 2) will be rounded down to 256.02 because the double value represented as 256.025d is somewhat smaller than the rational value 256.025 and hence will be rounded down.

Notes:

• This behaviour is very similar to that of the BigDecimal(double) constructor (but not to valueOf(double) which uses the string constructor).
• The problem can be circumvented with a double rounding step to a higher precision first, but it is complicated and I am not going into the details here

For those reasons and everything mentioned above in this post I cannot recommend to use DoubleRounder.

回答21:

The code snippet below shows how to display n digits. The trick is to set variable pp to 1 followed by n zeros. In the example below, variable pp value has 5 zeros, so 5 digits will be displayed.

double pp = 10000;

double myVal = 22.268699999999967;
String needVal = "22.2687";

double i = (5.0/pp);

String format = "%10.4f";
String getVal = String.format(format,(Math.round((myVal +i)*pp)/pp)-i).trim();

回答22:

If you're using DecimalFormat to convert double to String, it's very straightforward:

DecimalFormat formatter = new DecimalFormat("0.0##");
formatter.setRoundingMode(RoundingMode.HALF_UP);

double num = 1.234567;
return formatter.format(num);

There are several RoundingMode enum values to select from, depending upon the behaviour you require.

回答23:

I came here just wanting a simple answer on how to round a number. This is a supplemental answer to provide that.

How to round a number in Java

The most common case is to use Math.round().

Math.round(3.7) // 4

Numbers are rounded to the nearest whole number. A .5 value is rounded up. If you need different rounding behavior than that, you can use one of the other Math functions. See the comparison below.

round

As stated above, this rounds to the nearest whole number. .5 decimals round up. This method returns an int.

Math.round(3.0); // 3
Math.round(3.1); // 3
Math.round(3.5); // 4
Math.round(3.9); // 4

Math.round(-3.0); // -3
Math.round(-3.1); // -3
Math.round(-3.5); // -3 *** careful here ***
Math.round(-3.9); // -4

ceil

Any decimal value is rounded up to the next integer. It goes to the ceiling. This method returns a double.

Math.ceil(3.0); // 3.0
Math.ceil(3.1); // 4.0
Math.ceil(3.5); // 4.0
Math.ceil(3.9); // 4.0

Math.ceil(-3.0); // -3.0
Math.ceil(-3.1); // -3.0
Math.ceil(-3.5); // -3.0
Math.ceil(-3.9); // -3.0

floor

Any decimal value is rounded down to the next integer. This method returns a double.

Math.floor(3.0); // 3.0
Math.floor(3.1); // 3.0
Math.floor(3.5); // 3.0
Math.floor(3.9); // 3.0

Math.floor(-3.0); // -3.0
Math.floor(-3.1); // -4.0
Math.floor(-3.5); // -4.0
Math.floor(-3.9); // -4.0

rint

This is similar to round in that decimal values round to the closest integer. However, unlike round, .5 values round to the even integer. This method returns a double.

Math.rint(3.0); // 3.0
Math.rint(3.1); // 3.0
Math.rint(3.5); // 4.0 ***
Math.rint(3.9); // 4.0
Math.rint(4.5); // 4.0 ***
Math.rint(5.5); // 6.0 ***

Math.rint(-3.0); // -3.0
Math.rint(-3.1); // -3.0
Math.rint(-3.5); // -4.0 ***
Math.rint(-3.9); // -4.0
Math.rint(-4.5); // -4.0 ***
Math.rint(-5.5); // -6.0 ***

回答24:

If you're using a technology that has a minimal JDK. Here's a way without any Java libs:

double scale = 100000;    
double myVal = 0.912385;
double rounded = (int)((myVal * scale) + 0.5d) / scale;

回答25:

DecimalFormat is the best ways to output, but I don't prefer it. I always do this all the time, because it return the double value. So I can use it more than just output.

Math.round(selfEvaluate*100000d.0)/100000d.0;

OR

Math.round(selfEvaluate*100000d.0)*0.00000d1;

If you need large decimal places value, you can use BigDecimal instead. Anyways .0 is important. Without it the rounding of 0.33333d5 return 0.33333 and only 9 digits are allows. The second function without .0 has problems with 0.30000 return 0.30000000000000004.

回答26:

here is my answer:

double num = 4.898979485566356;
DecimalFormat df = new DecimalFormat("#.##");      
time = Double.valueOf(df.format(num));

System.out.println(num); // 4.89

回答27:

So after reading most of the answers, I realized most of them won't be precise, in fact using BigDecimal seems like the best choice, but if you don't understand how the RoundingMode works, you will inevitable lose precision. I figured this out when working with big numbers in a project and thought it could help others having trouble rounding numbers. For example.

BigDecimal bd = new BigDecimal("1363.2749");
bd = bd.setScale(2, RoundingMode.HALF_UP);
System.out.println(bd.doubleValue());

You would expect to get 1363.28 as an output, but you will end up with 1363.27, which is not expected, if you don't know what the RoundingMode is doing. So looking into the Oracle Docs, you will find the following description for RoundingMode.HALF_UP.

Rounding mode to round towards "nearest neighbor" unless both neighbors are equidistant, in which case round up.

So knowing this, we realized that we won't be getting an exact rounding, unless we want to round towards nearest neighbor. So, to accomplish an adequate round, we would need to loop from the n-1 decimal towards the desired decimals digits. For example.

private double round(double value, int places) throws IllegalArgumentException {

    if (places < 0) throw new IllegalArgumentException();

    // Cast the number to a String and then separate the decimals.
    String stringValue = Double.toString(value);
    String decimals = stringValue.split("\\.")[1];

    // Round all the way to the desired number.
    BigDecimal bd = new BigDecimal(stringValue);
    for (int i = decimals.length()-1; i >= places; i--) {
        bd = bd.setScale(i, RoundingMode.HALF_UP);
    }

    return bd.doubleValue();
}

This will end up giving us the expected output, which would be 1363.28.

回答28:

Where dp = decimal place you want, and value is a double.

    double p = Math.pow(10d, dp);

    double result = Math.round(value * p)/p;

回答29:

Keep in mind that String.format() and DecimalFormat produce string using default Locale. So they may write formatted number with dot or comma as a separator between integer and decimal parts. To make sure that rounded String is in the format you want use java.text.NumberFormat as so:

  Locale locale = Locale.ENGLISH;
  NumberFormat nf = NumberFormat.getNumberInstance(locale);
  // for trailing zeros:
  nf.setMinimumFractionDigits(2);
  // round to 2 digits:
  nf.setMaximumFractionDigits(2);

  System.out.println(nf.format(.99));
  System.out.println(nf.format(123.567));
  System.out.println(nf.format(123.0));

Will print in English locale (no matter what your locale is): 0.99 123.57 123.00

The example is taken from Farenda - how to convert double to String correctly.

回答30:

If you Consider 5 or n number of decimal. May be this answer solve your prob.

    double a = 123.00449;
    double roundOff1 = Math.round(a*10000)/10000.00;
    double roundOff2 = Math.round(roundOff1*1000)/1000.00;
    double roundOff = Math.round(roundOff2*100)/100.00;

    System.out.println("result:"+roundOff);

Output will be: 123.01
this can be solve with loop and recursive function.