四舍五入到三位有效数字

Question

Does anybody know how I can round a double value to 3 significant figures like the examples on this website

http://www.purplemath.com/modules/rounding2.htm

Answer 1

double d = ...;
BigDecimal bd = new BigDecimal(d);
bd = bd.round(new MathContext(3));
double rounded = bd.doubleValue();

Answer 2

    public String toSignificantFiguresString(BigDecimal bd, int significantFigures ){
    String test = String.format("%."+significantFigures+"G", bd);
    if (test.contains("E+")){
        test = String.format(Locale.US, "%.0f", Double.valueOf(String.format("%."+significantFigures+"G", bd)));
    }
    return test;
}

Answer 3

If you want to do it by hand:

import java.lang.Math;

public class SigDig {

  public static void main(String[] args) {
    System.out.println("   -123.456   rounded up   to 2 sig figures is " + sigDigRounder(-123.456, 2,  1));
    System.out.println("     -0.03394 rounded down to 3 sig figures is " + sigDigRounder(-0.03394, 3, -1));
    System.out.println("    474       rounded up   to 2 sig figures is " + sigDigRounder(474, 2,  1));
    System.out.println("3004001       rounded down to 4 sig figures is " + sigDigRounder(3004001, 4, -1));
  }

  public static double sigDigRounder(double value, int nSigDig, int dir) {

    double intermediate = value/Math.pow(10,Math.floor(Math.log10(Math.abs(value)))-(nSigDig-1));

    if(dir > 0)      intermediate = Math.ceil(intermediate);
    else if (dir< 0) intermediate = Math.floor(intermediate);
    else             intermediate = Math.round(intermediate);

    double result = intermediate * Math.pow(10,Math.floor(Math.log10(Math.abs(value)))-(nSigDig-1));

    return(result);

  }
}

The above method rounds a double to a desired number of significant figures, handles negative numbers, and can be explicitly told to round up or down

Answer 4

There's nothing wrong with the answer given by Sean Owen ( https://stackoverflow.com/a/7548871/274677 ). However depending on your use case you might want to arrive at a String representation. In that case, IMO it is best to convert while still in BigDecimal space using:

bd.toPlainString();

... if that's your use case then you might be frustrated that the code adapted from Owen's answer will produce the following:

d = 0.99, significantDigits = 3 ==> 0.99

... instead of the strictly more accurate 0.990 .

If such things are important in your use case then I suggest the following adaptation to Owen's answer ; you can obviously also return the BigDecimal itself instead of calling toPlainString() — I just provide it this way for completeness.

public static String setSignificanDigits(double value, int significantDigits) {
    if (significantDigits < 0) throw new IllegalArgumentException();

    // this is more precise than simply doing "new BigDecimal(value);"
    BigDecimal bd = new BigDecimal(value, MathContext.DECIMAL64);
    bd = bd.round(new MathContext(significantDigits, RoundingMode.HALF_UP));
    final int precision = bd.precision();
    if (precision < significantDigits)
    bd = bd.setScale(bd.scale() + (significantDigits-precision));
    return bd.toPlainString();
}    

Answer 5

double d=3.142568; System.out.printf("Answer : %.3f", d);

Answer 6

how I can round a double value to 3 significant figures

You can't. Doubles are representing in binary. They do not have decimal places to be rounded to. The only way you can get a specific number of decimal places is to convert it to a decimal radix and leave it there. The moment you convert it back to double you have lost the decimal precision again.

For all the fans, here and elsewhere, of converting to other radixes and back, or multiplying and dividing by powers of ten, please display the resulting double value % 0.001 or whatever the required precision dictates, and explain the result.

EDIT: Specifically, the proponents of those techniques need to explain the 92% failure rate of the following code:

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 % .0001)="+(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");
    }
}

Answer 7

I usually don't round the number itself but round the String representation of the number when I need to display it because usually it's the display that matters, that needs the rounding (although this may not be true in situations, and perhaps yours, but you need to elaborate on this if so). This way, my number retains its accuracy, but it's display is simplified and easier to read. To do this, one can use a DecimalFormat object, say initialzed with a "0.000" String ( new DecimalFormat("0.000") ), or use String.format("%.3f", myDouble) , or several other ways.

For example:

// yeah, I know this is just Math.PI.
double myDouble = 3.141592653589793;
DecimalFormat myFormat = new DecimalFormat("0.000");
String myDoubleString = myFormat.format(myDouble);
System.out.println("My number is: " + myDoubleString);

// or you can use printf which works like String.format:
System.out.printf("My number is: %.3f%n", myDouble);