rounding to an arbitrary number of significant digits

Question

How can you round any number (not just integers > 0) to N significant digits?

For example, if I want to round to 3 significant digits, I'm looking for a formula that could take:

1,239,451 and return 1,240,000

12.1257 and return 12.1

.0681 and return .0681

5 and return 5

Naturally the algorithm should not be hard-coded to only handle N of 3, although that would be a start.

Answer 1

SUMMARY:

double roundit(double num, double N)
{
    double d = log10(num);
    double power;
    if (num > 0)
    {
        d = ceil(d);
        power = -(d-N);
    }
    else
    {
        d = floor(d); 
        power = -(d-N);
    }

    return (int)(num * pow(10.0, power) + 0.5) * pow(10.0, -power);
}

---

So you need to find the decimal place of the first non-zero digit, then save the next N-1 digits, then round the Nth digit based on the rest.

We can use log to do the first.

log 1239451 = 6.09
log 12.1257 = 1.08
log 0.0681  = -1.16

So for numbers > 0, take the ceil of the log. For numbers < 0, take the floor of the log.

Now we have the digit d: 7 in the first case, 2 in the 2nd, -2 in the 3rd.

We have to round the (d-N)th digit. Something like:

double roundedrest = num * pow(10, -(d-N));

pow(1239451, -4) = 123.9451
pow(12.1257, 1)  = 121.257
pow(0.0681, 4)   = 681

Then do the standard rounding thing:

roundedrest = (int)(roundedrest + 0.5);

And undo the pow.

roundednum = pow(roundedrest, -(power))

Where power is the power calculated above.

---

About accuracy: Pyrolistical's answer is indeed closer to the real result. But note that you can't represent 12.1 exactly in any case. If you print the answers as follows:

System.out.println(new BigDecimal(n));

The answers are:

Pyro's: 12.0999999999999996447286321199499070644378662109375
Mine: 12.10000000000000142108547152020037174224853515625
Printing 12.1 directly: 12.0999999999999996447286321199499070644378662109375

So, use Pyro's answer!

Answer 2

Have you tried just coding it up the way you'd do it by hand?

1. Convert the number to a string
2. Starting at the beginning of the string, count digits - leading zeroes aren't significant, everything else is.
3. When you get to the "nth" digit, peek ahead at the next digit and if it's 5 or higher, round up.
4. Replace all of the trailing digits with zeroes.

Answer 3

Here's a short and sweet JavaScript implementation:

function sigFigs(n, sig) {
    var mult = Math.pow(10, sig - Math.floor(Math.log(n) / Math.LN10) - 1);
    return Math.round(n * mult) / mult;
}

alert(sigFigs(1234567, 3)); // Gives 1230000
alert(sigFigs(0.06805, 3)); // Gives 0.0681
alert(sigFigs(5, 3)); // Gives 5

Answer 4

Isn't the "short and sweet" JavaScript implementation

Number(n).toPrecision(sig)

e.g.

alert(Number(12345).toPrecision(3)

?

Sorry, I'm not being facetious here, it's just that using the "roundit" function from Claudiu and the .toPrecision in JavaScript gives me different results but only in the rounding of the last digit.

JavaScript:

Number(8.14301).toPrecision(4) == 8.143

.NET

roundit(8.14301,4) == 8.144

Answer 5

Here's the same code in Java without the 12.100000000000001 bug that the accepted answer has.

I also removed repeated code, changed power to a type integer to prevent floating issues when n - d is done, and made the long intermediate more clear

The bug was caused by multiplying a large number with a small number. Instead I divide two numbers of similar size.

EDIT
Fixed more bugs. Added check for 0 as it would result in NaN. Made the function actually work with negative numbers (The original code doesn't handle negative numbers because a log of a negative number is a complex number)

public static double roundToSignificantFigures(double num, int n) {
    if(num == 0) {
        return 0;
    }

    final double d = Math.ceil(Math.log10(num < 0 ? -num: num));
    final int power = n - (int) d;

    final double magnitude = Math.pow(10, power);
    final long shifted = Math.round(num*magnitude);
    return shifted/magnitude;
}

Answer 6

See my answer on this other thread

Answer 7

Here is a modified version of Ates' JavaScript that handles negative numbers.

function sigFigs(n, sig) {
    if ( n === 0 )
        return 0
    var mult = Math.pow(10,
        sig - Math.floor(Math.log(n < 0 ? -n: n) / Math.LN10) - 1);
    return Math.round(n * mult) / mult;
 }

Answer 8

/**
 * Set Significant Digits.
 * @param value value
 * @param digits digits
 * @return
 */
public static BigDecimal setSignificantDigits(BigDecimal value, int digits) {
    //# Start with the leftmost non-zero digit (e.g. the "1" in 1200, or the "2" in 0.0256).
    //# Keep n digits. Replace the rest with zeros.
    //# Round up by one if appropriate.
    int p = value.precision();
    int s = value.scale();
    if (p < digits) {
        value = value.setScale(s + digits - p); //, RoundingMode.HALF_UP
    }
    value = value.movePointRight(s).movePointLeft(p - digits).setScale(0, RoundingMode.HALF_UP)
        .movePointRight(p - digits).movePointLeft(s);
    s = (s > (p - digits)) ? (s - (p - digits)) : 0;
    return value.setScale(s);
}

Answer 9

How about this java solution :

double roundToSignificantFigure(double num, int precision){
 return new BigDecimal(num)
            .round(new MathContext(precision, RoundingMode.HALF_EVEN))
            .doubleValue(); 
}