By DougN


2008-10-14 18:37:17 8 Comments

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

For example, if I want to round to three 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.

17 comments

@Michael Hampton 2018-01-28 02:42:21

I needed this in Go, which was a bit complicated by the Go standard library's lack of math.Round() (before go1.10). So I had to whip that up too. Here is my translation of Pyrolistical's excellent answer:

// TODO: replace in go1.10 with math.Round()
func round(x float64) float64 {
    return float64(int64(x + 0.5))
}

// SignificantDigits rounds a float64 to digits significant digits.
// Translated from Java at https://stackoverflow.com/a/1581007/1068283
func SignificantDigits(x float64, digits int) float64 {
    if x == 0 {
        return 0
    }

    power := digits - int(math.Ceil(math.Log10(math.Abs(x))))
    magnitude := math.Pow(10, float64(power))
    shifted := round(x * magnitude)
    return shifted / magnitude
}

@Michael Hampton 2018-08-28 17:35:11

This got a mystery downvote! But I can't find an error or problem in it. What's going on here?

@Duncan Calvert 2017-03-27 16:22:43

return new BigDecimal(value, new MathContext(significantFigures, RoundingMode.HALF_UP)).doubleValue();

@Zaz 2015-05-04 22:56:25

JavaScript:

Number( my_number.toPrecision(3) );

The Number function will change output of the form "8.143e+5" to "814300".

@Pyrolistical 2009-10-17 00:20:15

Here's the same code in Java without the 12.100000000000001 bug other answers have

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;
}

@Claudiu 2009-12-14 23:41:53

Your code is indeed much nicer.

@Pyrolistical 2010-05-21 22:45:41

Thanks for accepting my answer. I just realized my answer is more than a year after the question. This is one of the reasons why stackoverflow is so cool. You can find useful information!

@cquezel 2014-11-22 17:10:15

Note that this may fail slightly for values close to the round limit. For example rounding 1.255 to 3 significant digits should return 1.26 but returns 1.25. That is because 1.255 * 100.0 is 125.499999... But this is to be expected when working with doubles

@Steve W 2016-01-20 15:04:38

Wow, I know this is old, but I'm trying to use it. I have a float that I want to display to 3 significant figures. If the float value is 1.0, I call your method but it still returns as 1.0, even if I cast the float as a double. I want it to return as 1. Any ideas?

@Curious Sam 2016-02-18 15:11:59

This java snippet ends up in official android example android.googlesource.com/platform/development/+/fcf4286/samp‌​les/…

@Pyrolistical 2016-03-08 22:40:21

lol nice find @CuriousSam

@Duncan Calvert 2017-03-24 18:01:23

Not perfect. For num = -7999999.999999992 and n = 2 returns -7999999.999999999 but should be -8000000.

@Eric Nicolas 2017-05-26 12:28:06

I tested on +7999999.999999992 and it gives +8000000 ok. I think negative numbers should be treated by a special case, as is zero, for safety.

@JackDev 2013-10-22 00:07:22

This came 5 years late, but though I'll share for others still having the same issue. I like it because it's simple and no calculations on the code side. See Built in methods for displaying Significant figures for more info.

This is if you just want to print it out.

public String toSignificantFiguresString(BigDecimal bd, int significantFigures){
    return String.format("%."+significantFigures+"G", bd);
}

This is if you want to convert it:

public BigDecimal toSignificantFigures(BigDecimal bd, int significantFigures){
    String s = String.format("%."+significantFigures+"G", bd);
    BigDecimal result = new BigDecimal(s);
    return result;
}

Here's an example of it in action:

BigDecimal bd = toSignificantFigures(BigDecimal.valueOf(0.0681), 2);

@matt 2019-03-10 09:27:45

This will display "big" numbers in scientific notation, e.g. 15k as 1.5e04.

@SomeGuy 2013-06-01 16:05:51

This is one that I came up with in VB:

Function SF(n As Double, SigFigs As Integer)
    Dim l As Integer = n.ToString.Length
    n = n / 10 ^ (l - SigFigs)
    n = Math.Round(n)
    n = n * 10 ^ (l - SigFigs)
    Return n
End Function

@Harikrishnan 2013-05-19 05:51:47

public static double roundToSignificantDigits(double num, int n) {
    return Double.parseDouble(new java.util.Formatter().format("%." + (n - 1) + "e", num).toString());
}

This code uses the inbuilt formatting function which is turned to a rounding function

@Michael Zlatkovsky - Microsoft 2011-06-21 16:10:45

Here is Pyrolistical's (currently top answer) code in Visual Basic.NET, should anyone need it:

Public Shared Function roundToSignificantDigits(ByVal num As Double, ByVal n As Integer) As Double
    If (num = 0) Then
        Return 0
    End If

    Dim d As Double = Math.Ceiling(Math.Log10(If(num < 0, -num, num)))
    Dim power As Integer = n - CInt(d)
    Dim magnitude As Double = Math.Pow(10, power)
    Dim shifted As Double = Math.Round(num * magnitude)
    Return shifted / magnitude
End Function

@Thomas Becker 2010-11-19 01:17:06

Pyrolistical's (very nice!) solution still has an issue. The maximum double value in Java is on the order of 10^308, while the minimum value is on the order of 10^-324. Therefore, you can run into trouble when applying the function roundToSignificantFigures to something that's within a few powers of ten of Double.MIN_VALUE. For example, when you call

roundToSignificantFigures(1.234E-310, 3);

then the variable power will have the value 3 - (-309) = 312. Consequently, the variable magnitude will become Infinity, and it's all garbage from then on out. Fortunately, this is not an insurmountable problem: it is only the factor magnitude that's overflowing. What really matters is the product num * magnitude, and that does not overflow. One way of resolving this is by breaking up the multiplication by the factor magintude into two steps:


 public static double roundToNumberOfSignificantDigits(double num, int n) {

    final double maxPowerOfTen = Math.floor(Math.log10(Double.MAX_VALUE));

    if(num == 0) {
        return 0;
    }

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

    double firstMagnitudeFactor = 1.0;
    double secondMagnitudeFactor = 1.0;
    if (power > maxPowerOfTen) {
        firstMagnitudeFactor = Math.pow(10.0, maxPowerOfTen);
        secondMagnitudeFactor = Math.pow(10.0, (double) power - maxPowerOfTen);
    } else {
        firstMagnitudeFactor = Math.pow(10.0, (double) power);
    }

    double toBeRounded = num * firstMagnitudeFactor;
    toBeRounded *= secondMagnitudeFactor;

    final long shifted = Math.round(toBeRounded);
    double rounded = ((double) shifted) / firstMagnitudeFactor;
    rounded /= secondMagnitudeFactor;
    return rounded;
}

@Pyrolistical 2012-05-07 18:30:42

Whoa, rounding at the edge of the universe!

@wolfgang grinfeld 2010-08-10 08:00:42

How about this java solution :

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

@Valeri Shibaev 2010-06-22 00:25:30

/**
 * 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);
}

@Jason Swank 2010-06-04 16:18:40

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;
 }

@Claudiu 2008-10-14 18:47:01

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!

@Ates Goral 2008-10-14 19:30:23

This algorithm seems prone to floating point errors. When implemented with JavaScript, I get: 0.06805 -> 0.06810000000000001 and 12.1 -> 12.100000000000001

@Claudiu 2008-10-15 01:28:48

12.1 by itself can't be represented accurately using floating point - it's not a result of this algorithm.

@Pyrolistical 2009-10-17 00:00:27

This code in Java produces 12.100000000000001 and this is using 64-bit doubles which can present 12.1 exactly.

@Claudiu 2009-10-17 17:02:00

It doesn't matter if it's 64 bit or 128 bit. You can't represent the fraction 1/10 using a finite sum of powers of 2, and that's how floating point numbers are represented

@Claudiu 2011-03-06 05:20:37

for those chiming in, basically Pyrolistical's answer is more precise than mine so that the floating point number printing algorithm prints '12.1' instead of '12.100000000000001'. his answer is better, even tho i was technically correct that you can't represent '12.1' exactly.

@David R Tribble 2009-10-17 00:41:19

[Corrected, 2009-10-26]

Essentially, for N significant fractional digits:

• Multiply the number by 10N
• Add 0.5
• Truncate the fraction digits (i.e., truncate the result into an integer)
• Divide by 10N

For N significant integral (non-fractional) digits:

• Divide the number by 10N
• Add 0.5
• Truncate the fraction digits (i.e., truncate the result into an integer)
• Multiply by 10N

You can do this on any calculator, for example, that has an "INT" (integer truncation) operator.

@Pyrolistical 2009-10-20 18:06:49

Nope. Read the question again. 1239451 with 3 sig figs using your algorithm would incorrectly yield 123951

@David R Tribble 2009-10-26 21:53:42

Yep, I corrected it to distinguish between rounding to a fractional number of digits (to the right of the decimal point) versus an integral number of digits (to the left).

@Justin Wignall 2009-04-08 15:26:57

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

@Zaz 2015-05-04 22:48:44

Number(814301).toPrecision(4) == "8.143e+5". Generally not what you want if you're showing this to users.

@Justin Wignall 2015-05-05 19:53:32

Very true Josh yes, I would generally recommend .toPrecision() only for decimal numbers and the accepted answer (with edit) should be used/reviewed as per your individual requirements.

@Ates Goral 2008-10-14 19:20:32

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

@Pyrolistical 2009-10-17 00:03:07

but 12.1257 gives 12.126

@Ates Goral 2009-10-17 22:47:47

Yes, that's valid rounding.

@sscirrus 2011-10-22 01:56:27

Nice answer Ates. Perhaps add a trigger to return 0 if n==0 :)

@Mahn 2014-04-01 20:20:55

PHP port for the lazy: gist.github.com/anonymous/9922298

@Lee 2015-10-23 01:03:26

is there a reason to do Math.log(n) / Math.LN10 rather than Math.log10(n)?

@Ates Goral 2015-10-23 03:58:01

@Lee developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… "This is a new technology, part of the ECMAScript 2015 (ES6) standard." So, basically, compatibility issues.

@Paul 2018-02-28 12:09:51

Am I missing something, or does this answer assume that Math.floor(x) == Math.ceil(x) - 1? Because it doesn't when x is an integer. I think the second argument of the pow function should be sig - Math.ceil(Math.log(n) / Math.LN10) (or just use Math.log10)

@Mark Bessey 2008-10-14 18:52:37

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.

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