Round a Double to X Significant Figures

Round a double to x significant figures

The framework doesn't have a built-in function to round (or truncate, as in your example) to a number of significant digits. One way you can do this, though, is to scale your number so that your first significant digit is right after the decimal point, round (or truncate), then scale back. The following code should do the trick:

static double RoundToSignificantDigits(this double d, int digits){
    if(d == 0)
        return 0;

    double scale = Math.Pow(10, Math.Floor(Math.Log10(Math.Abs(d))) + 1);
    return scale * Math.Round(d / scale, digits);
}

If, as in your example, you really want to truncate, then you want:

static double TruncateToSignificantDigits(this double d, int digits){
    if(d == 0)
        return 0;

    double scale = Math.Pow(10, Math.Floor(Math.Log10(Math.Abs(d))) + 1 - digits);
    return scale * Math.Truncate(d / scale);
}

Round a double to 3 significant figures

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

How to round a number to significant figures in Python

You can use negative numbers to round integers:

>>> round(1234, -3)
1000.0

Thus if you need only most significant digit:

>>> from math import log10, floor
>>> def round_to_1(x):
...   return round(x, -int(floor(log10(abs(x)))))
... 
>>> round_to_1(0.0232)
0.02
>>> round_to_1(1234243)
1000000.0
>>> round_to_1(13)
10.0
>>> round_to_1(4)
4.0
>>> round_to_1(19)
20.0

You'll probably have to take care of turning float to integer if it's bigger than 1.

Rounding to an arbitrary number of significant digits

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;
}

Rounding a double value to x number of decimal places in swift

You can use Swift's round function to accomplish this.

To round a Double with 3 digits precision, first multiply it by 1000, round it and divide the rounded result by 1000:

let x = 1.23556789
let y = Double(round(1000 * x) / 1000)
print(y) /// 1.236

Unlike any kind of printf(...) or String(format: ...) solutions, the result of this operation is still of type Double.

EDIT:

Regarding the comments that it sometimes does not work, please read this: What Every Programmer Should Know About Floating-Point Arithmetic

Formatting numbers with significant figures in C#

See: RoundToSignificantFigures by "P Daddy".

I've combined his method with another one I liked.

Rounding to significant figures is a lot easier in TSQL where the rounding method is based on rounding position, not number of decimal places - which is the case with .Net math.round. You could round a number in TSQL to negative places, which would round at whole numbers - so the scaling isn't needed.

Also see this other thread. Pyrolistical's method is good.

The trailing zeros part of the problem seems like more of a string operation to me, so I included a ToString() extension method which will pad zeros if necessary.

using System;
using System.Globalization;

public static class Precision
{
    // 2^-24
    public const float FLOAT_EPSILON = 0.0000000596046448f;

    // 2^-53
    public const double DOUBLE_EPSILON = 0.00000000000000011102230246251565d;

    public static bool AlmostEquals(this double a, double b, double epsilon = DOUBLE_EPSILON)
    {
        // ReSharper disable CompareOfFloatsByEqualityOperator
        if (a == b)
        {
            return true;
        }
        // ReSharper restore CompareOfFloatsByEqualityOperator

        return (System.Math.Abs(a - b) < epsilon);
    }

    public static bool AlmostEquals(this float a, float b, float epsilon = FLOAT_EPSILON)
    {
        // ReSharper disable CompareOfFloatsByEqualityOperator
        if (a == b)
        {
            return true;
        }
        // ReSharper restore CompareOfFloatsByEqualityOperator

        return (System.Math.Abs(a - b) < epsilon);
    }
}

public static class SignificantDigits
{
    public static double Round(this double value, int significantDigits)
    {
        int unneededRoundingPosition;
        return RoundSignificantDigits(value, significantDigits, out unneededRoundingPosition);
    }

    public static string ToString(this double value, int significantDigits)
    {
        // this method will round and then append zeros if needed.
        // i.e. if you round .002 to two significant figures, the resulting number should be .0020.

        var currentInfo = CultureInfo.CurrentCulture.NumberFormat;

        if (double.IsNaN(value))
        {
            return currentInfo.NaNSymbol;
        }

        if (double.IsPositiveInfinity(value))
        {
            return currentInfo.PositiveInfinitySymbol;
        }

        if (double.IsNegativeInfinity(value))
        {
            return currentInfo.NegativeInfinitySymbol;
        }

        int roundingPosition;
        var roundedValue = RoundSignificantDigits(value, significantDigits, out roundingPosition);

        // when rounding causes a cascading round affecting digits of greater significance, 
        // need to re-round to get a correct rounding position afterwards
        // this fixes a bug where rounding 9.96 to 2 figures yeilds 10.0 instead of 10
        RoundSignificantDigits(roundedValue, significantDigits, out roundingPosition);

        if (Math.Abs(roundingPosition) > 9)
        {
            // use exponential notation format
            // ReSharper disable FormatStringProblem
            return string.Format(currentInfo, "{0:E" + (significantDigits - 1) + "}", roundedValue);
            // ReSharper restore FormatStringProblem
        }

        // string.format is only needed with decimal numbers (whole numbers won't need to be padded with zeros to the right.)
        // ReSharper disable FormatStringProblem
        return roundingPosition > 0 ? string.Format(currentInfo, "{0:F" + roundingPosition + "}", roundedValue) : roundedValue.ToString(currentInfo);
        // ReSharper restore FormatStringProblem
    }

    private static double RoundSignificantDigits(double value, int significantDigits, out int roundingPosition)
    {
        // this method will return a rounded double value at a number of signifigant figures.
        // the sigFigures parameter must be between 0 and 15, exclusive.

        roundingPosition = 0;

        if (value.AlmostEquals(0d))
        {
            roundingPosition = significantDigits - 1;
            return 0d;
        }

        if (double.IsNaN(value))
        {
            return double.NaN;
        }

        if (double.IsPositiveInfinity(value))
        {
            return double.PositiveInfinity;
        }

        if (double.IsNegativeInfinity(value))
        {
            return double.NegativeInfinity;
        }

        if (significantDigits < 1 || significantDigits > 15)
        {
            throw new ArgumentOutOfRangeException("significantDigits", value, "The significantDigits argument must be between 1 and 15.");
        }

        // The resulting rounding position will be negative for rounding at whole numbers, and positive for decimal places.
        roundingPosition = significantDigits - 1 - (int)(Math.Floor(Math.Log10(Math.Abs(value))));

        // try to use a rounding position directly, if no scale is needed.
        // this is because the scale mutliplication after the rounding can introduce error, although 
        // this only happens when you're dealing with really tiny numbers, i.e 9.9e-14.
        if (roundingPosition > 0 && roundingPosition < 16)
        {
            return Math.Round(value, roundingPosition, MidpointRounding.AwayFromZero);
        }

        // Shouldn't get here unless we need to scale it.
        // Set the scaling value, for rounding whole numbers or decimals past 15 places
        var scale = Math.Pow(10, Math.Ceiling(Math.Log10(Math.Abs(value))));

        return Math.Round(value / scale, significantDigits, MidpointRounding.AwayFromZero) * scale;
    }
}

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