Rounding to an arbitrary number of significant digits

DougN picture DougN · Oct 14, 2008 · Viewed 39.6k times · Source

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.

Answer

Pyrolistical picture Pyrolistical · Oct 17, 2009

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;
}