问题
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.
回答1:
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;
}
回答2:
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
回答3:
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!
回答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
回答5:
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;
}
回答6:
How about this java solution :
double roundToSignificantFigure(double num, int precision){
return new BigDecimal(num)
.round(new MathContext(precision, RoundingMode.HALF_EVEN))
.doubleValue();
}
回答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;
}
回答8:
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);
回答9:
Have you tried just coding it up the way you'd do it by hand?
- Convert the number to a string
- Starting at the beginning of the string, count digits - leading zeroes aren't significant, everything else is.
- When you get to the "nth" digit, peek ahead at the next digit and if it's 5 or higher, round up.
- Replace all of the trailing digits with zeroes.
回答10:
[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.
回答11:
/**
* 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);
}
回答12:
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
回答13:
JavaScript:
Number( my_number.toPrecision(3) );
The Number function will change output of the form "8.143e+5" to "814300".
回答14:
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
回答15:
return new BigDecimal(value, new MathContext(significantFigures, RoundingMode.HALF_UP)).doubleValue();
回答16:
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
}
回答17:
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
来源：https://stackoverflow.com/questions/202302/rounding-to-an-arbitrary-number-of-significant-digits