What I would like to do is a method to round float number to N decimal places (N will be given in stdin), do some math operation with it and then print the result. I found this:
public BigDecimal round(float number, int decimal){
BigDecimal obj = new BigDecimal(number).setScale(decimal, BigDecimal.ROUND_HALF_UP);
return obj;
Which works pretty well, but not when number N (int decimal in this method) is high. For example: x = -10, y = -11.8814, N = 8 and it prints this:
-10.00000000 + -11.88140011 = -21.88140106
And this is what I would want:
-10.00000000 + -11.88140000 = -21.88140000
Thanks everybody for suggestions :)
I think the easiest thing to do is multiply your numbers by 10^N and use round() to turn them into integers. Run your math and then divide them back down
public static void main(String[] args) {
double a = -10.0;
double b = -11.88140011;
int n = 4;
long total = round(a, n) + round(b, n);
System.out.println(String.format("%.8f%n", total * Math.pow(10, -1*n)));
}
static Long round(double number, int decimal) {
return Math.round(number * Math.pow(10, decimal));
}
There is no float representation of 11.88140000 with all 8 decimal places intact, so at the moment you pass that number to your method regardless of the implementation it has no chance to return the wanted result, see article for float on Wikipedia:
... the total precision is 24 bits (equivalent to log10(224) ≈ 7.225 decimal digits)
And that precision is including all digits, not only those after the decimal separator.
Your question already has an answer, but to understand the problem float vs. decimal better, there is another post about the problem explaining it a bit more in depth (same for every programming language):
Just for completeness: 0.3 would be exactly stored as it is in a BigDecimal. In float (due to it's binary nature), it would be stored as 0.30000001192092896. You instantiate your BigDecimal using a float and directly step into this problem.
So, here is the answer. I didn't use function round I only printed the numbers with results:
System.out.printf("%." + n + "f + %." + n + "f = %." + n + "f\n", a, b, a + b);
Which I tried before I posted this question, BUT I used float and haven't tried using double (with function nextDouble() for scanning). So, double was the answer.
Thanks for your suggestions! :))
I'm trying to calculate the following:
(1 - (1/365)) * (1 - (2/365) = 0.99727528617
I would like to store the entire decimal. Here is my code, but it is giving me an answer of 1:
public BigDecimal probability(int t){
BigDecimal probT; // holds our probability of a single (1-(t/365))
BigDecimal result; // holds our result
result = BigDecimal.ONE; // initialize result to 1
// for 1 to t-1
for (int n = 1; n < t; n++){
int numerator = n; // numerator the value of n
int denominator = 365; // denominator 365
// numberator / denominator (round down)
probT = BigDecimal.valueOf(numerator).divide(BigDecimal.valueOf(denominator), RoundingMode.DOWN);
// 1-answer
probT = BigDecimal.ONE.subtract(probT);
// multiply the probabilities together
result = result.multiply(probT);
}
return result;
}
BigDecimal ans2 = bc.probability(3);
System.out.println("P(3) = " + ans2.toString());
Output:
P(3) = 1
That's because the division you are computing is made with a scale of 0. Quoting the method divide(divisor, roundingMode) Javadoc:
Returns a BigDecimal whose value is (this / divisor), and whose scale is this.scale().
In this case, this.scale() refers to the scale of the numerator, which is 0 because the numerator is BigDecimal.valueOf(n), with n being an integer.
You need to change this division to use divide(divisor, scale, roundingMode) instead and specify the scale you want.
From the java doc
When a MathContext object is supplied with a precision setting of 0
(for example, MathContext.UNLIMITED), arithmetic operations are exact,
as are the arithmetic methods which take no MathContext object. (This
is the only behavior that was supported in releases prior to 5.)
As a corollary of computing the exact result, the rounding mode
setting of a MathContext object with a precision setting of 0 is not
used and thus irrelevant. In the case of divide, the exact quotient
could have an infinitely long decimal expansion; for example, 1
divided by 3.
If the quotient has a nonterminating decimal expansion and the
operation is specified to return an exact result, an
ArithmeticException is thrown. Otherwise, the exact result of the
division is returned, as done for other operations.
To fix, you need to do something like this:
// numberator / denominator (round down)
probT = BigDecimal.valueOf(numerator).divide(BigDecimal.valueOf(denominator), 10,RoundingMode.DOWN);
where 10 is precision(decimal places precision) and RoundingMode.DOWN is rounding mode
public class doublePrecision {
public static void main(String[] args) {
double total = 0;
total += 5.6;
total += 5.8;
System.out.println(total);
}
}
The above code prints:
11.399999999999
How would I get this to just print (or be able to use it as) 11.4?
As others have mentioned, you'll probably want to use the BigDecimal class, if you want to have an exact representation of 11.4.
Now, a little explanation into why this is happening:
The float and double primitive types in Java are floating point numbers, where the number is stored as a binary representation of a fraction and a exponent.
More specifically, a double-precision floating point value such as the double type is a 64-bit value, where:
1 bit denotes the sign (positive or negative).
11 bits for the exponent.
52 bits for the significant digits (the fractional part as a binary).
These parts are combined to produce a double representation of a value.
(Source: Wikipedia: Double precision)
For a detailed description of how floating point values are handled in Java, see the Section 4.2.3: Floating-Point Types, Formats, and Values of the Java Language Specification.
The byte, char, int, long types are fixed-point numbers, which are exact representions of numbers. Unlike fixed point numbers, floating point numbers will some times (safe to assume "most of the time") not be able to return an exact representation of a number. This is the reason why you end up with 11.399999999999 as the result of 5.6 + 5.8.
When requiring a value that is exact, such as 1.5 or 150.1005, you'll want to use one of the fixed-point types, which will be able to represent the number exactly.
As has been mentioned several times already, Java has a BigDecimal class which will handle very large numbers and very small numbers.
From the Java API Reference for the BigDecimal class:
Immutable,
arbitrary-precision signed decimal
numbers. A BigDecimal consists of an
arbitrary precision integer unscaled
value and a 32-bit integer scale. If
zero or positive, the scale is the
number of digits to the right of the
decimal point. If negative, the
unscaled value of the number is
multiplied by ten to the power of the
negation of the scale. The value of
the number represented by the
BigDecimal is therefore (unscaledValue
× 10^-scale).
There has been many questions on Stack Overflow relating to the matter of floating point numbers and its precision. Here is a list of related questions that may be of interest:
Why do I see a double variable initialized to some value like 21.4 as 21.399999618530273?
How to print really big numbers in C++
How is floating point stored? When does it matter?
Use Float or Decimal for Accounting Application Dollar Amount?
If you really want to get down to the nitty gritty details of floating point numbers, take a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic.
When you input a double number, for example, 33.33333333333333, the value you get is actually the closest representable double-precision value, which is exactly:
33.3333333333333285963817615993320941925048828125
Dividing that by 100 gives:
0.333333333333333285963817615993320941925048828125
which also isn't representable as a double-precision number, so again it is rounded to the nearest representable value, which is exactly:
0.3333333333333332593184650249895639717578887939453125
When you print this value out, it gets rounded yet again to 17 decimal digits, giving:
0.33333333333333326
If you just want to process values as fractions, you can create a Fraction class which holds a numerator and denominator field.
Write methods for add, subtract, multiply and divide as well as a toDouble method. This way you can avoid floats during calculations.
EDIT: Quick implementation,
public class Fraction {
private int numerator;
private int denominator;
public Fraction(int n, int d){
numerator = n;
denominator = d;
}
public double toDouble(){
return ((double)numerator)/((double)denominator);
}
public static Fraction add(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop + bTop, a.denominator * b.denominator);
}
else{
return new Fraction(a.numerator + b.numerator, a.denominator);
}
}
public static Fraction divide(Fraction a, Fraction b){
return new Fraction(a.numerator * b.denominator, a.denominator * b.numerator);
}
public static Fraction multiply(Fraction a, Fraction b){
return new Fraction(a.numerator * b.numerator, a.denominator * b.denominator);
}
public static Fraction subtract(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop-bTop, a.denominator*b.denominator);
}
else{
return new Fraction(a.numerator - b.numerator, a.denominator);
}
}
}
Observe that you'd have the same problem if you used limited-precision decimal arithmetic, and wanted to deal with 1/3: 0.333333333 * 3 is 0.999999999, not 1.00000000.
Unfortunately, 5.6, 5.8 and 11.4 just aren't round numbers in binary, because they involve fifths. So the float representation of them isn't exact, just as 0.3333 isn't exactly 1/3.
If all the numbers you use are non-recurring decimals, and you want exact results, use BigDecimal. Or as others have said, if your values are like money in the sense that they're all a multiple of 0.01, or 0.001, or something, then multiply everything by a fixed power of 10 and use int or long (addition and subtraction are trivial: watch out for multiplication).
However, if you are happy with binary for the calculation, but you just want to print things out in a slightly friendlier format, try java.util.Formatter or String.format. In the format string specify a precision less than the full precision of a double. To 10 significant figures, say, 11.399999999999 is 11.4, so the result will be almost as accurate and more human-readable in cases where the binary result is very close to a value requiring only a few decimal places.
The precision to specify depends a bit on how much maths you've done with your numbers - in general the more you do, the more error will accumulate, but some algorithms accumulate it much faster than others (they're called "unstable" as opposed to "stable" with respect to rounding errors). If all you're doing is adding a few values, then I'd guess that dropping just one decimal place of precision will sort things out. Experiment.
You may want to look into using java's java.math.BigDecimal class if you really need precision math. Here is a good article from Oracle/Sun on the case for BigDecimal. While you can never represent 1/3 as someone mentioned, you can have the power to decide exactly how precise you want the result to be. setScale() is your friend.. :)
Ok, because I have way too much time on my hands at the moment here is a code example that relates to your question:
import java.math.BigDecimal;
/**
* Created by a wonderful programmer known as:
* Vincent Stoessel
* xaymaca#gmail.com
* on Mar 17, 2010 at 11:05:16 PM
*/
public class BigUp {
public static void main(String[] args) {
BigDecimal first, second, result ;
first = new BigDecimal("33.33333333333333") ;
second = new BigDecimal("100") ;
result = first.divide(second);
System.out.println("result is " + result);
//will print : result is 0.3333333333333333
}
}
and to plug my new favorite language, Groovy, here is a neater example of the same thing:
import java.math.BigDecimal
def first = new BigDecimal("33.33333333333333")
def second = new BigDecimal("100")
println "result is " + first/second // will print: result is 0.33333333333333
Pretty sure you could've made that into a three line example. :)
If you want exact precision, use BigDecimal. Otherwise, you can use ints multiplied by 10 ^ whatever precision you want.
As others have noted, not all decimal values can be represented as binary since decimal is based on powers of 10 and binary is based on powers of two.
If precision matters, use BigDecimal, but if you just want friendly output:
System.out.printf("%.2f\n", total);
Will give you:
11.40
You're running up against the precision limitation of type double.
Java.Math has some arbitrary-precision arithmetic facilities.
You can't, because 7.3 doesn't have a finite representation in binary. The closest you can get is 2054767329987789/2**48 = 7.3+1/1407374883553280.
Take a look at http://docs.python.org/tutorial/floatingpoint.html for a further explanation. (It's on the Python website, but Java and C++ have the same "problem".)
The solution depends on what exactly your problem is:
If it's that you just don't like seeing all those noise digits, then fix your string formatting. Don't display more than 15 significant digits (or 7 for float).
If it's that the inexactness of your numbers is breaking things like "if" statements, then you should write if (abs(x - 7.3) < TOLERANCE) instead of if (x == 7.3).
If you're working with money, then what you probably really want is decimal fixed point. Store an integer number of cents or whatever the smallest unit of your currency is.
(VERY UNLIKELY) If you need more than 53 significant bits (15-16 significant digits) of precision, then use a high-precision floating-point type, like BigDecimal.
private void getRound() {
// this is very simple and interesting
double a = 5, b = 3, c;
c = a / b;
System.out.println(" round val is " + c);
// round val is : 1.6666666666666667
// if you want to only two precision point with double we
// can use formate option in String
// which takes 2 parameters one is formte specifier which
// shows dicimal places another double value
String s = String.format("%.2f", c);
double val = Double.parseDouble(s);
System.out.println(" val is :" + val);
// now out put will be : val is :1.67
}
Use java.math.BigDecimal
Doubles are binary fractions internally, so they sometimes cannot represent decimal fractions to the exact decimal.
/*
0.8 1.2
0.7 1.3
0.7000000000000002 2.3
0.7999999999999998 4.2
*/
double adjust = fToInt + 1.0 - orgV;
// The following two lines works for me.
String s = String.format("%.2f", adjust);
double val = Double.parseDouble(s);
System.out.println(val); // output: 0.8, 0.7, 0.7, 0.8
Doubles are approximations of the decimal numbers in your Java source. You're seeing the consequence of the mismatch between the double (which is a binary-coded value) and your source (which is decimal-coded).
Java's producing the closest binary approximation. You can use the java.text.DecimalFormat to display a better-looking decimal value.
Short answer: Always use BigDecimal and make sure you are using the constructor with String argument, not the double one.
Back to your example, the following code will print 11.4, as you wish.
public class doublePrecision {
public static void main(String[] args) {
BigDecimal total = new BigDecimal("0");
total = total.add(new BigDecimal("5.6"));
total = total.add(new BigDecimal("5.8"));
System.out.println(total);
}
}
Multiply everything by 100 and store it in a long as cents.
Computers store numbers in binary and can't actually represent numbers such as 33.333333333 or 100.0 exactly. This is one of the tricky things about using doubles. You will have to just round the answer before showing it to a user. Luckily in most applications, you don't need that many decimal places anyhow.
Floating point numbers differ from real numbers in that for any given floating point number there is a next higher floating point number. Same as integers. There's no integer between 1 and 2.
There's no way to represent 1/3 as a float. There's a float below it and there's a float above it, and there's a certain distance between them. And 1/3 is in that space.
Apfloat for Java claims to work with arbitrary precision floating point numbers, but I've never used it. Probably worth a look.
http://www.apfloat.org/apfloat_java/
A similar question was asked here before
Java floating point high precision library
Use a BigDecimal. It even lets you specify rounding rules (like ROUND_HALF_EVEN, which will minimize statistical error by rounding to the even neighbor if both are the same distance; i.e. both 1.5 and 2.5 round to 2).
Why not use the round() method from Math class?
// The number of 0s determines how many digits you want after the floating point
// (here one digit)
total = (double)Math.round(total * 10) / 10;
System.out.println(total); // prints 11.4
Check out BigDecimal, it handles problems dealing with floating point arithmetic like that.
The new call would look like this:
term[number].coefficient.add(co);
Use setScale() to set the number of decimal place precision to be used.
If you have no choice other than using double values, can use the below code.
public static double sumDouble(double value1, double value2) {
double sum = 0.0;
String value1Str = Double.toString(value1);
int decimalIndex = value1Str.indexOf(".");
int value1Precision = 0;
if (decimalIndex != -1) {
value1Precision = (value1Str.length() - 1) - decimalIndex;
}
String value2Str = Double.toString(value2);
decimalIndex = value2Str.indexOf(".");
int value2Precision = 0;
if (decimalIndex != -1) {
value2Precision = (value2Str.length() - 1) - decimalIndex;
}
int maxPrecision = value1Precision > value2Precision ? value1Precision : value2Precision;
sum = value1 + value2;
String s = String.format("%." + maxPrecision + "f", sum);
sum = Double.parseDouble(s);
return sum;
}
You can Do the Following!
System.out.println(String.format("%.12f", total));
if you change the decimal value here %.12f
So far I understand it as main goal to get correct double from wrong double.
Look for my solution how to get correct value from "approximate" wrong value - if it is real floating point it rounds last digit - counted from all digits - counting before dot and try to keep max possible digits after dot - hope that it is enough precision for most cases:
public static double roundError(double value) {
BigDecimal valueBigDecimal = new BigDecimal(Double.toString(value));
String valueString = valueBigDecimal.toPlainString();
if (!valueString.contains(".")) return value;
String[] valueArray = valueString.split("[.]");
int places = 16;
places -= valueArray[0].length();
if ("56789".contains("" + valueArray[0].charAt(valueArray[0].length() - 1))) places--;
//System.out.println("Rounding " + value + "(" + valueString + ") to " + places + " places");
return valueBigDecimal.setScale(places, RoundingMode.HALF_UP).doubleValue();
}
I know it is long code, sure not best, maybe someone can fix it to be more elegant. Anyway it is working, see examples:
roundError(5.6+5.8) = 11.399999999999999 = 11.4
roundError(0.4-0.3) = 0.10000000000000003 = 0.1
roundError(37235.137567000005) = 37235.137567
roundError(1/3) 0.3333333333333333 = 0.333333333333333
roundError(3723513756.7000005) = 3.7235137567E9 (3723513756.7)
roundError(3723513756123.7000005) = 3.7235137561237E12 (3723513756123.7)
roundError(372351375612.7000005) = 3.723513756127E11 (372351375612.7)
roundError(1.7976931348623157) = 1.797693134862316
Do not waste your efford using BigDecimal. In 99.99999% cases you don't need it. java double type is of cource approximate but in almost all cases, it is sufficiently precise. Mind that your have an error at 14th significant digit. This is really negligible!
To get nice output use:
System.out.printf("%.2f\n", total);
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to iterate between 0.1f and 1.0f with 0.1f increments in Java?
Part of my program needs to use values inside a while loop as:
0.1
0.2
0.3
...
0.9
so I need to provide them inside that loop.
Here is the code:
double x = 0.0;
while ( x<=1 )
{
// increment x by 0.1 for each iteration
x += 0.1;
}
I need the output to be EXACTLY:
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
But it actually gives me something like:
0.1
0.2
0.300000000000000000000000004
0.4
0.5
0.6
0.79999999999999999999999999
0.89999999999999999999999999
0.99999999999999999999999999
Welcome to the world of floating point, where 0.1 isn't 0.1. The problem is that many numbers, including 0.1, cannot be represented exactly in a double. So you aren't really adding exactly 0.1 to x each time through the loop.
One approach is to use integer arithmetic and divide by 10:
int i = 0;
while (i <= 10) {
double x = i / 10.0;
. . .
i++;
}
Another approach is to make x a BigDecimal, where you can specify that you want a particular precision. It basically is doing what the above loop does (an integer plus a scale), but packaged up in a nice class with lots of bells and whistles. Oh, and it has arbitrary precision.
you need to use the decimal formatter to get the expected output.
Below is the code for generating the expected output:
import java.text.DecimalFormat;
public class FloatIncrement {
public static void main (String args[]){
double x= 0.0;
DecimalFormat form = new DecimalFormat("#.#");
while(x<0.9){
x= x+0.1;
System.out.println("X : "+Double.valueOf(form.format(x)));
}
}
}
To get output you want, you could use DecimalFormat. Here is some sample code.
import java.text.DecimalFormat;
public class DF {
public static void main(String [] args) {
double x = 0.1;
DecimalFormat form = new DecimalFormat("#.#");
while (x <= .9) {
System.out.println(Double.valueOf(form.format(x)));
x += 0.1;
}
}
}
As far as the implementation you have now, there is no guarantee as to the precision of what gets printed due to the nature of floating point numbers.
Using BigDecimal
double x = 0.0;
int decimalPlaces = 2;
while ( x<=1 )
{
x += 0.1;
BigDecimal bd = new BigDecimal(x);
bd = bd.setScale(decimalPlaces, BigDecimal.ROUND_HALF_UP);
x = bd.doubleValue();
System.out.println(x);
}
That's because you can use binary floating point to do precise decimal arithmetic because FP cannot precisely represent all decimal values.
You need to use an integer value representing some decimal fractional unit like hundredths or thousandths or use something like BigDecimal.
Double is stored in binary
float and double store numbers as a certain number of significant figures and a radix point (kind of like scientific notation). The significant figures part is not always perfect, because it's stored as a certain number of binary digits - so you can't expect it to perform the way you're expecting it to. (for a better explanation see http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html)
Consider using a class such as BigDecimal or a class that implements rational numbers, like the ones mentioned here - Is there a commonly used rational numbers library in Java?
You could also just turn i into an integer, and change it from 1 to 10, and compensate for this in your code.