Setting precision for double in java

Setting precision for double in java - java

DecimalFormat df = new DecimalFormat("#.000000");
int a[] = { 2, 2, 3, 3, 4, 4 };
double sum = 0.000000;
for (int i = 0; i < a.length; i++)
{
sum = sum + (double) a[i];
}
output1=Double.valueOf(df.format(sum / a.length));
where sum/a.length value is 3. output1 is double variable. Now the result I wanted is 3.000000 and it must be store in double variable output1 but I can't get it.

Although in certain cases it might work, in general there is no way to determine/force the decimal precision of a double value, or indeed any IEEE floating point number.
If you want decimal precision in Java, use BigDecimal. This is even more important if the numbers you work with represent money.
If an approximate result is good enough (and there are lots of calculations where it is), you can use double but be aware that it's a binary floating point number and accurate rounding to decimals might not always be possible.

The primitive type double is an approximation of a real number, with a sequence of (negative) powers of 2.
Hence the decimal notation 0.2 = 0*2-1 + ... + 1*2-4 + ... with an error as one would need an infinite sequence in base 2.
If one wants a precision with the value, one needs BigDecimal:
BigDecimal oneFifth = new BigDecimal("0.200"); // Precision/scale 3
BigDecimal hundredPlusOnefifth =
oneFifth.multiply(BigDecimal.valueOf(501)); // 100.200
Using a String in the constructor, BigDecimal can set the precision.
Not so nice writing expressions in BigDecimal though.
With double one might live, while carefully rounding at appropriate points in the code. There always will be a small error and, outputting needs a formatter as the number of digits is lost.

The value of 3.0 and 3.00000 are the same in a double variable. When you print it, format it the way you want:
System.out.println( df.format( output1 ) );

Looks like sum is int and you have the result of integer division (because a.length is int). Just multiply one of those values by 1.0:
output1 = Double.valueOf(df.format((sum * 1.0) / a.length));
With your edited code, your problem is not in obtaining the value of output1 but how you show it. Don't print output1 directly, instead use the DecimalFormat you used previously:
System.out.println(df.format(output1));

Related

Why this double operation is not precise in Java? [duplicate]

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);

Java - Numbers aren't subtracting correctly? [duplicate]

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);
}
}
}

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);

Java - Equation issues [duplicate]

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);
}
}
}

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);

Double Values Increases Randomly

The double Value increases randomly.
for(double i=-1;i<=1;i+=0.1)
{
for(double j=-1;j<=1;j+=0.1)
{
//logic
system.out.print(i);
system.out.print(j);
}
}
Here, the value comes like:
-1, -0.9, -0.8, -0.69, -0.51 ....-0.099 , 1.007 (WHY, U ARE GREATER THAN 1)
The output is not same but kind of this.
But, I want the exact values only. WHat should I do ??

You can use an integer counter, and multiply to get the double:
for(int i = -10; i <= 10; i++) {
double iDouble = 0.1 * i;
....
}
The double will still have rounding error - that is inevitable - but the rounding error will not affect the loop count.

You can't get exact values do to the limitations of doubles. They can't always represent exactly the decimal you want, and they have precision errors. In your case you may want to cast the double to an int for the double comparison, but as #alex said you shouldn't be doing this.

This is due to the way that doubles are stored in memory, they are only exact if the fractional part is a negative power of two, e.g. 0, 1/2, 1/4, etc. This is also why you should never use equality statements for doubles, but rather > and <. For exact calculations, you could use BigDecimal:
BigDecimal bigDecimal = new BigDecimal(123.45);
bigDecimal = bigDecimal.add(new BigDecimal(123.45));
System.out.println(bigDecimal.floatValue()); // prints 246.9

As said here, floating-point variables must not be used as loop counters. Limited-precision IEEE 754 floating-point types cannot represent:
all simple fractions exactly
all decimals precisely, even when the decimals can be represented in a small number of digits.
all digits of large values, meaning that incrementing a large floating-point value might not change that value within the available precision.
(...) Using floating-point loop counters can lead to unexpected behavior.
Instead, use integer loop counter and increment another variable inside this loop like this:
for (int count = 1; count <= 20; count += 1) {
double x = -1 + count * 0.1;
/* ... */
}

Round a double to 2 decimal places [duplicate]

This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 8 years ago.
If the value is 200.3456, it should be formatted to 200.34.
If it is 200, then it should be 200.00.

Here's an utility that rounds (instead of truncating) a double to specified number of decimal places.
For example:
round(200.3456, 2); // returns 200.35
Original version; watch out with this
public static double round(double value, int places) {
if (places < 0) throw new IllegalArgumentException();
long factor = (long) Math.pow(10, places);
value = value * factor;
long tmp = Math.round(value);
return (double) tmp / factor;
}
This breaks down badly in corner cases with either a very high number of decimal places (e.g. round(1000.0d, 17)) or large integer part (e.g. round(90080070060.1d, 9)). Thanks to Sloin for pointing this out.
I've been using the above to round "not-too-big" doubles to 2 or 3 decimal places happily for years (for example to clean up time in seconds for logging purposes: 27.987654321987 -> 27.99). But I guess it's best to avoid it, since more reliable ways are readily available, with cleaner code too.
So, use this instead
(Adapted from this answer by Louis Wasserman and this one by Sean Owen.)
public static double round(double value, int places) {
if (places < 0) throw new IllegalArgumentException();
BigDecimal bd = BigDecimal.valueOf(value);
bd = bd.setScale(places, RoundingMode.HALF_UP);
return bd.doubleValue();
}
Note that HALF_UP is the rounding mode "commonly taught at school". Peruse the RoundingMode documentation, if you suspect you need something else such as Bankers’ Rounding.
Of course, if you prefer, you can inline the above into a one-liner:
new BigDecimal(value).setScale(places, RoundingMode.HALF_UP).doubleValue()
And in every case
Always remember that floating point representations using float and double are inexact.
For example, consider these expressions:
999199.1231231235 == 999199.1231231236 // true
1.03 - 0.41 // 0.6200000000000001
For exactness, you want to use BigDecimal. And while at it, use the constructor that takes a String, never the one taking double. For instance, try executing this:
System.out.println(new BigDecimal(1.03).subtract(new BigDecimal(0.41)));
System.out.println(new BigDecimal("1.03").subtract(new BigDecimal("0.41")));
Some excellent further reading on the topic:
Item 48: "Avoid float and double if exact answers are required" in Effective Java (2nd ed) by Joshua Bloch
What Every Programmer Should Know About Floating-Point Arithmetic
If you wanted String formatting instead of (or in addition to) strictly rounding numbers, see the other answers.
Specifically, note that round(200, 0) returns 200.0. If you want to output "200.00", you should first round and then format the result for output (which is perfectly explained in Jesper's answer).

If you just want to print a double with two digits after the decimal point, use something like this:
double value = 200.3456;
System.out.printf("Value: %.2f", value);
If you want to have the result in a String instead of being printed to the console, use String.format() with the same arguments:
String result = String.format("%.2f", value);
Or use class DecimalFormat:
DecimalFormat df = new DecimalFormat("####0.00");
System.out.println("Value: " + df.format(value));

I think this is easier:
double time = 200.3456;
DecimalFormat df = new DecimalFormat("#.##");
time = Double.valueOf(df.format(time));
System.out.println(time); // 200.35
Note that this will actually do the rounding for you, not just formatting.

The easiest way, would be to do a trick like this;
double val = ....;
val = val*100;
val = Math.round(val);
val = val /100;
if val starts at 200.3456 then it goes to 20034.56 then it gets rounded to 20035 then we divide it to get 200.34.
if you wanted to always round down we could always truncate by casting to an int:
double val = ....;
val = val*100;
val = (double)((int) val);
val = val /100;
This technique will work for most cases because for very large doubles (positive or negative) it may overflow. but if you know that your values will be in an appropriate range then this should work for you.

Please use Apache commons math:
Precision.round(10.4567, 2)

function Double round2(Double val) {
return new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue();
}
Note the toString()!!!!
This is because BigDecimal converts the exact binary form of the double!!!
These are the various suggested methods and their fail cases.
// Always Good!
new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue()
Double val = 260.775d; //EXPECTED 260.78
260.77 - WRONG - new BigDecimal(val).setScale(2,RoundingMode.HALF_UP).doubleValue()
Double val = 260.775d; //EXPECTED 260.78
260.77 - TRY AGAIN - Math.round(val * 100.d) / 100.0d
Double val = 256.025d; //EXPECTED 256.03d
256.02 - OOPS - new DecimalFormat("0.00").format(val)
// By default use half even, works if you change mode to half_up
Double val = 256.025d; //EXPECTED 256.03d
256.02 - FAIL - (int)(val * 100 + 0.5) / 100.0;

double value= 200.3456;
DecimalFormat df = new DecimalFormat("0.00");
System.out.println(df.format(value));

If you really want the same double, but rounded in the way you want you can use BigDecimal, for example
new BigDecimal(myValue).setScale(2, RoundingMode.HALF_UP).doubleValue();

double d = 28786.079999999998;
String str = String.format("%1.2f", d);
d = Double.valueOf(str);

For two rounding digits. Very simple and you are basically updating the variable instead of just display purposes which DecimalFormat does.
x = Math.floor(x * 100) / 100;

Rounding a double is usually not what one wants. Instead, use String.format() to represent it in the desired format.

In your question, it seems that you want to avoid rounding the numbers as well? I think .format() will round the numbers using half-up, afaik?
so if you want to round, 200.3456 should be 200.35 for a precision of 2. but in your case, if you just want the first 2 and then discard the rest?
You could multiply it by 100 and then cast to an int (or taking the floor of the number), before dividing by 100 again.
200.3456 * 100 = 20034.56;
(int) 20034.56 = 20034;
20034/100.0 = 200.34;
You might have issues with really really big numbers close to the boundary though. In which case converting to a string and substring'ing it would work just as easily.

value = (int)(value * 100 + 0.5) / 100.0;

We Keep Coding

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