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Round Double value and exponential notation (java)

how to round "3.416436417734133 in "3.416436418" (nine positions after point) but also if i have "3.7578845854848E41" it round to "3.7578845855E41"? i'm trying to realyze a calculator..

You can use DecimalFormat, I am not sure about the other numbers but currently you have numbers which have single digit before the decimal point. So, check following example where you can format the double value. Note one more thing that you may need to change format pattern for your use case.
FOR EXAMPLE :
double d = 3.7578845854848E41;
double d2 = 3.416436417734133;
DecimalFormat f = new DecimalFormat("0.#########E0");
System.out.println(f.format(d));
System.out.println(f.format(d2));
OUTPUT :
3.757884585E41
3.416436418E0
//Replace E0 with space as format returns String
EDIT :
Because of your default locale. You can change local like this,
//Change locale
DecimalFormatSymbols decimalFormatSymbols = new DecimalFormatSymbols(Locale.US);
DecimalFormat f = new DecimalFormat("0.#########E0", decimalFormatSymbols);
//And than use decimal format

You may use BigDecimal to add a "scale" to your double value :
Double d = 3.416436417734133;
BigDecimal round = new BigDecimal(d);
round = round.setScale(9, BigDecimal.ROUND_CEILING);
System.out.println(round);

You can use this code.
BigDecimal aDecimal = new BigDecimal(3.416436417734133);
BigDecimal another = aDecimal.setScale(9, aDecimal.ROUND_HALF_UP);
System.out.println("another: " + another);
System.out.println(new BigDecimal(3.7578845854848E41,new
MathContext(11,RoundingMode.CEILING)));

Formatting a number with correct decimal scale

I need to format a number with scale of 2 decimal places. The original number may be a whole number or a number with three decimal places. However the result should be formatted to have commas and also two decimal places always regardless of whether the original number is whole number or having decimal places.
When original num = 56565656.342 ==> I need 56,565,656.34
When original num = 56565656 ==> I need 56,565,656.00
When original num = 56565656.7 ==> I need 56,565,656.70
I am using the following code which is formatting the code but its failing to add the two decimal places in the above 2 & 3 cases.
String originalNumber = "56565656.7";
BigDecimal b = new BigDecimal(originalNumber).setScale(2, BigDecimal.ROUND_HALF_UP);
String formattedNumber = NumberFormat.getInstance().format(b);
Please let me know if there is any way to accomplish this in efficeint way.
Thanks in advance.

Take a look at the DecimalFormat class.
Alternatively you can setScale method from the BigDecimal Class.
BigDecimal bg1 = new BigDecimal("56565656.342");
BigDecimal bg2 = new BigDecimal("56565656.00");
BigDecimal bg3 = new BigDecimal("56565656.70");
DecimalFormat df = new DecimalFormat("###,###.00");
System.out.println(df.format(bg1.doubleValue()));
System.out.println(df.format(bg2.doubleValue()));
System.out.println(df.format(bg3.doubleValue()));
System.out.println(bg1.setScale(2, BigDecimal.ROUND_HALF_UP));
System.out.println(bg2.setScale(2, BigDecimal.ROUND_HALF_UP));
System.out.println(bg3.setScale(2, BigDecimal.ROUND_HALF_UP));
Yields:
56,565,656.34
56,565,656.00
56,565,656.70
56565656.34
56565656.00
56565656.70
EDIT: Also forgot to mention: If you are after precision, I would recommend you use the setScale method, using the .doubleValue() method will yield a double which can cause loss of precision.

Just use NumberFormat and specify the fraction digits, and rounding method, to print :
String [] originalNumbers = new String[] {
"56565656.342",
"56565656.7",
"56565656"
};
NumberFormat df = NumberFormat.getInstance();
df.setMinimumFractionDigits(2);
df.setMaximumFractionDigits(2);
df.setRoundingMode(RoundingMode.HALF_UP);
for (String number : originalNumbers) {
String formattedNumber = df.format(new BigDecimal(number));
System.out.println(formattedNumber);
}
Will print
56,565,656.34
56,565,656.70
56,565,656.00
** Edit **
DecimalFormat df = new DecimalFormat("#,###.00");
Will produce the exact same result with the given code above.

DecimalFormat class would do it for you.... You will have to specify appropriate format.

round up to 2 decimal places in java? [duplicate]

This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 7 years ago.
I have read a lot of stackoverflow questions but none seems to be working for me. i am using math.round() to round off.
this is the code:
class round{
public static void main(String args[]){
double a = 123.13698;
double roundOff = Math.round(a*100)/100;
System.out.println(roundOff);
}
}
the output i get is: 123 but i want it to be 123.14. i read that adding *100/100 will help but as you can see i didn't manage to get it to work.
it is absolutely essential for both input and output to be a double.
it would be great great help if you change the line 4 of the code above and post it.

Well this one works...
double roundOff = Math.round(a * 100.0) / 100.0;
Output is
123.14
Or as #Rufein said
double roundOff = (double) Math.round(a * 100) / 100;
this will do it for you as well.

double d = 2.34568;
DecimalFormat f = new DecimalFormat("##.00");
System.out.println(f.format(d));

String roundOffTo2DecPlaces(float val)
{
return String.format("%.2f", val);
}

BigDecimal a = new BigDecimal("123.13698");
BigDecimal roundOff = a.setScale(2, BigDecimal.ROUND_HALF_EVEN);
System.out.println(roundOff);

Go back to your code, and replace 100 by 100.00 and let me know if it works.
However, if you want to be formal, try this:
import java.text.DecimalFormat;
DecimalFormat df=new DecimalFormat("0.00");
String formate = df.format(value);
double finalValue = (Double)df.parse(formate) ;

double roundOff = Math.round(a*100)/100;
should be
double roundOff = Math.round(a*100)/100D;
Adding 'D' to 100 makes it Double literal, thus result produced will have precision

I know this is 2 year old question but as every body faces a problem to round off the values at some point of time.I would like to share a different way which can give us rounded values to any scale by using BigDecimal class .Here we can avoid extra steps which are required to get the final value if we use DecimalFormat("0.00") or using Math.round(a * 100) / 100 .
import java.math.BigDecimal;
public class RoundingNumbers {
public static void main(String args[]){
double number = 123.13698;
int decimalsToConsider = 2;
BigDecimal bigDecimal = new BigDecimal(number);
BigDecimal roundedWithScale = bigDecimal.setScale(2, BigDecimal.ROUND_HALF_UP);
System.out.println("Rounded value with setting scale = "+roundedWithScale);
bigDecimal = new BigDecimal(number);
BigDecimal roundedValueWithDivideLogic = bigDecimal.divide(BigDecimal.ONE,decimalsToConsider,BigDecimal.ROUND_HALF_UP);
System.out.println("Rounded value with Dividing by one = "+roundedValueWithDivideLogic);
}
}
This program would give us below output
Rounded value with setting scale = 123.14
Rounded value with Dividing by one = 123.14

Try :
class round{
public static void main(String args[]){
double a = 123.13698;
double roundOff = Math.round(a*100)/100;
String.format("%.3f", roundOff); //%.3f defines decimal precision you want
System.out.println(roundOff); }}

This is long one but a full proof solution, never fails
Just pass your number to this function as a double, it will return you rounding the decimal value up to the nearest value of 5;
if 4.25, Output 4.25
if 4.20, Output 4.20
if 4.24, Output 4.20
if 4.26, Output 4.30
if you want to round upto 2 decimal places,then use
DecimalFormat df = new DecimalFormat("#.##");
roundToMultipleOfFive(Double.valueOf(df.format(number)));
if up to 3 places, new DecimalFormat("#.###")
if up to n places, new DecimalFormat("#.nTimes #")
public double roundToMultipleOfFive(double x)
{
x=input.nextDouble();
String str=String.valueOf(x);
int pos=0;
for(int i=0;i<str.length();i++)
{
if(str.charAt(i)=='.')
{
pos=i;
break;
}
}
int after=Integer.parseInt(str.substring(pos+1,str.length()));
int Q=after/5;
int R =after%5;
if((Q%2)==0)
{
after=after-R;
}
else
{
if(5-R==5)
{
after=after;
}
else after=after+(5-R);
}
return Double.parseDouble(str.substring(0,pos+1).concat(String.valueOf(after))));
}

seems like you are hit by integer arithmetic: in some languages (int)/(int) will always be evaluated as integer arithmetic.
in order to force floating-point arithmetic, make sure that at least one of the operands is non-integer:
double roundOff = Math.round(a*100)/100.f;

I just modified your code. It works fine in my system. See if this helps
class round{
public static void main(String args[]){
double a = 123.13698;
double roundOff = Math.round(a*100)/100.00;
System.out.println(roundOff);
}
}

public static float roundFloat(float in) {
return ((int)((in*100f)+0.5f))/100f;
}
Should be ok for most cases. You can still changes types if you want to be compliant with doubles for instance.

How do I round a double to two decimal places in Java? [duplicate]

This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 3 years ago.
This is what I did to round a double to 2 decimal places:
amount = roundTwoDecimals(amount);
public double roundTwoDecimals(double d) {
DecimalFormat twoDForm = new DecimalFormat("#.##");
return Double.valueOf(twoDForm.format(d));
}
This works great if the amount = 25.3569 or something like that, but if the amount = 25.00 or the amount = 25.0, then I get 25.0! What I want is both rounding as well as formatting to 2 decimal places.

Just use: (easy as pie)
double number = 651.5176515121351;
number = Math.round(number * 100);
number = number/100;
The output will be 651.52

Are you working with money? Creating a String and then converting it back is pretty loopy.
Use BigDecimal. This has been discussed quite extensively. You should have a Money class and the amount should be a BigDecimal.
Even if you're not working with money, consider BigDecimal.

Use a digit place holder (0), as with '#' trailing/leading zeros show as absent:
DecimalFormat twoDForm = new DecimalFormat("#.00");

Use this
String.format("%.2f", doubleValue) // change 2, according to your requirement.

You can't 'round a double to [any number of] decimal places', because doubles don't have decimal places. You can convert a double to a base-10 String with N decimal places, because base-10 does have decimal places, but when you convert it back you are back in double-land, with binary fractional places.

This is the simplest i could make it but it gets the job done a lot easier than most examples ive seen.
double total = 1.4563;
total = Math.round(total * 100);
System.out.println(total / 100);
The result is 1.46.

You can use org.apache.commons.math.util.MathUtils from apache common
double round = MathUtils.round(double1, 2, BigDecimal.ROUND_HALF_DOWN);

double amount = 25.00;
NumberFormat formatter = new DecimalFormat("#0.00");
System.out.println(formatter.format(amount));

You can use Apache Commons Math:
Precision.round(double x, int scale)
source: http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/util/Precision.html#round(double,%20int)

Your Money class could be represented as a subclass of Long or having a member representing the money value as a native long. Then when assigning values to your money instantiations, you will always be storing values that are actually REAL money values. You simply output your Money object (via your Money's overridden toString() method) with the appropriate formatting. e.g $1.25 in a Money object's internal representation is 125. You represent the money as cents, or pence or whatever the minimum denomination in the currency you are sealing with is ... then format it on output. No you can NEVER store an 'illegal' money value, like say $1.257.

Starting java 1.8 you can do more with lambda expressions & checks for null. Also, one below can handle Float or Double & variable number of decimal points (including 2 :-)).
public static Double round(Number src, int decimalPlaces) {
return Optional.ofNullable(src)
.map(Number::doubleValue)
.map(BigDecimal::new)
.map(dbl -> dbl.setScale(decimalPlaces, BigDecimal.ROUND_HALF_UP))
.map(BigDecimal::doubleValue)
.orElse(null);
}

You can try this one:
public static String getRoundedValue(Double value, String format) {
DecimalFormat df;
if(format == null)
df = new DecimalFormat("#.00");
else
df = new DecimalFormat(format);
return df.format(value);
}
or
public static double roundDoubleValue(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;
}

DecimalFormat df = new DecimalFormat("###.##");
double total = Double.valueOf(val);

First declare a object of DecimalFormat class. Note the argument inside the DecimalFormat is #.00 which means exactly 2 decimal places of rounding off.
private static DecimalFormat df2 = new DecimalFormat("#.00");
Now, apply the format to your double value:
double input = 32.123456;
System.out.println("double : " + df2.format(input)); // Output: 32.12
Note in case of double input = 32.1;
Then the output would be 32.10 and so on.

If you want the result to two decimal places you can do
// assuming you want to round to Infinity.
double tip = (long) (amount * percent + 0.5) / 100.0;
This result is not precise but Double.toString(double) will correct for this and print one to two decimal places. However as soon as you perform another calculation, you can get a result which will not be implicitly rounded. ;)

Math.round is one answer,
public class Util {
public static Double formatDouble(Double valueToFormat) {
long rounded = Math.round(valueToFormat*100);
return rounded/100.0;
}
}
Test in Spock,Groovy
void "test double format"(){
given:
Double performance = 0.6666666666666666
when:
Double formattedPerformance = Util.formatDouble(performance)
println "######################## formatted ######################### => ${formattedPerformance}"
then:
0.67 == formattedPerformance
}

Presuming the amount could be positive as well as negative, rounding to two decimal places may use the following piece of code snippet.
amount = roundTwoDecimals(amount);
public double roundTwoDecimals(double d) {
if (d < 0)
d -= 0.005;
else if (d > 0)
d += 0.005;
return (double)((long)(d * 100.0))/100);
}

where num is the double number
Integer 2 denotes the number of decimal places that we want to print.
Here we are taking 2 decimal palces
System.out.printf("%.2f",num);

Here is an easy way that guarantee to output the myFixedNumber rounded to two decimal places:
import java.text.DecimalFormat;
public class TwoDecimalPlaces {
static double myFixedNumber = 98765.4321;
public static void main(String[] args) {
System.out.println(new DecimalFormat("0.00").format(myFixedNumber));
}
}
The result is: 98765.43

int i = 180;
int j = 1;
double div= ((double)(j*100)/i);
DecimalFormat df = new DecimalFormat("#.00"); // simple way to format till any deciaml points
System.out.println(div);
System.out.println(df.format(div));

You can use this function.
import org.apache.commons.lang.StringUtils;
public static double roundToDecimals(double number, int c)
{
String rightPad = StringUtils.rightPad("1", c+1, "0");
int decimalPoint = Integer.parseInt(rightPad);
number = Math.round(number * decimalPoint);
return number/decimalPoint;
}

round decimal to nearest 10th

need to round my answer to nearest10th.
double finalPrice = everyMile + 2.8;
DecimalFormat fmt = new DecimalFormat("0.00");
this.answerField.setText("£" + fmt.format(finalPrice) + " Approx");
the above code rounds a whole number to the nearest 10th however it wont round a decimal. e.g 2.44 should be rounded to 2.40

Use BigDecimal instead.
You really, really don't want to use binary floating point for monetary values.
EDIT: round() doesn't let you specify the decimal places, only the significant figures. Here's a somewhat fiddly technique, but it works (assuming you want to truncate, basically):
import java.math.*;
public class Test
{
public static void main(String[] args)
{
BigDecimal bd = new BigDecimal("20.44");
bd = bd.movePointRight(1);
BigInteger floor = bd.toBigInteger();
bd = new BigDecimal(floor).movePointLeft(1);
System.out.println(bd);
}
}
I'd like to hope there's a simpler way of doing this...

Change a bit the pattern to hard-code the final zero:
double finalPrice = 2.46;
DecimalFormat fmt = new DecimalFormat("0.0'0'");
System.out.println("£" + fmt.format(finalPrice) + " Approx");
Now, if you're manipulating real-world money, you'd better not use double, but int or BigInteger.

This outputs 2.40
BigDecimal bd = new BigDecimal(2.44);
System.out.println(bd.setScale(1,RoundingMode.HALF_UP).setScale(2));

Try the following:
double finalPriceRoundedToNearestTenth = Math.round(10.0 * finalPrice) / 10.0;

EDIT
Try this:
double d = 25.642;
String s = String.format("£ %.2f", Double.parseDouble(String.format("%.1f", d).replace(',', '.')));
System.out.println(s);
I know this is a stupid way, but it works.