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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)));

How to convert a number to its exponential notation in Android?

There is a line, which contains a number:
String s = "12345678901234567890";
Also, there may be a floating point number.
How to display a number in exponential notation, if the obtained value of the number of characters greater than 10, and in normal form, if less than 10?

Try it like this
BigDecimal d = new BigDecimal("12345678901234567890");
DecimalFormat df = new DecimalFormat("0.###E0");
System.out.println(df.format(d));
See more about DecimalFormat in the docs

How to always round off upto 2 decimal places in java

I have tried the following code but it is not working in a particular case.
Eg: Suppose, I have a double value=2.5045 and i want it to be rounded off upto two decimal places using the below code.After rounding off, i get the answer as 2.5. But I want the answer to be 2.50 instead. In this case,zero is trimmed off. Is there any way to retain the zero so as to get the desired answer as 2.50 after rounding off.
private static DecimalFormat twoDForm = new DecimalFormat("#.##");
public static double roundTwoDecimals(double amount) {
return Double.valueOf(twoDForm.format(amount));
}

try this pattern
new DecimalFormat("0.00");
but this will change only formatting, double cannot hold number of digits after decimal poin, try BigDecimal
BigDecimal bd = new BigDecimal(2.5045).setScale(2, RoundingMode.HALF_UP);

Look at the documentation for DecimalFormat. For # it says:
Digit, zero shows as absent
0 is probably what you want:
Digit
So what you are looking for is either "0.00" or "#.00" as a format string, depending on whether you want the first digit before the period, to be visible in case the numbers absolute value is smalle than 0.

Try this
DecimalFormat format = new DecimalFormat("#");
format.setMinimumFractionDigits(2);
answer.setText(format.format(data2));

Try This
double d = 4.85999999999;
long l = (int)Math.round(d * 100); // truncates
d = l / 100.0;

You are returning a double. But double or Double are objects representing a number and don't carry any formatting information. Ìf you need to output two decimal places the point to do this is when you convert your double to a String.

use # if you want to ignore 0
new DecimalFormat("###,#0.00").format(d)

There is another way to achieve this . I have already posted answer in post
will just answer again here. As we will require rounding off values many times .
public class RoundingNumbers {
public static void main(String args[]){
double number = 2.5045;
int decimalsToConsider = 2;
BigDecimal bigDecimal = new BigDecimal(number);
BigDecimal roundedWithScale = bigDecimal.setScale(decimalsToConsider, 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);
}
}
Output we will get is
Rounded value with setting scale = 2.50
Rounded value with Dividing by one = 2.50

double kilobytes = 1205.6358;
double newKB = Math.round(kilobytes*100.0)/100.0;
DecimalFormat df = new DecimalFormat("###.##");
System.out.println("kilobytes (DecimalFormat) : " + df.format(kilobytes));
Try this if u are still getting the above problem

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.

Truncating a Float Value Without Using SetRoundingMode()

I am truncating a float here.But my value is getting rounded.I do not want that.E.g If my value is 12.989 -> it should be printed as 12.98 only. Can someone help
I cannot use decimal format's SetRoundingMode because that is supported from java 1.6 only.
Mine is 1.5 JDK. CAn someone help me out without using SetRoundingMode() Method????
String pattern = "##,##0.00";
NumberFormat nf = NumberFormat.getNumberInstance();
DecimalFormat df = (DecimalFormat)nf;
double fPart;
Float value=233.989f;
String dstr = String.valueOf(value);
dstr = dstr.substring(dstr.indexOf(".")+1);
Long db = Long.valueOf(dstr);
if(db > 0){
df.applyPattern(pattern);
System.out.println("input="+value+", fPart="+dstr);
}
String output = df.format(value);
System.out.println(output);

You can always use old school trick, multiply by 10^n, truncate, divide by 10^n:
float x = 233.989f;
x = (float)(Math.floor(x * 100) / 100);
I've also experimented with BigDecimal:
MathContext mc = new MathContext(5, RoundingMode.FLOOR)
BigDecimal decimal = new BigDecimal(233.989, mc);
System.out.println(decimal);
It does the job but you have to specify total number of digits. You can't just say I want 2 decimal places and I don't care about digits left of decimal point. That's way first parameter of MathContext is 5, not 2. If you opt for this approach, you can quickly calculate non decimal digits with Math.Ceil(Math.log10(x)).
Note:
When dividing (first approach) at least one of operands must be floating point (float or double)
When working with strings (you code), it's not safe to presume that '.' is decimal separator
Truncating decimals with Math.floor only works for positive values

Not sure If I understood you problem correclty. But If you want to truncate without rounding up or down, you can use just like
DecimalFormat df = new DecimalFormat("##.##");
df.format(12.912385);

You can use regular expressions to get the second digit after "." and then subtract the string from the beginning to that position and then transform the string into a float or double.
Pattern pattern = Pattern.compile("regular expression");
Matcher matcher = pattern.matcher("your string");
if(matcher.find())
{
int poz_begin = matcher.start();
int poz_end = matcher.end();
}