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how do i restric a float value to only one placess after decimal in kotlin android

I have a number with some number decimal places,
How can i round this float number with one number decimal places
for example
1.366565646 convert to 1.3

In your case I think you need to trim the number not round it,
You can use this:
double d = 1.366565646;
DecimalFormat df = new DecimalFormat("#.#");
double p = Double.parseDouble(df.format(d));

Keep up to 3 decimals without doing any rounding

I have a double that I want to keep only 3 decimal places but without applying any rounding at all.
E.g. 92.36699 should be 92.366
I tried the following:
DecimalFormat nf= new DecimalFormat("#0.000");
String number = nf.format(originalNumber);
But this results in 92.367
How can I do what I need?

This isn't "no rounding", it's DOWN rounding. Simply set the roundingMode.
DecimalFormat nf = new DecimalFormat("#0.000");
nf.setRoundingMode(RoundingMode.DOWN);
String number = nf.format(originalNumber);
Note the difference between FLOOR and DOWN - only relevant for negative numbers. FLOOR rounds towards negative infinity therefore -92.36699 would become "-92.367".

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

Print the number of digits before a decimal point

I know there are ways to get the number of digits after a decimal point, for instance the substring method, but how would I go about doing this for the number of digits before a decimal place?
I need to use this to convert US change (double) into Euro change(double). The way I would like to do this is by taking the number before a decimal (such as $1.) and times it by its euro equivalent (.7498) and take the number after a decimal (.16) and times that by its .01 euro coin value (.0075), add both values together to get the euro equivalent of $1.16 (.8698).

To get the number before decimal point,do this:
String str = new Double(your_double_number).toString().subString(0,str.indexOf('.'));
double v = Double.valueOf(str);
If you are using '$' sign then take 1 in place of 0.
Hope it will help you.

convert US change (double) into Euro change(double)
Please don't do that. Never, never, ever use double or float to represent money, because those datatypes cannot represent most decimal fractions, so you get rounding errors before you even start to do any calculations.
Instead, use BigDecimal.

First of all - just multiplying the Dollar value by the exchange rate will get you the euro value so there's no need to do that as far as i can see - you will just introduce rounding errors.
But if you did need to - just use substring
String dollarVal = "$1.16"
String justFullDollar = dollarVal.substring(1, dollarVal.indexOf("."));
String justCents = dollarVal.substring(dollarVal.indexOf(".")+1);
The Correct way would be to store all you money as integers or arbitrary precision objects that way you get no floating point errors too.
Convert to cents, multiply and convert back again.
e.g.
String dollarVal = "$1.16"
BigDecimal dollars = new BigDecimal(dollarVal.substring(1)); //1.16
BigDecimal cents = dollars.multiply(new BigDecimal(100)); //116
BigDecimal eurocents = cents.multiply(new BigDecimal(exchangeRate)); //86.9768
BigDecimal euros = eurocents.divide(new BigDecimal(100)); //0.869768
DecimalFormat formatter = new DecimalFormat("###.00");
String euroVal = "€" + formatter.format(euros);

You can use
String s[] = new Double(your number).toString().split(".");
The s[0] is the number before decimal point. and s[1] is the number after decimal point.
Both are in String, so you need to parse them into double using
double num1 = Double.parseDouble(s[0]);
double num2 = Double.parseDouble(s[1]);

Converting numeric value with currency symbol back to Decimal with NumberFormat

I would like to convert a possibly Decimal value prefixed with currency symbol into only numeric value.
For example -
The value can be like any of the following
String s1 = "£32,847,676.65";
String s2 = "£3,456.00";
String s3 = "£831,209";
I would like the result after conversion to be like - 32847676.65, 3456.00 and 831209.
I tried using the parse() method of the NumberFormat in this way -
NumberFormat nf = NumberFormat.getCurrencyInstance(Locale.UK);
numberFormat.setMinimumFractionDigits(2);
Number num = nf.parse(s1);
double dd = num.doubleValue();
BigDecimal gg = new BigDecimal(dd);
System.out.println(gg);
But the result is - 32847676.649999998509883880615234375 which is not quite exactly the correct one.
I need it to be numeric so that may be I can perform some kind of calculation.
Can you guys guide me with what else can I try

You already parse the value correctly. The problem is this:
BigDecimal gg = new BigDecimal(dd);
You covnert the value to BigDecimal, and the rounding problems of doubles account for the decimal places after the dot. Use:
BigDecimal gg = new BigDecimal(dd).setScale(2);
or
BigDecimal gg = new BigDecimal(dd).setScale(2,RoundingMode.HALF_UP);

When playing with BigDecimal, the appropriate constructor is BigDecimal(String val)
NumberFormat nf = NumberFormat.getCurrencyInstance(Locale.UK);
BigDecimal gg = new BigDecimal(nf.parse(s1).toString());
System.out.println(gg);
BigDecimal(double val) does construct an exact decimal representation of the double value, which is not the human readable value you expected.
"The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.
[...]
Therefore, it is generally recommended that the String constructor be used in preference to this one"
Source : BigDecimal javadoc

You can try the following without BigDecimal or NumberFormat. ;)
String s1 = "£32,847,676.65";
// remove the £ and ,
String s2 = s1.replaceAll("[£,]", "");
// then turn into a double
double d = Double.parseDouble(s2);
// and round up to two decimal places.
double value = (long) (d * 100 + 0.5) / 100.0;
System.out.printf("%.2f%n", value);
prints
32847676.65
If youw ant to avoid rounding error in your calculations but don't want the heavy weight BigDecimal you can use long cents.
// value in cents as an integer.
long value = (long) (d * 100 + 0.5);
// perform some calculations on value here
System.out.printf("%.2f%n", value / 100.0);

It is not guaranteed to work, but according to the NumberFormat API documentation, its getXyzInstance methods will return a DecimalFormat instance for "the vast majority of locales". This can probably be interpreted as "for all locales, unless proprietary locale service providers are installed".
If you can cast your NumberFormat to DecimalFormat, you can tell it to parse to a BigDecimal directly, reducing your code to:
DecimalFormat nf = (DecimalFormat) NumberFormat.getCurrencyInstance(Locale.UK);
nf.setParseBigDecimal(true);
BigDecimal gg = (BigDecimal) nf.parse(s1);
System.out.println(gg);
In this case, you will have no problem with the inaccuracy of binary floating point numbers.