I want to be able to make sure a number only has two decimal places.
E.G Area entered = 256.12345 so Area would be 256.12.
This is what I have:
DecimalFormat df = new DecimalFormat( "#,###,###,##0.00" );
double area = new Double(area.format(area)).doubleValue();
area = (double)(r*r);
You're not actually using the instance df.
Change your code to use it instead of calling methods on area (which won't work, since primitives don't have methods):
double area = new Double(df.format(area)).doubleValue();
However, the formality of precision is more for printing purposes than storing purposes (Double will store it in the IEEE floating point standard for doubles, which may lead to imprecise floating point values).
To get around that, use a BigDecimal instead, with a precision of 2:
BigDecimal decimal = new BigDecimal(area);
decimal.setScale(2);
System.out.println(decimal); // will print area to two decimal places
This is the way you do it.
//formatting numbers upto 2 decimal places in Java
DecimalFormat df = new DecimalFormat("#,###,##0.00");
System.out.println(df.format(364565.14));
System.out.println(df.format(364565.1454));
//formatting numbers upto 3 decimal places in Java
df = new DecimalFormat("#,###,##0.000");
System.out.println(df.format(364565.14));
System.out.println(df.format(364565.1454));
}
}
Output:
364,565.14
364,565.15
364,565.140
364,565.145
Related
val numberFormatter = NumberFormat.getNumberInstance(Locale.getDefault())
val conversionPattern = "#,##0.####"
val decimalFormatter = numberFormatter as DecimalFormat
decimalFormatter.applyPattern(conversionPattern)
decimalFormatter.format("9999999999999999".toDouble()) // Getting -> "10,000,000,000,000,000"
// Need -> "9,999,999,999,999,999"
What's going wrong? Is it overflow? I am dealing with really large numbers so I'm using BigDecimal for the underlying value but to format it with grouping separators there's no format() function that accepts BigDecimal. How can I format numbers with at least 20 digits?
Use BigDecimal or BigInteger for numbers as large as this. For example:
decimalFormatter.format(new BigDecimal("9999999999999999"))
A double precision floating point number has only around 15-16 decimal digits of precision. Your number has 16 nines. There is no double precision floating point number exactly equal to 9999999999999999, so it is rounded to the closest one - which happens to be 1016.
I'm basically trying to use DecimalFormat to get to two decimal places. I'm taking two integer values then dividing them and casting to a double I've put in sample values below. When I do as below I get a value that is no longer to two decimal places. It seems to be when multiply by the 3 it loses it's rounding.
DecimalFormat df = new DecimalFormat("#.00");
double d = Double.parseDouble(df.format((double)5/6))*3;
System.out.println(d);
Can you let me know why this occurs and how to fix this?
In the statemet:
double d = Double.parseDouble(df.format((double)5/6))*3;
the formatting is not preserved (Double returns a double).
You could do, e.g.:
System.out.println(df.format(d));
I require a DecimalFormat or a better equivalent of representing a Double value (in Java) which could be:
25 or 25.5
I need for that to be represented as either a whole number (25) or to two decimal places if it has any (25.50). This is because i'm printing it out as money.
I have the following format already:
DecimalFormat decFormat = new DecimalFormat("##,###.##");
This works perfectly if the Double is a whole number; I get the output $25,000. Except if the value is 25,000.5; it prints $25,000.5 when I need it to be printed as $25,000.50. The problem is as stated in the docs:
# a digit, zero shows as absent
So essentially the last zero is dropped off since it is optional.
I cannot do:
DecimalFormat decFormat = new DecimalFormat("##,###.#0");
as that is not allowed.
How can I achieve this?
Note:
These questions are related but do not cover what I need specifically with the DecimalFormat. Most of the answers suggest using a BigDecimal or printf. Is this the best thing to do? I don't have to use DecimalFormat but prefer to since i've started on that path (lots of code everywhere already using it).
Best way to Format a Double value to 2 Decimal places
How do I round a double to two decimal places in Java?
Round a double to 2 decimal places
This is definitely a bit of a hack, but I don't know if the DecimalFormat syntax allows for anything better. This simply checks to see if the number is real, and formats based on the spec you asked for.
double number = 25000.5;
DecimalFormat df;
if(number%1==0)
df = new DecimalFormat("##,###");
else
df = new DecimalFormat("##,###.00");
System.out.println(df.format(number));
When you need to return Decimal Format value this works
import java.text.DecimalFormat;
/**
* #return The weight of this brick in kg.
*/
public double getWeight()
{
DecimalFormat df = new DecimalFormat("#.##");
double number = ( getVolume() * WEIGHT_PER_CM3 ) / 1000;
//System.out.println(df.format(number));
return Double.valueOf ( df.format( number ) );
}
I'm trying to round a double to the nearest two decimal places however, it is just rounding to the nearest full number.
For example, 19634.0 instead of 19634.95.
This is the current code I use for the rounding
double area = Math.round(Math.PI*Radius()*Radius()*100)/100;
I can't see where i am going wrong.
Many thanks for any help.
Well, Math.round(Math.PI*Radius()*Radius()*100) is long. 100 is int.
So Math.round(Math.PI*Radius()*Radius()*100) / 100 will become long (19634).
Change it to Math.round(Math.PI*Radius()*Radius()*100) / 100.0. 100.0 is double, and the result will also be double (19634.95).
You can use a DecimalFormat object:
DecimalFormat df = new DecimalFormat ();
df.setMaximumFractionDigits (2);
df.setMinimumFractionDigits (2);
System.out.println (df.format (19634.95));
Do you actually want want to round the value to 2 places, which will cause snowballing rounding errors in your code, or simply display the number with 2 decimal places? Check out String.format(). Complex but very powerful.
You might want to take a look at the DecimalFormat class.
double x = 4.654;
DecimalFormat twoDigitFormat = new DecimalFormat("#.00");
System.out.println("x=" + twoDigitFormat.format());
This gives "x=4.65". The difference between # and 0 in the pattern is that the zeros are always displayed and # will not if the last ones are 0.
The following example came from this forum, but seems to be what you are looking for.
double roundTwoDecimals(double d) {
DecimalFormat twoDForm = new DecimalFormat("#.##");
return Double.valueOf(twoDForm.format(d));
}
I was going through the class decimal format as I was trying format a decimal number in Java upto 2 decimal places or 3 decimal places.
I come up with this solution as shown below but please also let me know are there any other alternative that java provides us to achieve the same thing..!!
import java.text.DecimalFormat;
public class DecimalFormatExample {
public static void main(String args[]) {
//formatting numbers upto 2 decimal places in Java
DecimalFormat df = new DecimalFormat("#,###,##0.00");
System.out.println(df.format(364565.14));
System.out.println(df.format(364565.1454));
//formatting numbers upto 3 decimal places in Java
df = new DecimalFormat("#,###,##0.000");
System.out.println(df.format(364565.14));
System.out.println(df.format(364565.1454));
}
}
Output:
364,565.14
364,565.15
364,565.140
364,565.145
Please advise what are other alternatives that java provide us to achieve the same thing..!!
If you are bothered by re-defining your DecimalFormat, or if you suspect you'll be needing to do redefine many times, you could also do inline formatting with String.format(). Check the syntax for Formatter especially the Numeric sub-title.
Here is an alternative to round off...
double a = 123.564;
double roundOff = Math.round(a * 10.0) / 10.0;
System.out.println(roundOff);
roundOff = Math.round(a * 100.0) / 100.0;
System.out.println(roundOff);
The output is
123.6
123.56
Number of 0s while multiplying and dividing decides the rounding off.
Here is one method.
float round(float value, int roundUpTo){
float x=(float) Math.pow(10,roundUpTo);
value = value*x; // here you will guard your decimal points from loosing
value = Math.round(value) ; //this returns nearest int value
return (float) value/p;
}