Right now I am doing this
DecimalFormat df = new DecimalFormat("#.##");
//usage: df.format(someDouble);
But I don't know if this is the correct way to "round off" the display of a double to 2 decimal points. Will this work even if the number of digits to the left of the decimal point > 1?
You should use:
DecimalFormat df = new DecimalFormat("#.00");
DecimalFormat df = new DecimalFormat("#.##"); means, that those decimal places are optional. You can see the difference when applying on 1.2 or 1.0.
--Documentation
To answer the question about the rounding, the documentation tell it :
Rounding
DecimalFormat provides rounding modes defined in RoundingMode for formatting. By default, it uses RoundingMode.HALF_EVEN.
So this is up to you to set the RoundingMode that you want with the setter.
public void setRoundingMode(RoundingMode roundingMode)
Will this work even if the number of digits to the left of the decimal point > 1?
Yes it will.
Related
I have a number with value
0.947
Now, I use DecimalFormat API of java.text with the following pattern and RoundingMode-
double numberToFormat = 0.947;
DecimalFormat restrictTo1DecimalPlace = new DecimalFormat("0.0%");
restrictTo1DecimalPlace.setRoundingMode(RoundingMode.DOWN);
String formattedString = restrictTo2DecimalPlace.format(numberToFormat);
Now, I was expecting the value of formattedString to be 94.7% but its 94.6%.
I know the value has been set to RoundMode.Down but then why does value of following are not rounded down -
0.9471 -> 94.7%
0.9447 -> 94.4%
The closest floating point number to 0.947 is actually
0.94699999999999995292654375589336268603801727294921875
This is what your computer stores as a double when you write 0.947.
Rounding that down gives you 94.6%.
That's life I'm afraid. If you want exact decimal behaviour then use a decimal type! See data type to represent a big decimal in java
I'm trying to prevent a java double from printing in scientific notation. I know I can use DecimalFormat for this, but this produces a String. How can I produce an actual double?
I've tried:
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(8);
double n = Double.parseDouble(df.format(z));
However, this still produces a double in scientific notation.
A format is not a property of a double, the data type. This defines just a set of values.
Printing always requires a conversion to a String, so DecimalFormat is fine.
As for using an entirely different set of decimal fractions as values, see java.math.BigDecimal. This is the set of "all" decimal fractions, with a dynamic number of fractional digits, operations and control of round-off.
I would like to print a BigDecimal in scientific notation so that the there are, say, 5 digits after the decimal point. For example: -3.12345E-51, 9.12345E100
.setScale() does not work in this case, because I don't know in advance the exponent of number in scientific notation. Moreover, BigDecimal doesn't seem to have a getExponent() method.
What's the best way to achieve this? Thank you in advance.
You should use DecimalFormat to print numbers in specified format:
BigDecimal bigDecimal = new BigDecimal(-3.12345E-51);
DecimalFormat format = new DecimalFormat("0.#####E0");
System.out.println(format.format(bigDecimal));
Also you can configure count of fraction digits using method DecimalFormat#setMaximumFractionDigits(int):
DecimalFormat format = new DecimalFormat("0E0");
format.setMaximumFractionDigits(6);
If I want a double to have 9 decimal places, do I have to convert it to a string and then back to a double to do this (string methods are the only methods I'm seeing for setting the precision)?. In any case, what is the conventional way for setting the precision for a double if, for example, I want my method to return a double with 9 decimal places.
You're getting confused between a double's internal representation and it's display. Internally double numbers are always stored the same way but you can of course use a formater like DecimalFormat to return 9 decimal points from your double number.
Use Big Decimal to get 9 decimal places
or
double d = 1.2345672626346;
DecimalFormat df = new DecimalFormat("#.#########");
System.out.print(df.format(d));
You can't control precision with primitives in Java. You need to use BigDecimal. Read this excellent tutorial..
You could also use:
DecimalFormat df = DecimalFormat.getInstance(Locale.getDefault());
df.setMinimumFractionDigits(9); //sets 9 digits after the '.'
String formatedString = String.format("%.2f");
it set the presition to 5 digit o any.
if you only want to print it use this way.
System.out.printf("%.5f",123.2342622467);
(Math.round(doubleValue*100))/100.0
Is there a better way to round decimals to 2 decimal places?
If you're interested in decimal places and therefore precise decimal values, you should typically be using java.math.BigDecimal to start with. You can then use Decimal.round or Decimal.setScale to round according to your exact needs.
DecimalFormat format=new DecimalFormat("#.##");
System.out.println(format.format(doubleValue));