CODEWITHSUNDEEP
java
user339108
user339108

Reputation: 13101

How do you trim the BigDecimal division results

BigDecimal numerator = new BigDecimal(numerator);
BigDecimal denominator = new BigDecimal(denominator);
double result = numerator.divide(denominator).doubleValue();

Division on certain conditions results in a zero at the end (e.g. 0.0060). Dividing equal numbers results in a zero at the end (e.g 1.0).

I would prefer to trim the trailing zero in both cases. How should I do this?

Upvotes: 3

Views: 2284

Answers (3)

Trevor Tippins
Trevor Tippins

Reputation: 2847

How about keeping the result as a BigDecimal and then you can set the scale on it to only represent the significant figures that you want.

An easy way to do this, for some numbers, is to use BigDecimal#stripTrailingZeros(). However, if the number is an integer with trailing zeros you'll get an engineering representation e.g. 600.0 will give you 6E+2. If this isn't what you want, you'll have to detect this condition and manually use BigDecimal#setScale() to set the scale appropriately.

If you need to keep to a restricted maximum number of decimal digits you'll need to use alternative formatting/rounding mechanisms before applying this technique.

It's also a good idea to only do this on values that you're going to display, not on the internal values of your model. Treat it as a view/presentation layer modification.

If you must convert to a double, then it's only the formatted representation you can alter. In this case, if you've got a variable number of decimal places that you want to format to, I'd just drop it into a string/character array, scan backwards for the first non-zero character and truncate it there. Not the most performant means, but simple and reliable.

You could even use a regex for this purpose.

Upvotes: 3

darioo
darioo

Reputation: 47183

Ok, so you've got this code:

BigDecimal numerator = new BigDecimal(numerator);
BigDecimal denominator = new BigDecimal(denominator);
double result = numerator.divide(denominator).doubleValue();

and you want more control over your output. Since result is a double, which is a primitive, you won't have much control.

From my understanding, you don't want to do any rounding to n decimal places, you want original precision paired with desired formatting.

You have few options.

BigDecimal numerator = new BigDecimal(numerator);
BigDecimal denominator = new BigDecimal(denominator);
BigDecimal div = numerator.divide(denominator);

If you stay with BigDecimal, your output will be better. If you put 10 as the numerator and denominator in above code, System.out.println(div) will yield 1.

Generally, be careful of using above code because some combinations of numerator and denominator will throw 

java.lang.ArithmeticException: "Non-terminating decimal expansion; no exact representable decimal result."

If you want to avoid such situations, and not worry about precision beyond double's internal representation, use double directly.

System.out.println(2312 / 2.543); //909.1624066063704
System.out.println(1.0 / 1.0);    //1.0
System.out.println(1 / 1);        //1

When using double numbers, you might get a 0 at the end, such as 0.0060 in your case. If you want to be sure what you're getting, you'll have to convert your result to a String using

String dec = String.valueOf(10.0/10.0); //1.0

and then using

String newDec = dec.endsWith("0") ? dec.substring(0, dec.length() - 1) : dec;

to eradicate that last 0. Of course, if your string ends with .0, you have a choice based on your preferences whether you want to leave that leading . or not.

Upvotes: 1

Andre
Andre

Reputation: 9170

you can do this by using DecimalFormat. something like:

DecimalFormat df = new DecimalFormat(".0");
double formatResult = df.format(result);

will create something of 1.0 if the result is 1.278494890. there are many possible patterns that could be used here

Upvotes: 1

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