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I am looking for the best way to round to even decimal values. For example I have a double = 4.267833399 and I want to round to the nearest even single decimal place, in this case 4.2 and not 4.3. And 4.3165656 would round to 4.4 not 4.3. I have searched hard and havent found a simple way to do this.

I would like the result as a double.

Thanks

DBell
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3 Answers3

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Assuming you want your answer in a double, this should work:

double result = Math.round(input * 5) / 5d;
BeeOnRope
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  • Probably best to avoid using a `float`, just divide by `5.0` – Peter Lawrey Dec 07 '16 at 21:19
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    Not to be confused with `0x5d` ;) – Peter Lawrey Dec 07 '16 at 21:21
  • This is perfect thanks. Now if only I could understand the maths of it. – DBell Dec 07 '16 at 22:11
  • It's easier to understand the rounding to a single decimal point case - basically you do `Math.round(v * 10) / 10d`. That first shifts the decimal place one point to the left, then rounds it off, and then shifts it back to the left. So you can use `Math.round()` by shifting the point around to where you want it, then shifting it back. The even case with a multiplier of 5 works similarly. – BeeOnRope Dec 07 '16 at 22:35
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You can round to one decimal place with

double d = Math.round(x * 10) / 10.0;

Note this will round 4.26 up to 4.3 as it is the closest, if you want to always round down you can do

double d = (long) (x * 10) / 10.0;

I want to round to the nearest even single decimal place

If you want to round to the nearest multiple of 0.2 you can do

double d = Math.round(x * 5) / 5.0;

This will round 4.299 to 4.2 and 4.3 to 4.4

Peter Lawrey
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You can use something like this

double myNumber = 12345.6789;
DecimalFormat df = new DecimalFormat("#.0");
df.format(myNumber);

This would return 12345.7

Edit: Noticed you're looking for rounding to an even number. You could use the above method of rounding and then conditionally manipulate your number after if the decimal is odd.

Fraser Graham
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