Java: Check if two double values match on specific no of decimal places

Question

Comparing those two values shall result in a "true":

53.9173333333333  53.9173

Thanks in advance!

Answer 1

If you want a = 1.00001 and b = 0.99999 be identified as equal:

return Math.abs(a - b) < 1e-4;

Otherwise, if you want a = 1.00010 and b = 1.00019 be identified as equal, and both a and b are positive and not huge:

return Math.floor(a * 10000) == Math.floor(b * 10000);
// compare by == is fine here because both sides are integral values.
// double can represent integral values below 2**53 exactly.

Otherwise, use the truncate method as shown in http://stackoverflow.com/questions/1976809/are-there-any-functions-for-truncating-a-double-in-java:

BigDecimal aa = new BigDecimal(a);
BigDecimal bb = new BigDecimal(b);
aa = aa.setScale(4, BigDecimal.ROUND_DOWN);
bb = bb.setScale(4, BigDecimal.ROUND_DOWN);
return aa.equals(bb);

Answer 2

Naively:

if(Math.abs(a-b) < 0.0001)

However, that's not going to work correctly for all values. It's actually impossible to get it to work as long as you're using double, because double is implemented as binary fractons and does not even have decimal places.

You'll have to convert your values to String or BigDecimal to make meaningful tests about their decimal places.

You may want to read the Floating-Point Guide to improve your understanding of how floating point values work.

Answer 3

Thanks. I did it this way:

double lon = 23.567889;
BigDecimal bdLon = new BigDecimal(lon);
bdLon = bdLon.setScale(4, BigDecimal.ROUND_HALF_UP);

System.out.println(bdLon.doubleValue());