To use a Comparator: Collections.sort(myList, new Comparator<Chromosome>() { @Override public int compare(Chromosome c1, Chromosome c2) { return Double.compare(c1.getScore(), c2.getScore()); } }); If you plan on sorting numerous Lists in this way I would suggest having Chromosome implement the Comparable interface (in which case you could simply call Collections.sort(myList), without the need of specifying an explicit … Read more
The C standard says that for the f and F floating point format specifiers: If a decimal-point character appears, at least one digit appears before it. I think that if you don’t want a zero to appear before the decimal point, you’ll probably have to do something like use snprintf() to format the number into … Read more
You really can’t compare floating point and integral values in a naive way; particularly, since there’s the classic floating point representation challenges. What you can do is subtract one from the other and see if the difference between them is less than some precision you care about, like so: int iValue = 0; double dValue … Read more
Use a fixed NumberFormat (specifically a DecimalFormat): double value = getValue(); String str = new DecimalFormat(“#”).format(value); alternatively simply cast to int (or long if the range of values it too big): String str = String.valueOf((long) value); But then again: why do you have an integer value (i.e. a “whole” number) in a double variable in … Read more
Use a DecimalFormatter: double number = 0.9999999999999; DecimalFormat numberFormat = new DecimalFormat(“#.00″); System.out.println(numberFormat.format(number)); Will give you “0.99”. You can add or subtract 0 on the right side to get more or less decimals. Or use ‘#’ on the right to make the additional digits optional, as in with #.## (0.30) would drop the trailing 0 … Read more
Are you asking about System.Math.Ceiling? Math.Ceiling(0.2) == 1 Math.Ceiling(0.8) == 1 Math.Ceiling(2.6) == 3 Math.Ceiling(-1.4) == -1
double is internally stored as a fraction in binary — like 1/4 + 1/8 + 1/16 + … The value 0.005 — or the value 1.435 — cannot be stored as an exact fraction in binary, so double cannot store the exact value 0.005, and the subtracted value isn’t quite exact. If you care about … Read more
To return a value when using try/catch you can use a temporary variable, e.g. public static double add(String[] values) { double sum = 0.0; try { int length = values.length; double arrayValues[] = new double[length]; for(int i = 0; i < length; i++) { arrayValues[i] = Double.parseDouble(values[i]); sum += arrayValues[i]; } } catch(NumberFormatException e) { … Read more
Example of using decimal, as Kibbee pointed out double d = 1.275; Math.Round(d, 2); // 1.27 Math.Round((decimal)d, 2); // 1.28