How to sort two arrays based on ratio in Java?

Question

I'd like to sort 2 arrays based on the ratios of their matching indices. So based on the ratio of a[x]/b[x], a[] and b[] are sorted accordingly. Here is the basic structure:

double[] a = {60.0, 100.0, 120.0};
double[] b = {20.0, 50.0, 30.0};

r0 = a[0]/b[0]
r1 = a[1]/b[1]
r2 = a[2]/b[2]
...

The Java code I'm working on doesn't work. Could someone give me some help?

Arrays.sort(ratio, new Comparator<double[]>() {
    @Override
    public double compare(double[] a, double[] b) {
        double r1 = (double)a[i]/b[i];
        double r2 = (double)a[i+1]/b[i+1];
        return r1 > r2;
    }

});

Answer 1

You can declare a Pair like this:

static class Pair {
    double a;
    double b;
    Pair(double a, double b) {
        this.a = a;
        this.b = b;
    }
}

and put arrays into an array of pairs. Then it's easy to sort them:

Pair[] p = new Pair[a.length];
for (int i = 0; i < p.length; i++)
    p[i] = new Pair(a[i], b[i]);

Arrays.sort(p, (p1, p2) -> Double.compare(p1.a/p1.b, p2.a/p2.b));

Of course, the results are in the new array. You can loop and put them back to a and b if you want.

Answer 2

Try this.

double[] a = new double[10];
double[] b = new double[10];
// fill data to a and b.
int[] indexes = IntStream.range(0, a.length)
    .boxed()
    .sorted((i, j) -> Double.compare(a[i] / b[i], a[j] / b[j]))
    .mapToInt(i -> i)
    .toArray();
double[] sortedA = IntStream.of(indexes)
    .mapToDouble(i -> a[i])
    .toArray();
double[] sortedB = IntStream.of(indexes)
    .mapToDouble(i -> b[i])
    .toArray();

If you do not use Java8.

double[] a = new double[10];
double[] b = new double[10];
// fill data
int length = a.length;
Integer[] indexes = new Integer[length];
for (int i = 0; i < length; ++i)
    indexes[i] = i;
Arrays.sort(indexes, new Comparator<Integer>() {
    @Override public int compare(Integer o1, Integer o2) {
        return Double.compare(a[o1] / b[o1], a[o2] / b[o2]);
    }
});
double[] sortedA = new double[length];
double[] sortedB = new double[length];
for (int i = 0; i < length; ++i) {
    sortedA[i] = a[indexes[i]];
    sortedB[i] = b[indexes[i]];
}

Answer 3

Let us see an example first:

Question: Which is the larger among the two? 8/6 or 9/7 ?

Answer: An easy way is to do the following for any two numbers of the form a/b and c/d . Find a*d and b*c . If a*d is bigger then a/b is greater than c/d , otherwise c/d is greater.

According to the example: 8*7=56 > 9*6=54 , hence 8/6 is greater than 9/7 .

So, use the above logic in the Comparator .