[java] java Arrays.sort 2d array

I am looking to sort the following array based on the values of [][0]

double[][] myArr = new double[mySize][2];

so for example, myArr contents is:

1      5
13     1.55
12     100.6
12.1   .85

I want it to get to:

1      5
12     100.6
12.1   .85
13     1.55

I am looking to do this without having to implement my own sort.

To sort in descending order you can flip the two parameters

int[][] array= {
    {1, 5},
    {13, 1},
    {12, 100},
    {12, 85} 
};
Arrays.sort(array, (b, a) -> Integer.compare(a[0], b[0]));

Output:

13, 5
12, 100
12, 85
1, 5

Simplified Java 8

IntelliJ suggests to simplify the top answer to the:

Arrays.sort(queries, Comparator.comparingDouble(a -> a[0]));

Welcome Java 8:

Arrays.sort(myArr, (a, b) -> Double.compare(a[0], b[0]));

much simpler code:

import java.util.Arrays; int[][] array = new int[][];

Arrays.sort(array, ( a, b) -> a[1] - b[1]);

Java 8 is now very common nowadays.

Arrays.sort(myArr,(double[] a,double[] b)->{
                //here multiple lines of code can be placed
                return a[0]-b[0]; 
            });

The simplest way:

Arrays.sort(myArr, (a, b) -> a[0] - b[0]);

For a general solution you can use the Column Comparator. The code to use the class would be:

Arrays.sort(myArr, new ColumnComparator(0));

Although this is an old thread, here are two examples for solving the problem in Java8.

sorting by the first column ([][0]):

double[][] myArr = new double[mySize][2];
// ...
java.util.Arrays.sort(myArr, java.util.Comparator.comparingDouble(a -> a[0]));

sorting by the first two columns ([][0], [][1]):

double[][] myArr = new double[mySize][2];
// ...
java.util.Arrays.sort(myArr, java.util.Comparator.<double[]>comparingDouble(a -> a[0]).thenComparingDouble(a -> a[1]));

You need to implement a Comparator<Double[]> like so:

public static void main(String[] args) throws IOException {
    final Double[][] doubles = new Double[][]{{5.0, 4.0}, {1.0, 1.0}, {4.0, 6.0}};
    final Comparator<Double[]> arrayComparator = new Comparator<Double[]>() {
        @Override
        public int compare(Double[] o1, Double[] o2) {
            return o1[0].compareTo(o2[0]);
        }
    };
    Arrays.sort(doubles, arrayComparator);
    for (final Double[] arr : doubles) {
        System.out.println(Arrays.toString(arr));
    }
}

Output:

[1.0, 1.0]
[4.0, 6.0]
[5.0, 4.0]

for decreasing order for integer array of 2 dimension you can use

Arrays.sort(contests, (a, b) -> Integer.compare(b[0],a[0]));//decreasing order

Arrays.sort(contests, (a, b) -> Integer.compare(a[0],b[0]);//decreasing order

import java.util.*;

public class Arrays2
{
    public static void main(String[] args)
    {
        int small, row = 0, col = 0, z;
        int[][] array = new int[5][5];

        Random rand = new Random();
        for(int i = 0; i < array.length; i++)
        {
            for(int j = 0; j < array[i].length; j++)
            {
                array[i][j] = rand.nextInt(100);
                System.out.print(array[i][j] + " ");
            }
            System.out.println();
        }

        System.out.println("\n");


        for(int k = 0; k < array.length; k++)
        {
            for(int p = 0; p < array[k].length; p++)
            {
                small = array[k][p];
                for(int i = k; i < array.length; i++)
                {
                    if(i == k)
                        z = p + 1;
                    else
                        z = 0;
                    for(;z < array[i].length; z++)
                    {
                        if(array[i][z] <= small)
                        {
                            small = array[i][z];
                            row = i;
                            col = z;
                        }
                    }
                }
            array[row][col] = array[k][p];
            array[k][p] = small;
            System.out.print(array[k][p] + " ");
            }
            System.out.println();
        }
    }
}

Good Luck

It is really simple, there are just some syntax you have to keep in mind.

Arrays.sort(contests, (a, b) -> Integer.compare(a[0],b[0]));//increasing order ---1

Arrays.sort(contests, (b, a) -> Integer.compare(b[0],a[0]));//increasing order ---2

Arrays.sort(contests, (a, b) -> Integer.compare(b[0],a[0]));//decreasing order ---3

Arrays.sort(contests, (b, a) -> Integer.compare(a[0],b[0]));//decreasing order ---4

If you notice carefully, then it's the change in the order of 'a' and 'b' that affects the result. For line 1, the set is of (a,b) and Integer.compare(a[0],b[0]), so it is increasing order. Now if we change the order of a and b in any one of them, suppose the set of (a,b) and Integer.compare(b[0],a[0]) as in line 3, we get decreasing order.