java Arrays.sort 2d array

问题:

SO,

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

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

so for ex, 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 have to implement my own sort. Any help is appreciated, thanks.

回答1:

Use Overloaded Arrays#Sort(T[] a, Comparator c) which takes Comparator as the second argument.

double[][] array= {
{1, 5},
{13, 1.55},
{12, 100.6},
{12.1, .85} };

java.util.Arrays.sort(array, new java.util.Comparator<double[]>() {
    public int compare(double[] a, double[] b) {
        return Double.compare(a[0], b[0]);
    }
});

回答2:

Welcome Java 8:

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

回答3:

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]

回答4:

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]));

回答5:

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

回答6:

Simplified Java 8

IntelliJ suggests to simplify the top answer to the:

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

回答7:

The simplest way:

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

回答8:

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

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