java Arrays.sort 2d array
JavaArraysSortingMatrixJava Problem Overview
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.
Java Solutions
Solution 1 - Java
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]);
}
});
JAVA-8: Instead of that big comparator, we can use lambda function as following-
Arrays.sort(array, Comparator.comparingDouble(o -> o[0]));
Solution 2 - Java
Welcome Java 8:
Arrays.sort(myArr, (a, b) -> Double.compare(a[0], b[0]));
Solution 3 - Java
The simplest way:
Arrays.sort(myArr, (a, b) -> a[0] - b[0]);
Solution 4 - Java
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]
Solution 5 - Java
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]));
Solution 6 - Java
Decreasing/increasing order for an 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]); //increasing order
Solution 7 - Java
Simplified Java 8
IntelliJ suggests to simplify the top answer to the:
Arrays.sort(queries, Comparator.comparingDouble(a -> a[0]));
Solution 8 - Java
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
Solution 9 - Java
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.
Solution 10 - Java
much simpler code:
import java.util.Arrays; int[][] array = new int[][];
Arrays.sort(array, ( a, b) -> a[1] - b[1]);
Solution 11 - Java
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
Solution 12 - Java
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];
});
Solution 13 - Java
You can use your own sort, it is very simple.
int[][] matrix = {
{2, 1, 3},
{5, 4, 6},
{8, 7, 9}
};
for (int k = 0; k < length; k++) {
for (int i= 0; i < matrix[k].length; i++) {
for (int j = 0; j < matrix[k].length; j++) {
if (matrix[k][i] < matrix[k][j]) {
int temp = matrix[k][i];
matrix[k][i] = matrix[k][j];
matrix[k][j] = temp;
}
}
}
}
System.out.println(Arrays.deepToString(matrix));
OUTPUT
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Solution 14 - Java
There are multiple approaches to do this, Here I'm sharing follow two methods by which it can be achieved.
- Using Comparator Arrays.sort : An built-in feature of Java.
- Using Merge Sort
> Using Comparator Arrays.sort : A built-in feature of Java.
import java.util.Arrays;
class Array2D {
public static void printTwoDimensionArray(int [][] arr) {
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println("");
}
}
public static void main(String [] args) {
int [][] arr = {
{1, 2},
{6, 8},
{4, 7},
{9, 11},
{7, 10},
{13, 16},
{5, 9},
{8, 9},
{10, 11}
};
Arrays.sort(arr, (a, b) -> Integer.compare(a[0], b[0]));
printTwoDimensionArray(arr);
}
}
> Using Merge Sort
import java.util.ArrayList;
class MergeSortComparator {
public static void printSingleDimensionArray(int [] arr) {
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + " ");
}
System.out.println("");
}
public static void printDoublyDimensionArray(int [][] arr) {
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println("");
}
}
public static void merge(int[][] arr, int start, int mid, int end, int index) {
int i, j, k;
int n1 = mid - start + 1;
int n2 = end - mid;
int columnLength = arr[0].length;
int [][] leftSubArray = new int [n1][columnLength];
int [][] rightSubArray = new int [n1][columnLength];
// Copy elements to Temp LeftSubArray
for (i = 0; i < n1; i++) {
for (j = 0; j < columnLength; j++) {
leftSubArray[i][j] = arr[start + i][j];
}
}
// Copy elements to Temp RightSubArray
for (i = 0; i < n2; i++) {
for (j = 0; j < columnLength; j++) {
rightSubArray[i][j] = arr[mid + 1 + i][j];
}
}
i = j = k = 0;
while(i < n1 && j < n2) {
if (leftSubArray[i][index] <= rightSubArray[j][index]) {
arr[start + k] = leftSubArray[i];
i++;
} else {
arr[start + k] = rightSubArray[j];
j++;
}
k++;
}
while(i < n1) {
arr[start + k] = leftSubArray[i];
i++;
k++;
}
while(j < n2 && (start + k) < end) {
arr[start + k] = rightSubArray[j];
j++;
k++;
}
}
public static void mergeSort(int[][] arr, int start, int end, int index) {
if (start >= end) {
return;
}
int mid = (start + end) / 2;
mergeSort(arr, start, mid, index);
mergeSort(arr, mid + 1, end, index);
merge(arr, start, mid, end, index);
return;
}
public static void main(String [] args) {
int [][] arr = {
{1, 2},
{6, 8},
{4, 7},
{9, 11},
{7, 10},
{13, 16},
{5, 9},
{8, 9},
{10, 11}
};
int m = arr.length;
int n = arr[0].length;
// Last argument as Index is set to 0,
mergeSort(arr, 0, m-1, 0);
printDoublyDimensionArray(arr);
}
}
Solution 15 - Java
For a general solution you can use the Column Comparator. The code to use the class would be:
Arrays.sort(myArr, new ColumnComparator(0));