I have a simple array, sort of like this:
1 2 3 4 5 6 7 8 9
6 2 7 2 9 6 8 10 5
2 6 4 7 8 4 3 2 5
9 8 7 5 9 7 4 1 10
5 3 6 8 2 7 3 7 2
So, let's call this matrix[5][9]. I wish to now remove every row within this matrix that contains a certain value, in this case 10, so I am left with...
1 2 3 4 5 6 7 8 9
2 6 4 7 8 4 3 2 5
5 3 6 8 2 7 3 7 2
Here's a sample class you can run that I believe does what you're looking for. Removing rows from 2D arrays is tricky business because like #KalebBrasee said, you can't really "remove" them, but rather you have to make a whole new 2D array instead. Hope this helps!
import java.util.ArrayList;
import java.util.List;
public class Matrix {
private double[][] data;
public Matrix(double[][] data) {
int r = data.length;
int c = data[0].length;
this.data = new double[r][c];
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
this.data[i][j] = data[i][j];
}
}
}
/* convenience method for getting a
string representation of matrix */
public String toString() {
StringBuilder sb = new StringBuilder(1024);
for (double[] row : this.data) {
for (double val : row) {
sb.append(val);
sb.append(" ");
}
sb.append("\n");
}
return (sb.toString());
}
public void removeRowsWithValue(final double value) {
/* Use an array list to track of the rows we're going to want to
keep...arraylist makes it easy to grow dynamically so we don't
need to know up front how many rows we're keeping */
List<double[]> rowsToKeep = new ArrayList<double[]>(this.data.length);
for (double[] row : this.data) {
/* If you download Apache Commons, it has built-in array search
methods so you don't have to write your own */
boolean found = false;
for (double testValue : row) {
/* Using == to compares doubles is generally a bad idea
since they can be represented slightly off their actual
value in memory */
if (Double.compare(value, testValue) == 0) {
found = true;
break;
}
}
/* if we didn't find our value in the current row,
that must mean its a row we keep */
if (!found) {
rowsToKeep.add(row);
}
}
/* now that we know what rows we want to keep, make our
new 2D array with only those rows */
this.data = new double[rowsToKeep.size()][];
for (int i = 0; i < rowsToKeep.size(); i++) {
this.data[i] = rowsToKeep.get(i);
}
}
public static void main(String[] args) {
double[][] test = {
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{6, 2, 7, 2, 9, 6, 8, 10, 5},
{2, 6, 4, 7, 8, 4, 3, 2, 5},
{9, 8, 7, 5, 9, 7, 4, 1, 10},
{5, 3, 6, 8, 2, 7, 3, 7, 2}};
//make the original array and print it out
Matrix m = new Matrix(test);
System.out.println(m);
//remove rows with the value "10" and then reprint the array
m.removeRowsWithValue(10);
System.out.println(m);
}
}
Use System.arraycopy or use java.util.List instead of arrays. ArrayList has fast access to random elements and a slow remove method, it's the opposite with LinkedList. You have to choose for yourself.
At the and you have to recreate the array and discard the old one. Changing the dimension of an existing array is not possible - if want this type of datastructure, then you should build the matrix based on Collections (ArrayList<ArrayList<Double>>), there you can remove a row easily.
Back to arrays - the idea is to collect all rows (double[] arrays) that you want to keep, create a result array with those rows and replace the old one with the new on on Matrix:
public void doSomethingWith(Matrix in) {
List<double[]> survivingRows = new ArrayList<double[]>();
for (double[] row:in.getRows()) {
if (isAGoodOne(row)) {
survivingRows.add(row);
}
}
double[][] result = new double[survivingRows][];
for (int i = 0; i < result.length; i++) {
result[i] = survivingRows.get(i);
}
in.setArray(result);
}
You can't remove elements from the Java built-in array data structure. You'll have to create a new array that has a length one less than the first array, and copy all the arrays into that array EXCEPT the one you want to remove.
My java syntax is a little rusty, but the following, if treated as pseudocode will work
public Matrix removeRows(Matrix input) {
int[][] output = new int[input.numRows][input.numColumns]();
int i = 0;
for (int[] row : input.rows()) { // Matrix.rows() is a method that returns an array of all the rows in the matrix
if (!row.contains(10)) {
output[i] = row;
}
}
return output
My take:
import java.util.Arrays;
public class RemoveArrayRow {
private static <T> T[] concat(T[] a, T[] b) {
final int alen = a.length;
final int blen = b.length;
if (alen == 0) {
return b;
}
if (blen == 0) {
return a;
}
final T[] result = (T[]) java.lang.reflect.Array.newInstance(a.getClass().getComponentType(), alen + blen);
System.arraycopy(a, 0, result, 0, alen);
System.arraycopy(b, 0, result, alen, blen);
return result;
}
public static void main(String[] args) {
double[][] d = { {11, 2, 3, 4, 5, 6, 7, 8, 9, 0},
{12, 2, 3, 4, 5, 6, 7, 8, 9, 1},
{13, 2, 3, 4, 5, 6, 7, 8, 9, 2},
{14, 2, 3, 4, 5, 6, 7, 8, 9, 3},
{15, 2, 3, 4, 5, 6, 7, 8, 9, 4} };
//remove the fourth row:
// (1)
double[][] d1 = concat(Arrays.copyOf(d, 3), Arrays.copyOfRange(d, 4, 5));
// (2)
double[][] d2 = new double[d.length - 1][d[0].length];
System.arraycopy(d, 0, d2, 0, 3);
System.arraycopy(d, 4, d2, 3, 1);
System.out.print(d1.length);
System.out.print(d2.length);
}
}
(1)
If you exclude the concat() function used for concatenating two arrays, it's done in one line:
double[][] d1 = concat(Arrays.copyOf(d, 3), Arrays.copyOfRange(d, 4, 5));
See this question as well. That's where the code for the concat() function comes from.
(2)
This method is faster and only uses already available functions.
Since it cannot avoid creating new 2D array to contain the after-removed data, firstly, create a new 2D int[][] b with same dimension as a[][]. secondly, loop through a[][], assign a to b and move b row up when a contain specific value. and sanity check the last row, which can contain specific data.
public static int[][] remove(int[][] a, int v) {
int r = a.length;
int c = a[0].length;
int[][] b = new int[r][c];
int red = 0;
boolean s = false;
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
b[i - red][j] = a[i][j];
if (a[i][j] == v) {
red += 1;
if(i==r-1){
s = true;
}
break;
}
}
}
//check last row
if(s){
for(int i = r-red;i <r-red +1; i++ )
for (int j = 0; j<c; j++){
b[i][j] = 0;
}
}
return b;
}
public static void main(String[] args){
int[][] a = { {1, 2, 3, 4, 5, 6, 7, 8, 1},
{6, 2, 7, 2, 9, 6, 8, 10, 5},
{2, 6, 4, 7, 8, 4, 2, 2, 5},
{9, 8, 7, 5, 9, 7, 4, 1, 1},
{5, 3, 6, 8, 2, 7, 3, 1, 1} };
print(remove(a, 10));
}
public static void print(int[][] a) {
int r = a.length;
int c = a[0].length;
int red = 0;
for (int i = 0; i < r; i++) {
System.out.printf("\nrow %d, \n", i);
for (int j = 0; j < c; j++) {
System.out.printf("%d, ", a[i][j]);
}
}
}
This may not be an exact solution but a concept of how you can achieve it using System.arraycopy.
In the example below, I want to copy all the rows except the first row. In your case, you can skip those rows which contain 10.
String[][] src = getSheetData(service, spreadSheetId, range);
String[][] dest = new String[src.length-1][src[0].length];
for (int i = 1; i < src.length; i++) {
System.arraycopy(src[i], 0, dest[i-1], 0, src[0].length-1);
}
Reference: https://docs.oracle.com/javase/6/docs/api/java/lang/System.html#arraycopy%28java.lang.Object,%20int,%20java.lang.Object,%20int,%20int%29
You can use IntStream.noneMatch method for this purpose:
int[][] arr1 = {
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{6, 2, 7, 2, 9, 6, 8, 10, 5},
{2, 6, 4, 7, 8, 4, 3, 2, 5},
{9, 8, 7, 5, 9, 7, 4, 1, 10},
{5, 3, 6, 8, 2, 7, 3, 7, 2}};
int[][] arr2 = Arrays.stream(arr1)
.filter(row -> Arrays.stream(row).noneMatch(i -> i == 10))
.toArray(int[][]::new);
// output
Arrays.stream(arr2).map(Arrays::toString).forEach(System.out::println);
Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[2, 6, 4, 7, 8, 4, 3, 2, 5]
[5, 3, 6, 8, 2, 7, 3, 7, 2]
Related
So what I want is this
int[][] arr=new int[2][8];
input:
1 1 3 1 5 3 7 1
5 2 4 8 3 7 5 2
output:
1 1 5 3 1 7 3 1
2 2 3 4 5 5 7 8
you can see that it is sorted by the second row in ascending order and the first row just follows,
how can I do this? help, please.
I tried doing below
Arrays.sort(arr[1]);
but I don't think it is working. It does sort the second row in ascending order but the first row is not matching the initial pair with the second row
Try this.
public static void main(String[] args) {
int[][] array = {
{1, 1, 3, 1, 5, 3, 7, 1},
{5, 2, 4, 8, 3, 7, 5, 2}
};
List<int[]> list = new AbstractList<int[]>() {
#Override
public int[] get(int index) {
return new int[] {array[1][index], array[0][index]};
}
#Override
public int[] set(int index, int[] value) {
int[] old = get(index);
array[1][index] = value[0];
array[0][index] = value[1];
return old;
}
#Override
public int size() {
return array[0].length;
}
};
Collections.sort(list, Arrays::compare);
for (int[] row : array)
System.out.println(Arrays.toString(row));
}
output:
[1, 1, 5, 3, 1, 7, 3, 1]
[2, 2, 3, 4, 5, 5, 7, 8]
Or
public static void main(String[] args) {
int[][] array = {
{1, 1, 3, 1, 5, 3, 7, 1},
{5, 2, 4, 8, 3, 7, 5, 2}
};
int[] sortedIndexes = IntStream.range(0, array[0].length)
.boxed()
.sorted(Comparator.comparing((Integer i) -> array[1][i])
.thenComparing(i -> array[0][i]))
.mapToInt(Integer::intValue)
.toArray();
int[][] output = IntStream.range(0, array.length)
.mapToObj(r -> IntStream.range(0, array[r].length)
.map(i -> array[r][sortedIndexes[i]])
.toArray())
.toArray(int[][]::new);
for (int[] r : output)
System.out.println(Arrays.toString(r));
}
It may be implemented using helper method(s) to transpose the input array, then transposed array may be sorted by column, and transposed again to restore the original rows/cols:
// create new array to store transposed
public static int[][] transpose(int[][] src) {
return transpose(src, new int[src[0].length][src.length]);
}
// use existing array to store the transposed
public static int[][] transpose(int[][] src, int[][] dst) {
for (int i = 0, n = src.length; i < n; i++) {
for (int j = 0, m = src[i].length; j < m; j++) {
dst[j][i] = src[i][j];
}
}
return dst;
}
Method sortByColumn (reusing the input array):
public static void sortByColumn(int[][] arr, Comparator<int[]> comparator) {
int[][] toSort = transpose(arr);
Arrays.sort(toSort, comparator);
transpose(toSort, arr);
}
Test:
int[][] arr = {
{7, 1, 3, 1, 5, 3, 1, 4, 4},
{5, 2, 4, 8, 3, 7, 5, 2, 5}
};
sortByColumn(arr, Comparator.comparingInt(col -> col[1]));
for (int[] row : arr) {
System.out.println(Arrays.toString(row));
}
Output:
in the first row values appear in the insertion order after sorting by the second element in each column.
[1, 4, 5, 3, 7, 1, 4, 3, 1]
[2, 2, 3, 4, 5, 5, 5, 7, 8]
Square arrays (width == height) may be transposed more efficiently without creating additional array:
public static int[][] transposeSquare(int[][] arr) {
for (int i = 0, n = arr.length; i < n; i++) {
// select the elements only above the main diagonal
for (int j = i + 1, m = arr[i].length; j < m; j++) {
int tmp = arr[i][j];
arr[i][j] = arr[j][i];
arr[j][i] = tmp;
}
}
return arr;
}
I am trying to create a for loop that prints numbers 1, 2, 3, 4, 5, 6, 7, 8. Once reaching the end the loop should reverse back starting from 8, 7, 6, 5, 4, 3, 2, 1. The output only goes through the elements and then ends, it doesn't reverse back. Is there a better way to code this, I am fairly new to programming and working with arrays and loops. Any help will be appreciated.
int num = 0;
int[] arrayNumber = new int[] { 1, 2, 3, 4, 5, 6, 7, 8 };
for (int i = 0; i < arrayNumber.length; ) {
if (i < 8) {
i++;
} else {
i--;
}
num = arrayNumber[i];
System.out.print(num);
}
You can use 2 Loops for each display like this below:
int[] numbers = {1, 2, 3, 4, 5, 6, 7, 8};
//this one fo
for (int n: numbers) {
System.out.println(n);
}
//this one for the Reverse display
for(int i = (numbers.length-1); i>=0;i--){
System.out.println(numbers[i]);
}
if you want to do an infinite loop (in python) :
i = 0;
test = [1, 2, 3, 4, 5, 6, 7, 8]
goesdown = False
while True:
print(test[i])
i += -1 if goesdown == True else 1
if(test[i] == test[-1]):
goesdown = True
if(test[i] == test[0]):
goesdown = False
I am trying to print out the 'middle' of the 2D array (a). For example, for given arrays in my code, I would like to print:
[3,4,5,6]
[4,5,6,7]
However I was only able to print out the 'middle' values. I would like to modify the 2D array (a) in the method inner and print it in main instead, and not use System.out.println in the nested for loop. How would I go about doing this?
Here is my code:
public static int[][] inner(int[][] a) {
int rowL = a.length - 1;
int colL = a[1].length - 1;
for (int row = 1; row < rowL; row++) {
for (int col = 1; col < colL; col++) {
//System.out.print(a[row][col]);
a = new int[row][col];
}
System.out.println();
}
return a;
}
public static void main(String[] args) {
int[][] a = {
{1, 2, 3, 4, 5, 6},
{2, 3, 4, 5, 6, 7},
{3, 4, 5, 6, 7, 8},
{4, 5, 6, 7, 8, 9}};
for (int[] row : a) {
System.out.println(Arrays.toString(row));
}
System.out.println();
for (int[] row : inner(a)) {
System.out.println(Arrays.toString(row));
}
}
Create a new array outside the loop and then fill that array inside the loop by translating the indices between the two arrays:
public static int[][] inner (int[][] a) {
int rowL = a.length - 1;
int colL = a[1].length -1;
int[][] ret = new int[rowL - 1][colL - 1];
for (int row = 1; row < rowL; row++) {
for (int col = 1; col < colL ; col++) {
ret[row - 1][col - 1] = a[row][col];
}
}
return ret;
}
If you just want to print the middle values (my definition for this code example is: middle = full array minus first and last element), you can make use of a StringBuilder:
public static void main(String[] args) {
int[][] a = {
{ 1, 2, 3, 4, 5, 6 },
{ 2, 3, 4, 5, 6, 7 },
{ 3, 4, 5, 6, 7, 8 },
{ 4, 5, 6, 7, 8, 9 }
};
for (int[] b : a) {
// create a String output for each inner array
StringBuilder outputBuilder = new StringBuilder();
// append an introducing bracket
outputBuilder.append("[");
// make the values to be printed ignore the first and last element
for (int i = 1; i < b.length - 1; i++) {
if (i < b.length - 2) {
/*
* append a comma plus whitespace
* if the element is not the last one to be printed
*/
outputBuilder.append(b[i]).append(", ");
} else {
// just append the last one without trailing comma plus whitespace
outputBuilder.append(b[i]);
}
}
// append a closing bracket
outputBuilder.append("]");
// print the result
System.out.println(outputBuilder.toString());
}
}
The output will be
[2, 3, 4, 5]
[3, 4, 5, 6]
[4, 5, 6, 7]
[5, 6, 7, 8]
You can use Arrays.stream(T[],int,int) method to iterate over a given range of an array:
int[][] arr = {
{1, 2, 3, 4, 5, 6},
{2, 3, 4, 5, 6, 7},
{3, 4, 5, 6, 7, 8},
{4, 5, 6, 7, 8, 9}};
int[][] middle = Arrays.stream(arr, 1, arr.length - 1)
.map(row -> Arrays.stream(row, 1, row.length - 1)
.toArray())
.toArray(int[][]::new);
// output
Arrays.stream(middle).map(Arrays::toString).forEach(System.out::println);
[3, 4, 5, 6]
[4, 5, 6, 7]
import java.util.Random;
public class Sudoku {
int[][] SquareNumbers = {
{ 4, 3, 5, 8, 7, 6, 1, 2, 9 }, { 8, 7, 6, 2, 1, 9, 3, 4, 5 }, { 2, 1, 9, 4, 3, 5, 7, 8, 6 },
{ 5, 2, 3, 6, 4, 7, 8, 9, 1 }, { 9, 8, 1, 5, 2, 3, 4, 6, 7 }, { 6, 4, 7, 9, 8, 1, 2, 5, 3 },
{ 7, 5, 4, 1, 6, 8, 9, 3, 2 }, { 3, 9, 2, 7, 5, 4, 6, 1, 8 }, { 1, 6, 8, 3, 9, 2, 5, 7, 4 } };
Random Digits = new Random(); // random numbers to exchange Rows
Random HiddenNumbers = new Random();
int Grid[][] = new int[9][9];
public int[][] Generator() {
for (int x = 0; x < Digits.nextInt(); x++) {
for (int da = 0; da < 3; da++) {
}
}
return SquareNumbers;
}
int[][] Hide() {
for (int i = 0; i < 9; i++)
for (int j = 0; j < 9; j++)
Grid[i][j] = SquareNumbers[i][j];
int Row, Columns, Concealer;
Concealer = 55 + Digits.nextInt(1);
for (int i = 0; i < Concealer; i++) {
Row = HiddenNumbers.nextInt(9);
Columns = HiddenNumbers.nextInt(9);
Grid[Row][Columns] = -1;
}
return Grid;
}
public int[][] getSquareNumbers() {
return SquareNumbers;
}
public void setSquareNumbers(int[][] SquareNumbers) {
this.SquareNumbers = SquareNumbers;
}
private static Sudoku instance = null;
protected Sudoku() {
// Exists only to defeat instantiation.
}
public static Sudoku getInstance() {
if (instance == null) {
instance = new Sudoku();
}
return instance;
}
}
Instead of a double array list, is there a way to randomize it so it works like a Sudoku game? As in it has no duplicates within columns or rows and every three by three smaller grid uses a number once?
One method for generating random sudoku puzzles would be as follows:
Generate a random puzzle satisfying the constraints of a sudoku puzzle
Randomly swap digits (e.g. replace 2s with 3s, 7s with 1s, etc)
Randomly swap columns or rows within their set of 3 (e.g. swap the first and third column, or the 5th and 6th)
Randomly switch sets of 3 columns or rows
with another set of columns.
While these will look like different puzzles, they will in fact, all be isotropic latin squares (i.e. all equivalent since they can be reduced to the same puzzle by reordering rows/columns).
If you want a more random solution, this may have already been answered here: https://stackoverflow.com/a/6964044/2471910
If I have an array:
int[] myArray = new int[10]
for (int i = 0; i < myArray.length; i++) {
myArray[i] = i;
}
//resulting array: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
How can I move everything behing the 4 up one space, and send the 4 to the back? Example:
this:
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
into this:
{0, 1, 2, 3, 5, 6, 7, 8, 9, 4}
How about this:
int[] myArray = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
System.arraycopy(myArray, 5, myArray, 4, 5);
myArray[myArray.length-1] = 4;
In the above code, I'm using the arraycopy method to copy a range of 5 numbers starting from index 5, to index 4 in the array, and then simply set a 4 in the last position.
Notice that using arraycopy is much faster than copying the values in a loop, since it's usually implemented as a native operation which copies memory positions.
EDIT :
A more generic solution, a method for sending to the back a given position in the array:
public static void sendBack(int[] array, int idx) {
int value = array[idx];
System.arraycopy(array, idx+1, array, idx, array.length-idx-1);
array[array.length-1] = value;
}
For your example, call it like this:
sendBack(myArray, 4);
// now myArray is {0, 1, 2, 3, 5, 6, 7, 8, 9, 4}
Like this?
int start = 4;
int temp = myArray[start];
for(int i = start; i < myArray.length - 1; i++) {
myArray[i] = myArray[i+1];
}
myArray[myArray.length-1] = temp;
It's the fastest way I can guess...
Use System.arraycopy.