I have an array like this one-
{1, 2, 3, 4, 5, 6}
I want to sort it in the order of multiples of 3 with remainders 0, 1 and 2. (the first group is multiples of 3, the second one is multiples of 3 with remainder 1 and the last one is multiples of 3 with remainder 2) and I want to preserve the order in which elements appear in the array.
The result should be -
{3, 6, 1, 4, 2, 5}
I have this code-
int current = 0;
int b = 0;
for (int i = 0; i < 3; i++) { //3 groups
for (int j = current; j < numbers.length; j++) {
if (numbers[j] % 3 == i) { //reminder should be 0,1 or 2
b = numbers[j];
numbers[j] = numbers[current];
numbers[current] = b;
current++;
}
}
}
But this code does not preserve the order in which elements appear in the array. The result I got is-
{3, 6, 1, 4, 5, 2}
But I want the result to be like {3, 6, 1, 4, 2, 5}. How can I achieve this?
Using stream and comparator
int[] array = {1, 2, 3, 4, 5, 6};
List<Integer> lst = Arrays.stream(array)
.boxed()
.sorted(Comparator.comparingInt(o -> o % 3))
.collect(Collectors.toList());
System.out.println(lst);
In your solution you are swapping the elements in place, which shuffles them from the initial order. That's why you don't have the same ordering at the end. I'm not sure if there is another way apart from having a second array to keep the sorted elements, while at the same time iterating over the original one like so:
public static void main(String[] args) {
int[] numbers = new int[]{1, 2, 3, 4, 5, 6};
int[] result = new int[numbers.length];
int b = 0;
int current = 0;
for (int i = 0; i < 3; i++) { //3 groups
for (int j = 0; j < numbers.length; j++) {
if (numbers[j] % 3 == i) { //reminder should be 0,1 or 2
result[current] = numbers[j];
current++;
}
}
}
System.out.println(Arrays.toString(result));
}
Output: [3, 6, 1, 4, 2, 5]
You can use an IntStream and a Comparator to sort the stream:
int[] arr = {1, 2, 3, 4, 5, 6};
int[] arrSorted = IntStream.of(arr).boxed()
.sorted(Comparator.comparingInt(i -> i % 3))
.mapToInt(Integer::intValue)
.toArray();
System.out.println(Arrays.toString(arrSorted));
Output:
[3, 6, 1, 4, 2, 5]
Note: From IntStream.of() javadoc:
Returns a sequential ordered stream whose elements are the specified
values.
I would create a new array of the same size and then place the elements in the correct order. For example like this:
int[] array = {1, 2, 3, 4, 5, 6};
int[] sorted = new int[array.length];
int counter = 0;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < array.length; j++) {
if (array[j] % 3 == i) {
sorted[counter] = array[j];
counter++;
}
}
}
System.out.println(Arrays.toString(sorted));
Output:
[3, 6, 1, 4, 2, 5]
Alternatively, you can use Java 8 features to reduce the amount of code like this:
int[] array = {1, 2, 3, 4, 5, 6};
int[] sorted = Arrays.stream(array).boxed().sorted(Comparator.comparingInt(a -> (a % 3))).mapToInt(i -> i).toArray();
Output:
[3, 6, 1, 4, 2, 5]
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 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]