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
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.
Use System.arraycopy or use java.util.List instead of arrays. ArrayList have fast access to random element and slow remove method, it's opposite with LinkedList. You have to choose for yourself.
EDIT:
Rant: I had to use bit.ly for this link, because stackoverflow sucks at its core buisness of parsing html with regexps (it chokes on this:
<a href="http://java.sun.com/j2se/1.4.2/docs/api/java/lang/System.html#arraycopy(java.lang.Object, int, java.lang.Object, int, int)">System.arraycopy</a>
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);
}
}
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.
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);
}
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