PHP - How to push an array into specific position?

Question

I hope I can explain my problem

I have on array(10x10) filled with 1 and 0 e.g.

0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000111100
0000000000
1111000000
0000000000

Now I'd like to push another array

e.g.

11111
11111

into it on a specific position point (3,3)
I've highlighted the position in bold

0000000000
0000000000
0011111000
0011111000
0000000000
0000000000
0000111100
0000000000
1111000000
0000000000

Futhermore, if there is already a value on the field add it, so if I repeat the process the matrix will become this.

0000000000
0000000000
0022222000
0022222000
0000000000
0000000000
0000111100
0000000000
1111000000
0000000000

I hope someone can point me in the right direction. I've already played with some array functions, but I can't get it to work.

What do you suggest?

Could you show us what array functions you have tried, and what exactly didn't work? — ohaal
– ohaal, Commented Jul 5, 2013 at 15:20
Just start at $arr[$i][$j], loop through your array-to-push while incrementing $j and $i as appropriate. It's pretty simple, really. — deceze
– deceze ♦, Commented Jul 5, 2013 at 15:21

deceze · Accepted Answer · 2013-07-05 15:30:37Z

2

Something along these lines should do it:

/**
 * Push a small matrix into a larger matrix.
 * Note: no size restrictions are applied, if $push
 * exceeds the size of $matrix or the coordinates are
 * too large you'll get a jaggy array.
 *
 * @param array $matrix A 2D array of ints.
 * @param array $push   A 2D array <= $matrix to push in.
 * @param int   $x      The starting x coordinate.
 * @param int   $y      The starting y coordinate.
 * @return array The modified $matrix.
 */
function matrix_push(array $matrix, array $push, $x, $y) {
    list($i, $j) = array($x, $y);

    foreach ($push as $row) {
        foreach ($row as $int) {
            $matrix[$i][$j++] += $int;
        }
        $i++;
        $j = $y;
    }

    return $matrix;
}

edited Jul 5, 2013 at 15:30

answered Jul 5, 2013 at 15:25

deceze♦

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4 Comments

jcsanyi Over a year ago

Can you explain the first line of the function? Why not just use x and y?

deceze Over a year ago

@jcsanyi Because I need the original $y later on. Admittedly $i is redundant, but more explicit.

funktioneer Over a year ago

Thanks for that fast answer

Sumurai8 Over a year ago

Maybe you can edit in the safety check from my answer, as you can extend rows, or create an index error by calling a non-existing row with your function.

Sumurai8 · Accepted Answer · 2013-07-05 15:30:22Z

1

For a 2-dimensional array you could do something like this. Please note that what you define as (3,3) in your question, is actually (2,2) from an array perspective. The input $x and $y would be 2 and 2 in this case.

function replaceArray( $orig, $replace, $x, $y ) {
  //Some safety checks
  //Return false if the array that we replace with exceeds the boundaries of the original array
  if( $x + count( $replace[0] ) > count( $orig[0] ) || $y + count( $replace ) > count( $orig ) ) {
    return false;
  }

  for( $i = 0; $i < count($replace[0]); $i++ ) {
    for( $j = 0; $j < count( $replace ); $j++ ) {
      $orig[ $x + $i ][ $x + $j ] += $replace[ $i ][ $j ];
    }
  }

  return $orig;
}

answered Jul 5, 2013 at 15:30

Sumurai8

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Comments

funktioneer · Accepted Answer · 2013-07-05 15:33:20Z

Wow thats was fast :)

Nice way you've implemented it. Here is my code that solve it too, but with some line more

$array1 = array(0=>array(0,0,0,0,0,0,0,0,0,0,0),
        1=>array(0,0,0,0,0,0,0,0,0,0,0),
        2=>array(0,0,0,0,0,0,0,0,0,0,0),
        3=>array(0,0,0,0,0,0,0,0,0,0,0),
        4=>array(0,0,0,0,0,0,0,0,0,0,0),
        5=>array(0,0,0,0,0,0,0,0,0,0,0),
        6=>array(0,0,0,0,0,0,0,0,0,0,0),
        7=>array(0,0,0,0,0,0,0,0,0,0,0));



$array2 = array(0=>array(1,1,1),
                1=>array(1,1,1),
                2=>array(1,1,1));

$row = 3;
$col = 3;
foreach($array2 AS $valuesToReplace)
{   $col = 3;
    foreach($valuesToReplace AS $value)
    {
        $array1[$row][$col] = $value;
        $col++;
    }
    $row++;
}

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