using array_search for multi dimensional array

Question 1

using array_search in a 1 dimensional array is simple

$array = array("apple", "banana", "cherry");
$searchValue = "cherry";
$key = array_search($searchValue, $array);

echo $key;

but how about an multi dimensional array?

    #RaceRecord

    [CarID] [ColorID] [Position]
[0]    1        1         3
[1]    2        1         1
[2]    3        2         4
[3]    4        2         2
[4]    5        3         5

for example i want to get the index of the car whose position is 1. How do i do this?

Question 2

function find_car_with_position($cars, $position) {
    foreach($cars as $index => $car) {
        if($car['Position'] == $position) return $index;
    }
    return FALSE;
}

Question 3

You need to iterate over your results, adding a new entry to the output when you encounter a new team, or updating the points value when you find the same team again. This is most easily done by initially indexing the output by the team name, and then using array_values to re-index the array numerically:

$teams = array();
foreach ($results as $result) {
    $team = $result['team'];
    if (!isset($teams[$team])) {
        $teams[$team] = array('team' => $team, 'points' => $result['punti']);
    }
    else {
        $teams[$team]['points'] += $result['punti'];
    }
}
$teams = array_values($teams);
print_r($teams);

Output (for your sample data):

Array
(
    [0] => Array
        (
            [team] => Red Bull Racing
            [points] => 418
        )
    [1] => Array
        (
            [team] => Scuderia Ferrari
            [points] => 353
        )
    [2] => Array
        (
            [team] => Mercedes-AMG
            [points] => 516
        )
    [3] => Array
        (
            [team] => Racing Point F1
            [points] => 147
        )
    [4] => Array
        (
            [team] => Haas F1
            [points] => 127
        )
)

Demo on 3v4l.org

Question 4

Option 1: Calling numpy from MATLAB

Assuming your system is set up according to the documentation, and you have the numpy package installed, you could do (in MATLAB):

np = py.importlib.import_module('numpy');

M = 2;
N = 4;
I = 2000;
J = 300;

A = matpy.mat2nparray( randn(M, M, I) );
B = matpy.mat2nparray( randn(M, M, N, J, I) );
C = matpy.mat2nparray( randn(M, J, I) );

D = matpy.nparray2mat( np.einsum('mki, klnji, lji -> mnji', A, B, C) );

Where matpy can be found here.

Option 2: Native MATLAB

Here the most important part is to get the permutations right, so we need to keep track of our dimensions. We'll be using the following order:

I(1) J(2) K(3) L(4) M(5) N(6)

Now, I'll explain how I got the correct permute order (let's take the example of A): einsum expects the dimension order to be mki, which according to our numbering is 5 3 1. This tells us that the 1^st dimension of A needs to be the 5^th, the 2^nd needs to be 3^rd and the 3^rd needs to be 1^st (in short 1->5, 2->3, 3->1). This also means that the "sourceless dimensions" (meaning those that have no original dimensions becoming them; in this case 2 4 6) should be singleton. Using ipermute this is really simple to write:

pA = ipermute(A, [5,3,1,2,4,6]);

In the above example, 1->5 means we write 5 first, and the same goes for the other two dimensions (yielding [5,3,1]). Then we just add the singletons (2,4,6) at the end to get [5,3,1,2,4,6]. Finally:

A = randn(M, M, I);
B = randn(M, M, N, J, I);
C = randn(M, J, I);

% Reference dim order: I(1) J(2) K(3) L(4) M(5) N(6)
pA = ipermute(A, [5,3,1,2,4,6]); % 1->5, 2->3, 3->1; 2nd, 4th & 6th are singletons
pB = ipermute(B, [3,4,6,2,1,5]); % 1->3, 2->4, 3->6, 4->2, 5->1; 5th is singleton
pC = ipermute(C, [4,2,1,3,5,6]); % 1->4, 2->2, 3->1; 3rd, 5th & 6th are singletons

pD = sum( ...
  permute(pA .* pB .* pC, [5,6,2,1,3,4]), ... 1->5, 2->6, 3->2, 4->1; 3rd & 4th are singletons
  [5,6]);

(see note regarding sum at the bottom of the post.)

Another way to do it in MATLAB, as mentioned by @AndrasDeak, is the following:

rD = squeeze(sum(reshape(A, [M, M, 1, 1, 1, I]) .* ...
                 reshape(B, [1, M, M, N, J, I]) .* ...
... % same as:   reshape(B, [1, size(B)]) .* ...
... % same as:   shiftdim(B,-1) .* ...
                 reshape(C, [1, 1, M, 1, J, I]), [2, 3]));

See also: squeeze, reshape, permute, ipermute, shiftdim.

Here's a full example that shows that tests whether these methods are equivalent:

function q55913093
M = 2;
N = 4;
I = 2000;
J = 300;

mA = randn(M, M, I);
mB = randn(M, M, N, J, I);
mC = randn(M, J, I);

%% Option 1 - using numpy:
np = py.importlib.import_module('numpy');

A = matpy.mat2nparray( mA );
B = matpy.mat2nparray( mB );
C = matpy.mat2nparray( mC );

D = matpy.nparray2mat( np.einsum('mki, klnji, lji -> mnji', A, B, C) );

%% Option 2 - native MATLAB:
%%% Reference dim order: I(1) J(2) K(3) L(4) M(5) N(6)

pA = ipermute(mA, [5,3,1,2,4,6]); % 1->5, 2->3, 3->1; 2nd, 4th & 6th are singletons
pB = ipermute(mB, [3,4,6,2,1,5]); % 1->3, 2->4, 3->6, 4->2, 5->1; 5th is singleton
pC = ipermute(mC, [4,2,1,3,5,6]); % 1->4, 2->2, 3->1; 3rd, 5th & 6th are singletons

pD = sum( permute( ...
  pA .* pB .* pC, [5,6,2,1,3,4]), ... % 1->5, 2->6, 3->2, 4->1; 3rd & 4th are singletons
  [5,6]);

rD = squeeze(sum(reshape(mA, [M, M, 1, 1, 1, I]) .* ...
                 reshape(mB, [1, M, M, N, J, I]) .* ...
                 reshape(mC, [1, 1, M, 1, J, I]), [2, 3]));

%% Comparisons:
sum(abs(pD-D), 'all')
isequal(pD,rD)

Running the above we get that the results are indeed equivalent:

>> q55913093
ans =
   2.1816e-10 
ans =
  logical
   1

Note that these two methods of calling sum were introduced in recent releases, so you might need to replace them if your MATLAB is relatively old:

S = sum(A,'all')   % can be replaced by ` sum(A(:)) `
S = sum(A,vecdim)  % can be replaced by ` sum( sum(A, dim1), dim2) `

As requested in the comments, here's a benchmark comparing the methods:

function t = q55913093_benchmark(M,N,I,J)

if nargin == 0
  M = 2;
  N = 4;
  I = 2000;
  J = 300;
end

% Define the arrays in MATLAB
mA = randn(M, M, I);
mB = randn(M, M, N, J, I);
mC = randn(M, J, I);

% Define the arrays in numpy
np = py.importlib.import_module('numpy');
pA = matpy.mat2nparray( mA );
pB = matpy.mat2nparray( mB );
pC = matpy.mat2nparray( mC );

% Test for equivalence
D = cat(5, M1(), M2(), M3());
assert( sum(abs(D(:,:,:,:,1) - D(:,:,:,:,2)), 'all') < 1E-8 );
assert( isequal (D(:,:,:,:,2), D(:,:,:,:,3)));

% Time
t = [ timeit(@M1,1), timeit(@M2,1), timeit(@M3,1)]; 

function out = M1()
  out = matpy.nparray2mat( np.einsum('mki, klnji, lji -> mnji', pA, pB, pC) );
end

function out = M2()
  out = permute( ...
          sum( ...
            ipermute(mA, [5,3,1,2,4,6]) .* ...
            ipermute(mB, [3,4,6,2,1,5]) .* ...
            ipermute(mC, [4,2,1,3,5,6]), [3,4]...
          ), [5,6,2,1,3,4]...
        );  
end

function out = M3()
out = squeeze(sum(reshape(mA, [M, M, 1, 1, 1, I]) .* ...
                  reshape(mB, [1, M, M, N, J, I]) .* ...
                  reshape(mC, [1, 1, M, 1, J, I]), [2, 3]));
end

end

On my system this results in:

>> q55913093_benchmark
ans =
    1.3964    0.1864    0.2428

Which means that the 2^nd method is preferable (at least for the default input sizes).

Question 5

There's a tonne of different approaches to this, so why not have some fun with it.

If you MUST use a for loop

No idea why you would, unless it's for a school assignment:

for($i=0;$i<count($data);$i++) {
  echo('<tr>');
  echo('<td>' . $data[$i][0] . '</td>');
  echo('<td>' . $data[$i][1] . '</td>');
  echo('<td>' . $data[$i][2] . '</td>');
  echo('</tr>');
}

But then that's kinda stupid directly accessing the ID's, lets use another for loop in the row:

for($i=0;$i<count($data);$i++) {
  echo('<tr>');
  for($j=0;$j<count($data[$i]);$j++) {
    echo('<td>' . $data[$i][$j] . '</td>');
  } 
  echo('</tr>');
}

Replace it with an equally as boring foreach loop:

<table>
<?php foreach($items as $row) {
  echo('<tr>');
  foreach($row as $cell) {
    echo('<td>' . $cell . '</td>');
  }
  echo('</tr>');
} ?>
</table>

Why not implode the array:

<table>
<?php foreach($items as $row) {
  echo('<tr>');
  echo('<td>');
  echo(implode('</td><td>', $row);
  echo('</td>');
  echo('</tr>');
} ?>
</table>

Mix it up, screw the foreach, and go for a walk; and implode stuff along the way:

<?php
function print_row(&$item) {
  echo('<tr>');
  echo('<td>');
  echo(implode('</td><td>', $item);
  echo('</td>');
  echo('</tr>');
}
?>

<table>
  <?php array_walk($data, 'print_row');?>
</table>

Double walking... OMG

Yeah, it looks a little silly now, but when you grow the table and things get more complex, things are a little better broken out and modularized:

<?php
function print_row(&$item) {
  echo('<tr>');
  array_walk($item, 'print_cell');
  echo('</tr>');
}

function print_cell(&$item) {
  echo('<td>');
  echo($item);
  echo('</td>');
}
?>

<table>
  <?php array_walk($data, 'print_row');?>
</table>

Question 6

You can do it like this, do the calculation from the innermost of the array. Check the demo.

<?php
function f(&$array)
{
    foreach($array as $k => &$v)
    {
        if(is_array($v))
        {
            if(count($v) == count($v, 1))
            {
                unset($array[$k]);
                if($k == 'sum')
                {
                    $v =  array_sum($v);
                    $array[] = $v;

                }elseif($k == 'multiply'){
                    $v = array_product($v);
                    $array[] = $v;
                }else{

                    foreach($v as $vv)
                        $array[] = $vv;
                }
            }else
                f($v);
        }
    }
}

while(count($array) != count($array, 1))
{
    f($array);
}

print_r($array);

Note：

traverse array from outer to inner
traverse array from inner to outer