How to find every possible combination of an arbitrary number of arrays in PHP

Question

I have an arbitrary number of nested arrays in php. For example:

Array
(
    [0] => Array
        (
            [0] => 36
            [0] => 2
            [0] => 9
        )

    [1] => Array
        (
            [0] => 95
            [1] => 21
            [2] => 102
            [3] => 38
        )

    [2] => Array
        (
            [0] => 3
            [1] => 5
        )

)

I want to find the most efficient way to combine all possible combinations of each of these nested arrays. I'd like to end up with something that looks like this...

Array
(
    [0] => "36,95,3"
    [1] => "36,95,5"
    [2] => "36,21,3"
    [3] => "36,21,5"

    etc...

)

The order the results are combined in is not important. That is to say, there is no difference between "3,95,36", "36,95,3", and "95,36,3". I would like to omit these redundant combinations.

Any suggestions on how to go about this would be much appreciated.

Thanks in advance,

Is the depth arbitrary too? or just the number at a given depth? If the depth is known, you can loop like Josh Curren says - if not you'll have to do it recursively. — Austin Fitzpatrick
– Austin Fitzpatrick, Commented Mar 3, 2010 at 17:42
Do you always want triples or can this be 36,95 and 36,21,95,5 as well? — Gordon
– Gordon, Commented Mar 3, 2010 at 17:55

Chirag Darji · Accepted Answer · 2010-04-14 10:03:54Z

1

<?php
$a = array("01", "02");
$b = array("white", "green");
$c = array("one", "two", "three");
$aG = array($a, $b, $c);
$codes = array();
$pos = 0;
generateCodes($aG);

function generateCodes($arr) {
    global $codes, $pos;
    if(count($arr)) {
    for($i=0; $i<count($arr[0]); $i++) {
        $tmp = $arr;
        $codes[$pos] = $arr[0][$i];
        $tarr = array_shift($tmp);
        $pos++;
        generateCodes($tmp);

    }
    } else {
    echo join(", ", $codes)."<br/>";
    }
    $pos--;
}
?>

answered Apr 14, 2010 at 10:03

Chirag Darji

111 bronze badge

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Slava Nadvorny · Accepted Answer · 2010-05-16 13:50:58Z

0

General permutation solution for any number of groups (in Groovy to keep it simple).

list = [["A", "B", "W"], ["C", "D"], ["X", "Y"]]

r = []
void p(List<List<String>> l, int n) {
    for(t in l[n]) {
        r[n] = t
        if (n < l.size() - 1) {
            p(l, n + 1)
        } else {
            println(r)
        }
    }
}

p(list, 0);

answered May 16, 2010 at 13:50

Slava Nadvorny

1513 bronze badges

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