多维数组中的数组置换保持键PHP [英] Array permutations in multidimensional array keeping the keys PHP

问题描述

两天来，我一直在疯狂地试图做到这一点，也许您可​​以启发我.这是针对马匹投注的排列.每次用户玩耍时，我都会得到一个多维数组(2个级别).第一级包含比赛ID，第二级包含用户为该比赛选择的马.看起来像这样:

For two days I've been running crazy trying to accomplish this, maybe you can enlighten me. This is for a horse betting permutation. Every time a user plays, I get a multidimensional array (2 levels). The first level contains the race ID, the the second level contains thee horses selected by the user for that race. It looks like this:

$play = array
(
    '4' => array(7, 32),
    '8' => array(4),
    '2' => array(9),
    '12' => array('5'),
    '83' => array('10', '11', '12', ''),
    '9' => array('3'),
);

我需要知道该剧的所有可能组合.使用此功能很容易做到:

I need to know what are all the possible combinations for that play. Which is easily done with this function:

function permutations(array $array)
{
    switch (count($array)) {
        case 1:
            return $array[0];
            break;
        case 0:
            throw new InvalidArgumentException('Requires at least one array');
            break;
    }

    $a = array_shift($array);
    $b = permutations($array);

    $return = array();
    foreach ($a as $key => $v) {
        if(is_numeric($v))
        {
            foreach ($b as $key2 => $v2) {
                $return[] = array_merge(array($v), (array) $v2);
            }
        }
    }

    return $return;
}

这将返回一个包含所有可能组合的数组.到目前为止一切顺利，结果如下所示:

This returns an array with all the possible combinations beautifully. So far so good, and the result looks like this:

Array
(
    [0] => Array
        (
            [0] => 7
            [1] => 4
            [2] => 9
            [3] => 5
            [4] => 10
            [5] => 3
        )

    [1] => Array
        (
            [0] => 7
            [1] => 4
            [2] => 9
            [3] => 5
            [4] => 11
            [5] => 3
        )

    [2] => Array
        (
            [0] => 7
            [1] => 4
            [2] => 9
            [3] => 5
            [4] => 12
            [5] => 3
        )

    [3] => Array
        (
            [0] => 32
            [1] => 4
            [2] => 9
            [3] => 5
            [4] => 10
            [5] => 3
        )

    [4] => Array
        (
            [0] => 32
            [1] => 4
            [2] => 9
            [3] => 5
            [4] => 11
            [5] => 3
        )

    [5] => Array
        (
            [0] => 32
            [1] => 4
            [2] => 9
            [3] => 5
            [4] => 12
            [5] => 3
        )

)

我的问题:我需要为每匹马设置数组"key"作为比赛ID"，而不是0、1、2、3. 我需要这样的结果:

My problem: I need the array "key" for every horse to be the "race ID", not 0,1,2,3. I need the result to be like this:

Array
(
    [0] => Array
        (
            [4] => 7
            [8] => 4
            [2] => 9
            [12] => 5
            [83] => 10
            [9] => 3
        )

    [1] => Array
        (
            [4] => 7
            [8] => 4
            [2] => 9
            [12] => 5
            [83] => 11
            [9] => 3
        )

    [2] => Array
        (
            [4] => 7
            [8] => 4
            [2] => 9
            [12] => 5
            [83] => 12
            [9] => 3
        )

    [3] => Array
        (
            [4] => 32
            [8] => 4
            [2] => 9
            [12] => 5
            [83] => 10
            [9] => 3
        )

    [4] => Array
        (
            [4] => 32
            [8] => 4
            [2] => 9
            [12] => 5
            [83] => 11
            [9] => 3
        )

    [5] => Array
        (
            [4] => 32
            [8] => 4
            [2] => 9
            [12] => 5
            [83] => 12
            [9] => 3
        )

)

我该怎么做?我知道这是一篇很长的文章，但我需要对此进行图解.我在将函数的头绕到函数递归时遇到了问题，在每个循环中我都迷路了.

How can I accomplish this? I know its a long post but I needed to graph this. I am having problems to wrap my head around the function recursion and I get totally lost in each loop.

推荐答案

这就是您所需要的.我已根据需要发表评论:

Here's what you need. I have commented as necessary:

function permutations(array $array)
{
    switch (count($array)) {
        case 1:
            // Return the array as-is; returning the first item
            // of the array was confusing and unnecessary
            return $array;
            break;
        case 0:
            throw new InvalidArgumentException('Requires at least one array');
            break;
    }

    // We 'll need these, as array_shift destroys them
    $keys = array_keys($array);

    $a = array_shift($array);
    $k = array_shift($keys); // Get the key that $a had
    $b = permutations($array);

    $return = array();
    foreach ($a as $v) {
        if(is_numeric($v))
        {
            foreach ($b as $v2) {
                // array($k => $v) re-associates $v (each item in $a)
                // with the key that $a originally had
                // array_combine re-associates each item in $v2 with
                // the corresponding key it had in the original array
                // Also, using operator+ instead of array_merge
                // allows us to not lose the keys once more
                $return[] = array($k => $v) + array_combine($keys, $v2);
            }
        }
    }

    return $return;
}

查看实际操作.

顺便说一句，递归计算所有排列很简单，但是您可能不想在生产环境中进行.您绝对应该考虑进行健全性检查，以计算有多少种排列，并且如果它们超过一定限制，则至少不允许处理继续进行.

By the way, calculating all the permutations recursively is neat, but you might not want to do it in a production environment. You should definitely consider a sanity check that calculates how many permutations there are and doesn't allow processing to continue if they are over some limit, at the very least.

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