您可以使用http://stereofrog.com/blok/on/070910中的解决方案。
如果链接中断，这是密码...

class Combinations implements Iterator
{
    protected $c = null;
    protected $s = null;
    protected $n = 0;
    protected $k = 0;
    protected $pos = 0;

    function __construct($s, $k) {
        if(is_array($s)) {
            $this->s = array_values($s);
            $this->n = count($this->s);
        } else {
            $this->s = (string) $s;
            $this->n = strlen($this->s);
        }
        $this->k = $k;
        $this->rewind();
    }
    function key() {
        return $this->pos;
    }
    function current() {
        $r = array();
        for($i = 0; $i < $this->k; $i++)
            $r[] = $this->s[$this->c[$i]];
        return is_array($this->s) ? $r : implode('', $r);
    }
    function next() {
        if($this->_next())
            $this->pos++;
        else
            $this->pos = -1;
    }
    function rewind() {
        $this->c = range(0, $this->k);
        $this->pos = 0;
    }
    function valid() {
        return $this->pos >= 0;
    }

    protected function _next() {
        $i = $this->k - 1;
        while ($i >= 0 && $this->c[$i] == $this->n - $this->k + $i)
            $i--;
        if($i < 0)
            return false;
        $this->c[$i]++;
        while($i++ < $this->k - 1)
            $this->c[$i] = $this->c[$i - 1] + 1;
        return true;
    }
}

foreach(new Combinations("1234567", 5) as $substring)
    echo $substring, ' ';

12345 12346 12347 12356 12357 12367 12456 12457 12467 12567 13456 13457 13567 14567 23456 23457 23467 23567 24567 34567

<?php

echo "<pre>";
$test = array("test_1","test_2","test_3");

// Get Combination
$return = uniqueCombination($test);

//Sort
sort($return);

//Pretty Print
print_r(array_map(function($v){ return implode(",", $v); }, $return));

function uniqueCombination($in, $minLength = 1, $max = 2000) {
    $count = count($in);
    $members = pow(2, $count);
    $return = array();
    for($i = 0; $i < $members; $i ++) {
        $b = sprintf("%0" . $count . "b", $i);
        $out = array();
        for($j = 0; $j < $count; $j ++) {
            $b{$j} == '1' and $out[] = $in[$j];
        }

        count($out) >= $minLength && count($out) <= $max and $return[] = $out;
        }
    return $return;
}

?>

输出

Array
(
    [0] => test_1
    [1] => test_2
    [2] => test_3
    [3] => test_1,test_2
    [4] => test_1,test_3
    [5] => test_2,test_3
    [6] => test_1,test_2,test_3
)

PEAR存储库中的Math_Combinatorics完全符合您的要求：
返回给定集合和子集大小的所有组合和排列的包，没有重复。保留关联数组。

require_once 'Math/Combinatorics.php';
$combinatorics = new Math_Combinatorics;

$input = array(1, 2, 3, 4, 5, 6, 7);
$output = $combinatorics->combinations($input, 5); // 5 is the subset size

// 1,2,3,4,5
// 1,2,3,4,6
// 1,2,3,4,7
// 1,2,3,5,6
// 1,2,3,5,7
// 1,2,3,6,7
// 1,2,4,5,6
// 1,2,4,5,7
// 1,2,4,6,7
// 1,2,5,6,7
// 1,3,4,5,6
// 1,3,4,5,7
// 1,3,4,6,7
// 1,3,5,6,7
// 1,4,5,6,7
// 2,3,4,5,6
// 2,3,4,5,7
// 2,3,4,6,7
// 2,3,5,6,7
// 2,4,5,6,7
// 3,4,5,6,7

另一个基于堆栈的解决方案。它速度很快，但占用大量内存。
希望这能帮上忙。
详细内容：

function _combine($numbers, $length)
{
    $combinations = array();
    $stack = array();

    // every combinations can be ordered
    sort($numbers);

    // startup
    array_push($stack, array(
        'store' => array(),
        'options' => $numbers,
    ));

    while (true) {
        // pop a item
        $item = array_pop($stack);

        // end of stack
        if (!$item) {
            break;
        }

        // valid store
        if ($length <= count($item['store'])) {
            $combinations[] = $item['store'];
            continue;
        }

        // bypass when options are not enough
        if (count($item['store']) + count($item['options']) < $length) {
            continue;
        }

        foreach ($item['options'] as $index => $n) {
            $newStore = $item['store'];
            $newStore[] = $n;

            // every combine can be ordered
            // so accept only options which is greater than store numbers
            $newOptions = array_slice($item['options'], $index + 1);

            // push new items
            array_push($stack, array(
                'store' => $newStore,
                'options' => $newOptions,
            ));
        }
    }

    return $combinations;
}

改进了此answer，使其也可以使用关联数组：

function uniqueCombination($values, $minLength = 1, $maxLength = 2000) {
    $count = count($values);
    $size = pow(2, $count);
    $keys = array_keys($values);
    $return = [];

    for($i = 0; $i < $size; $i ++) {
        $b = sprintf("%0" . $count . "b", $i);
        $out = [];

        for($j = 0; $j < $count; $j ++) {
            if ($b[$j] == '1') {
                $out[$keys[$j]] = $values[$keys[$j]];
            }
        }

        if (count($out) >= $minLength && count($out) <= $maxLength) {
             $return[] = $out;
        }
    }

    return $return;
}

例如：

print_r(uniqueCombination([
    'a' => 'xyz',
    'b' => 'pqr',
]);

结果：

Array
(
    [0] => Array
        (
            [b] => pqr
        )

    [1] => Array
        (
            [a] => xyz
        )

    [2] => Array
        (
            [a] => xyz
            [b] => pqr
        )

)

它仍然适用于非关联数组：

print_r(uniqueCombination(['a', 'b']);

结果：

Array
(
    [0] => Array
        (
            [1] => b
        )

    [1] => Array
        (
            [0] => a
        )

    [2] => Array
        (
            [0] => a
            [1] => b
        )

)

优化组合算法速度和内存的新方案
思维模式：生成K个数组的组合。新的解决方案将使用K个“for”语句。一个“for”一个数字。例如：$K = 5表示使用了5个“for”语句

$total = count($array);
$i0 = -1;
for ($i1 = $i0 + 1; $i1 < $total; $i1++) {
    for ($i2 = $i1 + 1; $i2 < $total; $i2++) {
        for ($i3 = $i2 + 1; $i3 < $total; $i3++) {
            for ($i4 = $i3 + 1; $i4 < $total; $i4++) {
                for ($i5 = $i4 + 1; $i5 < $total; $i5++) {
                    $record = array();
                    for ($i = 1; $i <= $k; $i++) {
                        $t = "i$i";
                        $record[] = $array[$$t];
                    }
                    $callback($record);
                }
            }
        }
    }
}

以及生成将由eval（）函数执行的真实的代码的代码细节

function combine($array, $k, $callback)
{
    $total = count($array);
    $init = '
        $i0 = -1;
    ';
    $sample = '
        for($i{current} = $i{previous} + 1; $i{current} < $total; $i{current}++ ) {
            {body}
        }
    ';

    $do = '
        $record = array();
        for ($i = 1; $i <= $k; $i++) {
            $t = "i$i";
            $record[] = $array[$$t];
        }
        $callback($record);
    ';
    $for = '';
    for ($i = $k; $i >= 1; $i--) {
        switch ($i) {
            case $k:
                $for = str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $do], $sample);
                break;
            case 1:
                $for = $init . str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $for], $sample);
                break;
            default:
                $for = str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $for], $sample);
                break;
        }
    }

    // execute
    eval($for);
}

如何合并K个数组

$k = 5;
$array = array(1, 2, 3, 4, 5, 6, 7);
$callback = function ($record) {
    echo implode($record) . "\n";
};
combine($array, $k, $callback);

我发现这里的其他答案令人困惑或过于复杂，所以我写了自己的答案。我认为这是一个递归方法的简单解决方案。基本思想是你遍历你的数组，并为每一项确定它是否在组合中（实际上，你不能决定，你递归地尝试两种方式）。你为第一项做这个选择，然后将它与数组其余部分递归生成的组合组合在一起。此解决方案使用数组的每个组合作为子数组填充结果数组。它按顺序使用项并保留关联，包括与数字键的关联。

function combinations(array $items, int $numToChoose, array &$results, $comb = []): void {
  if (count($items) < $numToChoose) {
    throw new \Exception("Asked to choose $numToChoose items from an array of length ". count($items));
  }

  // if nothing left to choose, we have a complete combination
  if ($numToChoose === 0) {
    $results[] = $comb;
    return;
  }

  // if we have to choose everything at this point, then we know what to do
  if (count($items) == $numToChoose) {
    $results[] = $comb + $items;
    return;
  }

  // The recursive cases: either use the first element or not and find combinations of the rest
  $val = reset($items);
  $key = key($items);
  unset($items[$key]);

  // not using it
  combinations($items, $numToChoose, $results, $comb);

  // using it
  $comb[$key] = $val;
  combinations($items, $numToChoose - 1, $results, $comb);
}

// Do a test run
$combs = [];
combinations([1=>1, 2=>2, 3=>3], 2, $combs);
var_dump($perms);

这将产生以下输出：

array(3) {
  [0]=>
  array(2) {
    [2]=>
    int(2)
    [3]=>
    int(3)
  }
  [1]=>
  array(2) {
    [1]=>
    int(1)
    [3]=>
    int(3)
  }
  [2]=>
  array(2) {
    [1]=>
    int(1)
    [2]=>
    int(2)
  }
}

我需要一个包含子集的组合函数，所以我采用了@Nguyen Van Vinh的答案，并根据需要进行了修改。
如果将[1,2,3,4]传递给函数，它将返回每个唯一组合和子集，排序如下：

[
  [1,2,3,4], [1,2,3], [1,2,4], [1,3,4], [2,3,4], [1,2], [1,3], [1,4], [2,3], [2,4], [3,4], [1], [2], [3], [4]
]

函数如下：

function get_combinations_with_length( $numbers, $length ){
    $result = array();
    $stack = array();
    // every combinations can be ordered
    sort($numbers);
    // startup
    array_push($stack, array(
        'store' => array(),
        'options' => $numbers,
    ));
    while (true) {
        // pop a item
        $item = array_pop($stack);
        // end of stack
        if (!$item) break;
        // valid store
        if ($length <= count($item['store'])) {
            $result[] = $item['store'];
            continue;
        }
        // bypass when options are not enough
        if (count($item['store']) + count($item['options']) < $length) {
            continue;
        }
        foreach ($item['options'] as $i=>$n) {
            $newStore = $item['store'];
            $newStore[] = $n;
            // every combine can be ordered, so accept only options that are greater than store numbers
            $newOptions = array_slice($item['options'], $i + 1);
            // array_unshift to sort numerically, array_push to reverse
            array_unshift($stack, array(
                'store' => $newStore,
                'options' => $newOptions,
            ));
        }
    }
    return $result;
}

function get_all_combinations( $numbers ){
    $length = count($numbers);
    $result = [];
    while ($length > 0) {
        $result = array_merge($result, get_combinations_with_length( $numbers, $length ));
        $length--;
    }
    return $result;
}

$numbers = [1,2,3,4];
$result = get_all_combinations($numbers);

echo 'START: '.json_encode( $numbers ).'<br><br>';
echo 'RESULT: '.json_encode( $result ).'<br><br>';
echo '('.count($result).' combination subsets found)';