如何在PHP中生成多个数组中的所有项目组合

时间：2011-12-19 20:25:23

标签： php arrays combinations iteration

我试图找到几个数组中所有项目的组合。 数组的数量是随机的（可以是2,3,4,5 ......）。 每个数组中的元素数量也是随机的......

例如，我有3个数组：

$arrayA = array('A1','A2','A3');
$arrayB = array('B1','B2','B3');
$arrayC = array('C1','C2');

我想生成一个3 x 3 x 2 = 18种组合的数组：

  • A1，B1，C1
  • A1，B1，C2
  • A1，B2，C1
  • A1，B2，C2
  • A1，B3，C1
  • A1，B3，C2
  • A2，B1，C1
  • A2，B1，C2 ...

问题是创建一个具有可变数量的源数组的函数......

9 个答案:

答案 0 :(得分：55)

这是递归解决方案：

function combinations($arrays, $i = 0) {
    if (!isset($arrays[$i])) {
        return array();
    }
    if ($i == count($arrays) - 1) {
        return $arrays[$i];
    }

    // get combinations from subsequent arrays
    $tmp = combinations($arrays, $i + 1);

    $result = array();

    // concat each array from tmp with each element from $arrays[$i]
    foreach ($arrays[$i] as $v) {
        foreach ($tmp as $t) {
            $result[] = is_array($t) ? 
                array_merge(array($v), $t) :
                array($v, $t);
        }
    }

    return $result;
}

print_r(
    combinations(
        array(
            array('A1','A2','A3'), 
            array('B1','B2','B3'), 
            array('C1','C2')
        )
    )
);

答案 1 :(得分：13)

这是一个笛卡尔积，我只是asked the same question not too long ago。这是algorithm that is posted on the PHP website。

function array_cartesian_product($arrays)
{
    $result = array();
    $arrays = array_values($arrays);
    $sizeIn = sizeof($arrays);
    $size = $sizeIn > 0 ? 1 : 0;
    foreach ($arrays as $array)
        $size = $size * sizeof($array);
    for ($i = 0; $i < $size; $i ++)
    {
        $result[$i] = array();
        for ($j = 0; $j < $sizeIn; $j ++)
            array_push($result[$i], current($arrays[$j]));
        for ($j = ($sizeIn -1); $j >= 0; $j --)
        {
            if (next($arrays[$j]))
                break;
            elseif (isset ($arrays[$j]))
                reset($arrays[$j]);
        }
    }
    return $result;
}

答案 2 :(得分：5)

此代码除了简单之外，还可以获取多个数组的所有组合并保留密钥。

function get_combinations($arrays) {
    $result = array(array());
    foreach ($arrays as $property => $property_values) {
        $tmp = array();
        foreach ($result as $result_item) {
            foreach ($property_values as $property_key => $property_value) {
                $tmp[] = $result_item + array($property_key => $property_value);
            }
        }
        $result = $tmp;
    }
    return $result;
}

<强>例： 

Array
(
    Array
    (
        '1' => 'White',
        '2' => 'Green',
        '3' => 'Blue'
    ),
    Array
    (
        '4' =>' Small',
        '5' => 'Big'
    )
)

将返回： 

Array
(
    [0] => Array
    (
        [1] => White
        [4] =>  Small
    )
    [1] => Array
    (
        [1] => White
        [5] => Big
    )
    [2] => Array
    (
        [2] => Green
        [4] =>  Small
    )
    [3] => Array
    (
        [2] => Green
        [5] => Big
    )
    [4] => Array
    (
        [3] => Blue
        [4] =>  Small
    )
    [5] => Array
    (
        [3] => Blue
        [5] => Big
    )
)

答案 3 :(得分：4)

我知道这个问题已经过时了，但今天我遇到了同样的问题并决定尝试新的发电机：

function generateCombinations(array $array) {
    foreach (array_pop($array) as $value) {
        if (count($array)) {
            foreach (generateCombinations($array) as $combination) {
                yield array_merge([$value], $combination);
            };
        } else {
            yield [$value];
        }
    }
}

foreach (generateCombinations(['a' => ['A'], 'b' => ['B'], 'c' => ['C', 'D'], 'd' => ['E', 'F', 'G']]) as $c) {
        var_dump($c);
    }

结果：

array(4) {
[0]=>
string(1) "E"
[1]=>
string(1) "C"
[2]=>
string(1) "B"
[3]=>
string(1) "A"
}
array(4) {
[0]=>
string(1) "E"
[1]=>
string(1) "D"
[2]=>
string(1) "B"
[3]=>
string(1) "A"
}
array(4) {
[0]=>
string(1) "F"
[1]=>
string(1) "C"
[2]=>
string(1) "B"
[3]=>
string(1) "A"
}
array(4) {
[0]=>
string(1) "F"
[1]=>
string(1) "D"
[2]=>
string(1) "B"
[3]=>
string(1) "A"
}
array(4) {
[0]=>
string(1) "G"
[1]=>
string(1) "C"
[2]=>
string(1) "B"
[3]=>
string(1) "A"
}
array(4) {
[0]=>
string(1) "G"
[1]=>
string(1) "D"
[2]=>
string(1) "B"
[3]=>
string(1) "A"
}

答案 4 :(得分：4)

我喜欢这种解决方案： https://stackoverflow.com/a/33259643/3163536 但是要回答实际的问题（假设每个组合的元素数应等于传入数组的数量），则应修改函数：

function getCombinations(...$arrays)
    {
        $result = [[]];
        foreach ($arrays as $property => $property_values) {
            $tmp = [];
            foreach ($result as $result_item) {
                foreach ($property_values as $property_value) {
                    $tmp[] = array_merge($result_item, [$property => $property_value]);
                }
            }
            $result = $tmp;
        }
        return $result;
    }

用法：

$arrayA = array('A1','A2','A3');
$arrayB = array('B1','B2','B3');
$arrayC = array('C1','C2');

print_r(getCombinations($arrayA, $arrayB, $arrayC));

结果：

Array
(
    [0] => Array
        (
            [0] => A1
            [1] => B1
            [2] => C1
        )

    [1] => Array
        (
            [0] => A1
            [1] => B1
            [2] => C2
        )

    [2] => Array
        (
            [0] => A1
            [1] => B2
            [2] => C1
        )

    [3] => Array
        (
            [0] => A1
            [1] => B2
            [2] => C2
        )

    [4] => Array
        (
            [0] => A1
            [1] => B3
            [2] => C1
        )

    [5] => Array
        (
            [0] => A1
            [1] => B3
            [2] => C2
        )

    [6] => Array
        (
            [0] => A2
            [1] => B1
            [2] => C1
        )

    [7] => Array
        (
            [0] => A2
            [1] => B1
            [2] => C2
        )

    [8] => Array
        (
            [0] => A2
            [1] => B2
            [2] => C1
        )

    [9] => Array
        (
            [0] => A2
            [1] => B2
            [2] => C2
        )

    [10] => Array
        (
            [0] => A2
            [1] => B3
            [2] => C1
        )

    [11] => Array
        (
            [0] => A2
            [1] => B3
            [2] => C2
        )

    [12] => Array
        (
            [0] => A3
            [1] => B1
            [2] => C1
        )

    [13] => Array
        (
            [0] => A3
            [1] => B1
            [2] => C2
        )

    [14] => Array
        (
            [0] => A3
            [1] => B2
            [2] => C1
        )

    [15] => Array
        (
            [0] => A3
            [1] => B2
            [2] => C2
        )

    [16] => Array
        (
            [0] => A3
            [1] => B3
            [2] => C1
        )

    [17] => Array
        (
            [0] => A3
            [1] => B3
            [2] => C2
        )

)

答案 5 :(得分：3)

还有一个想法：

$ar = [
    'a' => [1,2,3],
    'b' => [4,5,6],
    'c' => [7,8,9]
];

$counts = array_map("count", $ar);
$total = array_product($counts);
$res = [];

$combinations = [];
$curCombs = $total;

foreach ($ar as $field => $vals) {
    $curCombs = $curCombs / $counts[$field];
    $combinations[$field] = $curCombs;
}

for ($i = 0; $i < $total; $i++) {
    foreach ($ar as $field => $vals) {
        $res[$i][$field] = $vals[($i / $combinations[$field]) % $counts[$field]];
    }
}

var_dump($res);

答案 6 :(得分：1)

这是一个从一组数字生成唯一组合的代码。

如果您有一组数字，例如1,3,4,7,12，则可以生成X个数字集，这些数字都是唯一的，没有重复。

第一个函数在PHP 7.4或更高版本中起作用，第二个函数使用键存储值。两者都基于基准运行良好。

function get_combos74($map, $size, &$generated = [], $loop = 1, $i = 0, $prefix = [])
{
    if ($loop == 1) {
        sort($map);
    }

    for (; $i < count($map); $i++) {
        if ($loop < $size) {
            get_combos74($map, $size, $generated, $loop + 1, $i + 1, [...$prefix, $map[$i]]);
        } else {
            $generated[] = [...$prefix, $map[$i]];
        }
    }

    return $generated;
}

function get_combosSTR($map, $size, &$generated = [], $loop = 1, $i = 0, $prefix = '')
{
    if ($loop == 1) {
        sort($map);
    }

    for (; $i < count($map); $i++) {
        if ($loop < $size) {
            get_combosSTR($map, $size, $generated, $loop + 1, $i + 1, "$prefix{$map[$i]}:");
        } else {
            $generated["$prefix{$map[$i]}"] = 0;
        }
    }

    return $generated;
}

答案 7 :(得分：0)

我改进了@mr1031011 提供的解决方案，允许选择性地保留密钥

/**
 * Generate an array of permutations from a given array[array[options]]
 * e.g.
 * ['foo' => [null, 'a'], 'bar' => ['b']]
 *
 * will generate:
 * [null, 'b']
 * ['a', 'b']
 *
 * or with preserveKeys = true:
 * ['foo' => null,'bar' => 'b']
 * ['foo' => 'a','bar' => 'b']
 *
 * @param array $array
 * @param bool  $preserveKeys
 *
 * @return Generator
 */
function generatePermutations(array $array, bool $preserveKeys = false): Generator
{
    if ($preserveKeys) {
        end($array);
        $key = key($array);
    }

    foreach (array_pop($array) as $value) {
        $item = isset($key) ? [$key => $value] : [$value];

        if (!count($array)) {
            yield $item;
            continue;
        }

        foreach (generatePermutations($array, $preserveKeys) as $combination) {
            yield array_merge($item, $combination);
        }
    }
}

答案 8 :(得分：-1)

我是从O'Relly（https://www.oreilly.com/library/view/php-cookbook/1565926811/ch04s25.html）那里得到的，

function pc_array_power_set($array) {
// initialize by adding the empty set
$results = array(array( ));

foreach ($array as $element)
    foreach ($results as $combination)
        array_push($results, array_merge(array($element), $combination));



   return $results;
}

然后使用

对其进行调用 
$set = array('A', 'B', 'C');
$power_set = pc_array_power_set($set);

那应该可以解决问题！

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