php 如何在PHP中生成多个数组中项目的所有组合
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How to generate in PHP all combinations of items in multiple arrays
提问by Yohan
I'im trying to find all combinations of items in several arrays. The number of arrays is random (this can be 2, 3, 4, 5...). The number of elements in each array is random too...
我试图在几个数组中找到所有项目的组合。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......
For exemple, I have the 3 arrays :
例如，我有 3 个数组：
$arrayA = array('A1','A2','A3');
$arrayB = array('B1','B2','B3');
$arrayC = array('C1','C2');
I would like to generate an array with 3 x 3 x 2 = 18 combinations :
我想生成一个包含 3 x 3 x 2 = 18 种组合的数组：
- A1, B1, C1
- A1, B1, C2
- A1, B2, C1
- A1, B2, C2
- A1, B3, C1
- A1, B3, C2
- A2, B1, C1
- A2, B1, C2 ...
- A1、B1、C1
- A1、B1、C2
- A1、B2、C1
- A1、B2、C2
- A1、B3、C1
- A1、B3、C2
- A2、B1、C1
- A2, B1, C2 ...
The problem is to create a function with a variable number of source arrays...
问题是创建一个具有可变数量的源数组的函数......
回答by Krzysztof
Here is recursive solution:
这是递归解决方案：
function combinations($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 $result;
}
print_r(
combinations(
array(
array('A1','A2','A3'),
array('B1','B2','B3'),
array('C1','C2')
)
)
);
回答by barfoon
This is a cartesian product, and I just asked the same question not too long ago. Here is the algorithm that is posted on the PHP website.
这是一个笛卡尔积，不久前我刚问过同样的问题。这是在PHP 网站上发布的算法。
function array_cartesian_product($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 $result;
}
回答by Guilhermo Luna
This code besides simplicity, get all combinations of multiple arrays and preserves keys.
这段代码除了简单之外，还能获取多个数组的所有组合并保留键。
function get_combinations($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;
}
Exemple:
例子：
Array
(
Array
(
'1' => 'White',
'2' => 'Green',
'3' => 'Blue'
),
Array
(
'4' =>' Small',
'5' => 'Big'
)
)
Will return:
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
)
)
回答by mr1031011
I know this question is old, but I got the same issue today and decided to give the new Generator a try:
我知道这个问题很老，但我今天遇到了同样的问题，并决定尝试一下新的 Generator：
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);
}
Result:
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"
}
回答by Eugene Fedorenko
One more idea:
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($res);
回答by Alexander Ivashchenko
I like this solution: https://stackoverflow.com/a/33259643/3163536but to answer the actual question (which assumes that the number of elements of each combination should be equal to the number of incoming arrays) the function should be modified:
我喜欢这个解决方案：https: //stackoverflow.com/a/33259643/3163536但要回答实际问题（假设每个组合的元素数应等于传入数组的数量），应修改函数：
function getCombinations(...$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;
}
The usage:
用法：
$arrayA = array('A1','A2','A3');
$arrayB = array('B1','B2','B3');
$arrayC = array('C1','C2');
print_r(getCombinations($arrayA, $arrayB, $arrayC));
The result:
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
)
)
