Yoh*_*han 38 php arrays iteration combinations
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......
例如,我有3个数组:
$arrayA = array('A1','A2','A3');
$arrayB = array('B1','B2','B3');
$arrayC = array('C1','C2');
我想生成一个3 x 3 x 2 = 18种组合的数组:
问题是创建一个具有可变数量的源数组的函数...
Krz*_*tof 61
这是递归解决方案:
function combinations($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 $result;
}
print_r(
combinations(
array(
array('A1','A2','A3'),
array('B1','B2','B3'),
array('C1','C2')
)
)
);
bar*_*oon 13
这是一个笛卡儿的产品,我不久前就问了同样的问题.这是PHP网站上发布的算法.
function array_cartesian_product($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 $result;
}
小智 8
此代码除了简单之外,还可以获得多个数组的所有组合并保留密钥.
function get_combinations($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 $result;
}
例:
Array
(
Array
(
'1' => 'White',
'2' => 'Green',
'3' => 'Blue'
),
Array
(
'4' =>' Small',
'5' => 'Big'
)
)
将返回:
Array
(
[0] => Array
(
[1] => White
[4] => Small
)
[1] => Array
(
[1] => White
[5] => Big
)
[2] => Array
(
[2] => Green
[4] => Small
)
[3] => Array
(
[2] => Green
[5] => Big
)
[4] => Array
(
[3] => Blue
[4] => Small
)
[5] => Array
(
[3] => Blue
[5] => Big
)
)
我喜欢这个解决方案：https : //stackoverflow.com/a/33259643/3163536 但要回答实际问题（假设每个组合的元素数应等于传入数组的数量），应修改函数：
function getCombinations(...$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;
}
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
)
)
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);
}
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"
}
小智 5
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);