假設(shè)我有一個(gè)數(shù)組，如下所示：Array(
[arm] => Array
(
[0] => A [1] => B [2] => C )
[gender] => Array
(
[0] => Female
[1] => Male
)
[location] => Array
(
[0] => Vancouver
[1] => Calgary
))如何在保留外部關(guān)聯(lián)數(shù)組的鍵并在內(nèi)部數(shù)組中使用它們的同時(shí)，找到笛卡兒積？算法的結(jié)果應(yīng)該是：Array(
[0] => Array
(
[arm] => A [gender] => Female
[location] => Vancouver
)
[1] => Array
(
[arm] => A [gender] => Female
[location] => Calgary
)
[2] => Array
(
[arm] => A [gender] => Male
[location] => Vancouver
)...etc.我已經(jīng)查閱了很多笛卡爾積算法，但我仍然停留在如何保存關(guān)聯(lián)鍵的細(xì)節(jié)上。我目前使用的算法只給出數(shù)字索引： $result = array();
foreach ($map as $a) {
if (empty($result)) {
$result = $a;
continue;
}
$res = array();
foreach ($result as $r) {
foreach ($a as $v) {
$res[] = array_merge((array)$r, (array)$v);
}
}
$result = $res;
}
print_r($result);任何幫助都將不勝感激。用PHP關(guān)聯(lián)數(shù)組求笛卡兒積
3 回答
ibeautiful
TA貢獻(xiàn)1993條經(jīng)驗(yàn) 獲得超6個(gè)贊
function inject($elem, $array) {
return array_map(function ($n) use ($elem) { return array_merge((array)$elem, (array)$n); }, $array);}function zip($array1, $array2) {
return array_reduce($array1, function ($v, $n) use ($array2) { return array_merge($v, inject($n, $array2)); }, array());}
function cartesian_product($array) {
$keys = array_keys($array);
$prod = array_shift($array);
$prod = array_reduce($array, 'zip', $prod);
return array_map(function ($n) use ($keys) { return array_combine($keys, $n); }, $prod);}(下面使用偽數(shù)組/列表/字典表示法，因?yàn)镻HP對(duì)于這類事情太冗長(zhǎng)了。)
injecta, [b][(a,b)]ab
inject('a', ['foo', 'bar'])
=> [('a', 'foo'), ('b', 'bar')]zipinject
zip(['a', 'b'], ['foo', 'bar'])
=> [('a', 'foo'), ('a', 'bar'), ('b', 'foo'), ('b', 'bar')]zip
zip(zip(['a', 'b'], ['foo', 'bar']), ['42', '76'])
=> [('a', 'foo', '42'), ('a', 'foo', '76'), ('a', 'bar', '42'), …]array_combine(['key1', 'key2', 'key3'], ['a', 'foo', '42']) => [ key1 : 'a', key2 : 'foo', key3 : '42' ]
function inject($elem, $array) {
$elem = (array)$elem;
foreach ($array as &$a) {
$a = array_merge($elem, (array)$a);
}
return $array;}function zip($array1, $array2) {
$prod = array();
foreach ($array1 as $a) {
$prod = array_merge($prod, inject($a, $array2));
}
return $prod;}function cartesian_product($array) {
$keys = array_keys($array);
$prod = array_shift($array);
$prod = array_reduce($array, 'zip', $prod);
foreach ($prod as &$a) {
$a = array_combine($keys, $a);
}
return $prod;}添加回答
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