Question

我有一个这样的数组：

$a = array(
    array(2 => 1, 4 => 2, 9 => 3),
    array(3 => 7, 4 => 5, 7 => 3),
    array(1 => 6, 4 => 5),
    ...
);

因此该数组包含大量具有整数键=＆gt;的子数组。整数值。现在我想找到不共享密钥的子数组，或者如果它们共享一个密钥，则该密钥的值必须相同。

示例：$ a [1]和$ a [2]匹配，因为$ a [1] [4] == $ a [2] [4]并且没有其他键匹配。但$ a [0]和$ a [1]不匹配，因为$ a [0] [4]！= $ a [1] [4]。

子阵列中的元素数量可能会有所不同。

有一种有效的方法吗？我能想到的唯一方法是在嵌套循环中检查每个可能的对，得到O（n ^ 2）。

如果有人想要更有意义的标题，请随时编辑它。

也许代码更清晰:(天真的实现）

$pairs = array();
for($i = 0; $i < count($a); $i++)
    for($j = $i+1; $j < count($a); $j++)
        if(array_intersect_key($a[$i], $a[$j]) == array_intersect_assoc($a[$i], $a[$j]))
            $pairs[] = array($i, $j);

替代：

$matching = array();
for($i = 0; $i < count($a); $i++)
    for($j = $i+1; $j < count($a); $j++)
        if(array_intersect_key($a[$i], $a[$j]) == array_intersect_assoc($a[$i], $a[$j]))
            list($matching[$i][], $matching[$j][]) = array($j, $i);

Answer 1

可能有办法实现，但这在某种程度上取决于您是否知道可能有多少匹配（或者您的数据的一般'匹配'）。如果有更多匹配，那么从假设所有匹配和消除开始可能更好。

无论如何，我认为您可以预先处理数据。我不确定这是否更快 - 这实际上取决于您的数据分布，但我首先尝试这样的事情并从那里开始工作：

$a = array(
    array(2 => 1, 4 => 2, 9 => 3),
    array(3 => 7, 4 => 5, 7 => 3),
    array(1 => 6, 4 => 5),
    array(1 => 6, 4 => 5, 7 => 5),
    array(2 => 1, 4 => 2, 9 => 3)   
);
// 1 and 2 match, 2 and 3 match, 0 and 4 match

$keyData = array();
for ($i = 0; $i < count($a); $i++) {
    foreach($a[$i] as $k => $v) {
        if (!isset($keyData[$k])) { 
            $keyData[$k] = array();
        }
        if (!isset($keyData[$k][$v])) {
            $keyData[$k][$v] = array();   
        }
        $keyData[$k][$v][] = $i;
    }
}

$potentialMatches = array();
foreach ($keyData as $key => $values) {
    // Ignore single key/value pairs
    if (count($values) > 1) {
        foreach ($values as $value => $arrayIndices) {
            for ($i = 0; $i < count($arrayIndices); $i ++) {
                for ($j = $i + 1; $j < count($arrayIndices); $j ++) { 
                    $potentialMatches[] = array($arrayIndices[$i], $arrayIndices[$j]);
                }
            }
        }
    }
}

// You might need to do this ...
/*
foreach ($potentialMatches as &$m) {
    array_unique($m);
}
*/

$pairs = array();
foreach ($potentialMatches as $m) { 
     if(array_intersect_key($a[$m[0]], $a[$m[1]]) 
        == array_intersect_assoc($a[$m[0]], $a[$m[1]])) {
         $pairs[] = $m;
    }
}

print_r($pairs);

输出：

Array
(
    [0] => Array
        (
            [0] => 0
            [1] => 4
        )

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

    [2] => Array
        (
            [0] => 2
            [1] => 3
        )

)

修改

正如我在评论中所说，这不会捕获不共享任何键的数组 - 您认为这是匹配的。下面的代码执行此操作，虽然我不确定它是否比嵌套解决方案更快（并且它将使用大量内存）

// New test data to cover the case I missed $a = array( array(2 => 1, 4 => 2, 9 => 3), array(3 => 7, 4 => 5, 7 => 3), array(1 => 6, 4 => 5), array(1 => 6, 4 => 5, 7 => 5), array(2 => 1, 4 => 2, 9 => 3), array(8 => 3) ); // 1 and 2 match, 2 and 3 match, 0 and 4 match, 5 matches all // First assume everything is a match, build an array of: // indicies => array of potential matches $potentialMatches = array_fill(0, count($a), array_keys($a)); // Build data about each key, the indicies that contain that key // and the indicies for each value of that key $keyData = array(); for ($i = 0; $i < count($a); $i++) { foreach($a[$i] as $k => $v) { if (!isset($keyData[$k])) { $keyData[$k] = array(); } if (!isset($keyData[$k][$v])) { $keyData[$k][$v] = array(); } $keyData[$k]['all'][] = $i; $keyData[$k][$v][] = $i; } } // print_r($keyData); // Now go through the key data and eliminate indicies that // can't match foreach ($keyData as $key => $values) { if (count($values) > 2) { // Ignore single key/value pairs // Two indecies do not match if they appear in seperate value lists // First get the list of all indicies that have this key $all = array_unique($values['all']); unset($values['all']); // Now go through the value lists foreach ($values as $value => $arrayIndices) { // The indicies for this value cannot match the other // indices in the system, i.e. this list $cantMatch = array_diff($all, $arrayIndices); // So remove the indicies that can't match from the potentials list foreach ($arrayIndices as $index) { $potentialMatches[$index] = array_diff($potentialMatches[$index], $cantMatch); } } } } //print_r($potentialMatches); // You said you didn't mind the output format, so that's probably enough // but that array contains (x,x) which is pointless and both (x,y) and (y,x) // so we can do one final bit of processing to print it out in a nicer way $pairs = array(); foreach ($potentialMatches as $x => $matches) { foreach ($matches as $y) { if ( ($x < $y) ) { $pairs[] = array($x, $y); } } } print_r($pairs);

<强>输出

Array ( [0] => Array ( [0] => 0 [1] => 4 ) [1] => Array ( [0] => 0 [1] => 5 ) [2] => Array ( [0] => 1 [1] => 2 ) [3] => Array ( [0] => 1 [1] => 5 ) [4] => Array ( [0] => 2 [1] => 3 ) [5] => Array ( [0] => 2 [1] => 5 ) [6] => Array ( [0] => 3 [1] => 5 ) [7] => Array ( [0] => 4 [1] => 5 ) )

Answer 2

如果您正在寻找相邻的匹配，

$temp = null;
$last_result = array();
foreach($a as $key => $value){
    if(is_null($temp)){
        $temp = $value;
    } else{
        $result = array_intersect_assoc($temp, $value);
        if(!empty($result))
            array_push($last_result, $result);
        $temp = $value;
    }
}
print_r($last_result);

否则只需使用array_intersect_assoc

您可以这样做的例子

$res = array_intersect_assoc($a[0],$a[1],$a[2]);
print_r($res);

PHP中数组的交集。寻找匹配的对

2 个答案: