Php 合并两个阵列的更有效方法

我将数组的每个元素与数组的每个其他元素进行比较，如果两个元素具有相同的源/目标、目标/源，我将内部数组与员工合并，例如

0=> source - 3 target - 4 officers => 0 - 'Aberdeen Asset Management PLC'
1=> source - 3 target - 4 officers => 0 - 'whatever'

它将被合并为

0=> source - 3 target - 4 officers => 0 - 'Aberdeen Asset Management PLC', 1 - 'whatever'

以下是数据的外观：

我的代码效率很低，有1000多行要执行，执行大约需要90秒，这对于此类事情来说是不可接受的

   foreach ($edges as $i => &$edge) {
        for ($j = $i + 1; $j < count($edges); $j++) {
            if ($edge['source'] == $edges[$j]['source'] && $edge['target'] == $edges[$j]['target']) {
                foreach ($edges[$j]['officers'] as $officer) {
                    array_push($edge['officers'], $officer);
                }
                array_splice($edges, $j, 1);
            }
        }
    }

foreach（$i=>&$edge）{
对于（$j=$i+1；$j

我认为你应该这样做（更新了）：

---

我认为你应该这样做（更新了）：

---

使用

array\u search

、

array\u keys

、

array\u slice

和

array\u merge

功能的解决方案：

// an exemplary array
$edges = [
    0 => ['source' => 3, 'target' => 4, 'officers' => ['Aberdeen Asset Management PLC']],
    1 => ['source' => 3, 'target' => 4, 'officers' => ['whatever']],
    3 => ['source' => 4, 'target' => 7, 'officers' => ['Jason']],
    4 => ['source' => 4, 'target' => 5, 'officers' => ['John']],
    5 => ['source' => 4, 'target' => 7, 'officers' => ['Bourne']]
];

foreach ($edges as $k => &$v) {
    $next_slice = array_slice($edges, array_search($k, array_keys($edges)) + 1);
    foreach ($next_slice as $key => $item) {
        if ($item['source'] == $v['source'] && $item['target'] == $v['target']) {
            $v['officers'] = array_merge($v['officers'], $item['officers']);
            unset($edges[$k + $key + 1]);
        }
    }
}

print_r($edges);

---

使用

数组搜索

，

数组键

，

数组切片

和

数组合并

功能的解决方案：

// an exemplary array
$edges = [
    0 => ['source' => 3, 'target' => 4, 'officers' => ['Aberdeen Asset Management PLC']],
    1 => ['source' => 3, 'target' => 4, 'officers' => ['whatever']],
    3 => ['source' => 4, 'target' => 7, 'officers' => ['Jason']],
    4 => ['source' => 4, 'target' => 5, 'officers' => ['John']],
    5 => ['source' => 4, 'target' => 7, 'officers' => ['Bourne']]
];

foreach ($edges as $k => &$v) {
    $next_slice = array_slice($edges, array_search($k, array_keys($edges)) + 1);
    foreach ($next_slice as $key => $item) {
        if ($item['source'] == $v['source'] && $item['target'] == $v['target']) {
            $v['officers'] = array_merge($v['officers'], $item['officers']);
            unset($edges[$k + $key + 1]);
        }
    }
}

print_r($edges);

---

@u\u mulder如果他告诉你他需要什么，然后复制粘贴，那就容易多了。毕竟，你这样做是免费的，因为你没有什么比编写代码更好的事情，这样其他人就可以从中获益，为什么不在你做的时候放纵一下他们的愿望呢？：）@嗯，如果他告诉你他需要什么，然后只是复制粘贴，那就容易多了。毕竟，你这样做是免费的，因为你没有什么比编写代码更好的事情，这样其他人就可以从中获益，为什么不在你做的时候放纵一下他们的愿望呢？：）