php数组求差集问题

Question

最近在写一个项目 项目中 涉及到数组差集问题 代码如下 <?php $arr1 = [ [ "cardId" => 1010284722, "beforeAmounts" => 100, "afterAmounts" => 20, "diffAmounts" => -80, "diffStatus" => 1 ], [ "cardId" => 1010284722, "beforeAmounts" => 100, "afterAmounts" => 200, "diffAmounts" => 100, "diffStatus" => 0 ], [ "cardId" => 177561410, "beforeAmounts" => 2000, "afterAmounts" => 1000, "diffAmounts" => -1000, "diffStatus" => 1 ], [ "cardId" => 177561410, "beforeAmounts" => 2000, "afterAmounts" => 1000, "diffAmounts" => -1000, "diffStatus" => 1 ], [ "cardId" => 1077060068, "beforeAmounts" => 789, "afterAmounts" => 100, "diffAmounts" => -689, "diffStatus" => 1 ], [ "cardId" => 1077060068, "beforeAmounts" => 789, "afterAmounts" => 100, "diffAmounts" => -689, "diffStatus" => 1 ] ]; $arr2 = [ [ "cardId" => 177561410, "beforeAmounts" => 2000, "afterAmounts" => 1000, "diffAmounts" => -1000, "diffStatus" => 1 ], [ "cardId" => 1077060068, "beforeAmounts" => 789, "afterAmounts" => 100, "diffAmounts" => -689, "diffStatus" => 1 ] ]; echo 'arr1个数:'.count($arr1).'<br /><br />'; echo 'arr1集合: '; var_dump($arr1); echo '<hr />'; echo 'arr2个数:'.count($arr2).'<br /><br />'; echo 'arr2集合: '; var_dump($arr2); echo '<hr />'; //这是我求差集的方法 foreach($arr1 as $k=>$v) if(in_array($v, $arr2)) unset($arr1[$k]); echo '差集个数:'.count($arr1).'<br /><br />'; echo '差集集合：'; var_dump($arr1); 因为有重复的数据 这样求出来之后 差集的个数 + $arr2 的个数 不等于 $arr1 的个数 问 ： 怎么样求差集才能 差集的个数 + $arr2 的个数 = $arr1 的个数 而且最后求出的差集 集合也包含那两个重复的数组元素 "cardId" => 177561410 和 "cardId" => 1077060068

Answer 1

如果是按需求（不是差集）来实现，可以将in_array换成array_search，如果查找成功，该函数会返回命中的key，如果在arr2中能找到，在删除$arr1[$k]时，同时删除$arr2[$idx]就行了，如果要保留$arr2，就copy一个数组来操作
foreach($arr1 as $k=>$v){

$idx = array_search($v,$arr2);
if($idx !== false){ //找到了
    unset($arr1[$k]);
    unset($arr2[$idx]);
}

}