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php數(shù)組求差集問(wèn)題

最近在寫(xiě)一個(gè)項(xiàng)目 項(xiàng)目中 涉及到數(shù)組差集問(wèn)題 代碼如下

<?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個(gè)數(shù):'.count($arr1).'<br /><br />';
echo 'arr1集合: ';
var_dump($arr1);
echo '<hr />';

echo 'arr2個(gè)數(shù):'.count($arr2).'<br /><br />';
echo 'arr2集合: ';
var_dump($arr2);
echo '<hr />';

//這是我求差集的方法
foreach($arr1 as $k=>$v) if(in_array($v, $arr2)) unset($arr1[$k]);

echo '差集個(gè)數(shù):'.count($arr1).'<br /><br />';
echo '差集集合：';
var_dump($arr1);

因?yàn)橛兄貜?fù)的數(shù)據(jù) 這樣求出來(lái)之后 差集的個(gè)數(shù) + $arr2 的個(gè)數(shù) 不等于 $arr1 的個(gè)數(shù)

問(wèn) ： 怎么樣求差集才能  
 差集的個(gè)數(shù) + $arr2 的個(gè)數(shù) = $arr1 的個(gè)數(shù)  
 而且最后求出的差集 集合也包含那兩個(gè)重復(fù)的數(shù)組元素 "cardId" => 177561410 和 "cardId" => 1077060068
回答
編輯回答
溫衫

如果是按需求（不是差集）來(lái)實(shí)現(xiàn)，可以將in_array換成array_search，如果查找成功，該函數(shù)會(huì)返回命中的key，如果在arr2中能找到，在刪除$arr1[$k]時(shí)，同時(shí)刪除$arr2[$idx]就行了，如果要保留$arr2，就copy一個(gè)數(shù)組來(lái)操作
foreach($arr1 as $k=>$v){

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

}

2017年11月7日 20:10
編輯回答
念初

樓主1對(duì)于2的差集，就是1中不包含2的部分。不包含2的只有兩個(gè)，這就是差集。如果說(shuō)你需要查處的差集大于2，那和差集概念不符，也就不是差集的范疇了

2017年10月22日 05:11