Question

我有两组这样的数组，例如。

$Arr1['uid'][]='user 1'; $Arr1['weight'][]=1;
$Arr1['uid'][]='user 2'; $Arr1['weight'][]=10;
$Arr1['uid'][]='user 3'; $Arr1['weight'][]=5;

$Arr2['uid'][]='user 1'; $Arr2['weight'][]=3;
$Arr2['uid'][]='user 4'; $Arr2['weight'][]=20;
$Arr2['uid'][]='user 5'; $Arr2['weight'][]=15;
$Arr2['uid'][]='user 2'; $Arr2['weight'][]=2;

当然，两个阵列的大小可能不同。 $ Arr1 的系数为0.7， $ Arr2 的系数为0.3。我需要计算以下公式

$result=$Arr1['weight'][$index]*$Arr1Coeff+$Arr2['weight'][$index]*$Arr2Coeff;

其中$Arr1['uid']=$Arr2['uid']。因此，当$Arr1['uid']中不存在$Arr2时，我们需要省略$Arr2，反之亦然。
而且，这是我现在使用的算法。

foreach($Arr1['uid'] as $index=>$arr1_uid){
    $pos=array_search($arr1_uid, $Arr2['uid']);
    if ($pos===false){
        $result=$Arr1['weight'][$index]*$Arr1Coeff;
        echo "<br>$arr1_uid has not found and RES=".$result;
    }else{
        $result=$Arr1['weight'][$index]*$Arr1Coeff+$Arr2['weight'][$pos]*$Arr2Coeff;
        echo "<br>$arr1_uid has found on $pos and RES=".$result;
    }
}

foreach($Arr2['uid'] as $index=>$arr2_uid){
    if (!in_array($arr2_uid, $Arr1['uid'])){
        $result=$Arr2['weight'][$index]*$Arr2Coeff;
        echo "<br>$arr2_uid has not found and RES=".$result;
    }else{
        echo "<br>$arr2_uid has found somewhere";
    }
}

问题是如何优化此算法？你能为这个问题提供更好的解决方案吗？
谢谢。

Answer 1

由于您整理数组的方式，您可以使用array_combine($keys, $values)使用$Arr1中的键和来自{{$Arr2的值将['uid']和['weight']汇集到关联数组中1}}。使用关联数组可以简化计算：

$combi1 = array_combine($Arr1['uid'], $Arr1['weight']);
$combi2 = array_combine($Arr2['uid'], $Arr2['weight']);

// loop through the keys from both arrays
foreach (array_keys($combi1+$combi2) as $uid) {
    // use the value from $combi1, or 0 if it isn't set
    $value1 = isset($combi1[$uid]) ? $combi1[$uid] : 0;
    // use the value from $combi2, or 0 if it isn't set
    $value2 = isset($combi2[$uid]) ? $combi2[$uid] : 0;
    // calculate our final weight
    $result = $value1 * $Arr1Coeff + $value2 * $Arr2Coeff;
    echo "<br>$uid final weight: ".$result."\n";
}

结果比较

您的代码：

user 1 has found on 0 and RES=1.6
user 2 has found on 3 and RES=7.6
user 3 has not found and RES=3.5
user 1 has found somewhere
user 4 has not found and RES=6
user 5 has not found and RES=4.5
user 2 has found somewhere

我的代码：

user 1 final weight: 1.6
user 2 final weight: 7.6
user 3 final weight: 3.5
user 4 final weight: 6
user 5 final weight: 4.5

Answer 2

如果您将用户用作数组键会更容易。像这样：

$Arr1['user 1'] => array('weight'=>1);
$Arr1['user 2'] => array('weight'=>10);
...

然后你可以使用array_diff_assoc和array_intersect_assoc来找出哪些元素在哪个元素中而不在另一个元素中。

Answer 3

您可以在线性O(log(n)) O(n)的{{1}}复杂性内容中阻止二进制搜索。但您必须先从此创建树或在array_search中对此数组进行排序。

有关二进制搜索的更多信息，请访问：

Answer 4

首先，您应该将数组初始化为关联数组。计算会更容易。

如何优化此算法？

4 个答案:

结果比较