php - 处理复杂的数据结构值

Question

我有两个坐标值。我正在计算每个坐标的欧几里得距离。我想在这里应用 K- 最近邻域算法。

1. 对于每个坐标，得到与所有其他点的欧氏距离

2. 选择最接近最大点的点(从所有点中，选择最接近最大坐标数的点)

       $result2 = mysqli_query($con,"SELECT pcount, ncount from test");
       $result1 = mysqli_query($con,"SELECT pcount, ncount from test");
       $i = 0;
                   $k = 0;
       $min = 0;
       while ($row1 = @mysqli_fetch_array($result1))
       {
           $pcount = $row1['pcount'];
           $ncount = $row1['ncount'];
           echo "pcount is $pcount<br/>";
           echo "ncount is $ncount<br/></br>";
           $a[$i] = $pcount ;
           $b[$i] = $pcount ;
           $j = 0;
           $result2 = mysqli_query($con,"SELECT pcount, ncount from test");
           while ($row2 = @mysqli_fetch_array($result2))
           {
           //echo "j is $j <br/>";
           $a[$j] = $row2['pcount'];   
           $b[$j] = $row2['ncount'];
           $diff = ($a[$i] - $a[$j])^2 + ($b[$i] - $b[$j])^2;
           if( $diff != 0)
               $diff = sqrt( $diff );
               $j= $j + 1;
               echo "$diff <br>";
                           arr[$k] = $diff;
                           $k = $k + 1;
           }               
           //echo "i is $i <br/>";
               $i = $i + 1;    
       }
   

这里，arr[$k] 包含所有点的距离差。我怎样才能得到最接近这里最大其他点的点。可能有超过 1 个这样的点。

Answer 1

我建议使用下面的方法。

0) 将数据库数据提取到二维数组中如下所示:

array(
    array($x1,$y1),
    array($x2,$y2),
    ...
)

1) 为每个点到所有其他点的(平方)距离之和构建数组存储。

2) 找到该数组中的最低值(最小值)。

3) 获取所有具有该值的点。

所以代码应该是这样的:

// $a_points = array( array($x1,$y1), array($x2,$y2), ... ); // array of point coordinates

$a_sum_dists = array();
foreach ($a_points as $i => $pt1) {
    list($x1, $y1) = $pt1;
    $sum = 0;
    foreach ($a_points as $j => $pt2) {
        if ($j == $i) continue;
        list($x2, $y2) = $pt2;
        $sum += pow($x2- $x1, 2) + pow($y2- $y1, 2);
    }
    $a_sum_dists[$i] = $sum;
}
$min_sum = min($a_sum_dists);
$a_result_points = array_intersect_key(
    $a_points, 
    array_filter($a_sum_dists, function ($v) use ($min_sum) {
        return $v == $min_sum;
    })
);

Demo