Question

我使用Jpgraph（http://www.jpgraph.com）创建了一个图表。

Jpgraph有一个样条线类，可以平滑线图上的线条。但它没有像我预期的那样工作。如您所见，该线的最高峰值高于实际标记（红色方块）。星期五晚上10点是2.0，但它似乎高于周五早上6点左右。

一个糟糕的笔记本电脑触摸板绘制了我期望的例子如下:)

我已设法将该类追踪到jpgraph_regstat.php。知道如何让峰值的中心成为实际的标记吗？

//------------------------------------------------------------------------
// CLASS Spline
// Create a new data array from an existing data array but with more points.
// The new points are interpolated using a cubic spline algorithm
//------------------------------------------------------------------------
class Spline {
    // 3:rd degree polynom approximation

    private $xdata,$ydata;   // Data vectors
    private $y2;   // 2:nd derivate of ydata
    private $n=0;

    function __construct($xdata,$ydata) {
        $this->y2 = array();
        $this->xdata = $xdata;
        $this->ydata = $ydata;

        $n = count($ydata);
        $this->n = $n;
        if( $this->n !== count($xdata) ) {
            JpGraphError::RaiseL(19001);
            //('Spline: Number of X and Y coordinates must be the same');
        }

        // Natural spline 2:derivate == 0 at endpoints
        $this->y2[0]    = 0.0;
        $this->y2[$n-1] = 0.0;
        $delta[0] = 0.0;

        // Calculate 2:nd derivate
        for($i=1; $i < $n-1; ++$i) {
            $d = ($xdata[$i+1]-$xdata[$i-1]);
            if( $d == 0  ) {
                JpGraphError::RaiseL(19002);
                //('Invalid input data for spline. Two or more consecutive input X-values are equal. Each input X-value must differ since from a mathematical point of view it must be a one-to-one mapping, i.e. each X-value must correspond to exactly one Y-value.');
            }
            $s = ($xdata[$i]-$xdata[$i-1])/$d;
            $p = $s*$this->y2[$i-1]+2.0;
            $this->y2[$i] = ($s-1.0)/$p;
            $delta[$i] = ($ydata[$i+1]-$ydata[$i])/($xdata[$i+1]-$xdata[$i]) -
            ($ydata[$i]-$ydata[$i-1])/($xdata[$i]-$xdata[$i-1]);
            $delta[$i] = (6.0*$delta[$i]/($xdata[$i+1]-$xdata[$i-1])-$s*$delta[$i-1])/$p;
        }

        // Backward substitution
        for( $j=$n-2; $j >= 0; --$j ) {
            $this->y2[$j] = $this->y2[$j]*$this->y2[$j+1] + $delta[$j];
        }
    }

    // Return the two new data vectors
    function Get($num=50) {
        $n = $this->n ;
        $step = ($this->xdata[$n-1]-$this->xdata[0]) / ($num-1);
        $xnew=array();
        $ynew=array();
        $xnew[0] = $this->xdata[0];
        $ynew[0] = $this->ydata[0];
        for( $j=1; $j < $num; ++$j ) {
            $xnew[$j] = $xnew[0]+$j*$step;
            $ynew[$j] = $this->Interpolate($xnew[$j]);
        }
        return array($xnew,$ynew);
    }

    // Return a single interpolated Y-value from an x value
    function Interpolate($xpoint) {

        $max = $this->n-1;
        $min = 0;

        // Binary search to find interval
        while( $max-$min > 1 ) {
            $k = ($max+$min) / 2;
            if( $this->xdata[$k] > $xpoint )
            $max=$k;
            else
            $min=$k;
        }

        // Each interval is interpolated by a 3:degree polynom function
        $h = $this->xdata[$max]-$this->xdata[$min];

        if( $h == 0  ) {
            JpGraphError::RaiseL(19002);
            //('Invalid input data for spline. Two or more consecutive input X-values are equal. Each input X-value must differ since from a mathematical point of view it must be a one-to-one mapping, i.e. each X-value must correspond to exactly one Y-value.');
        }


        $a = ($this->xdata[$max]-$xpoint)/$h;
        $b = ($xpoint-$this->xdata[$min])/$h;
        return $a*$this->ydata[$min]+$b*$this->ydata[$max]+
        (($a*$a*$a-$a)*$this->y2[$min]+($b*$b*$b-$b)*$this->y2[$max])*($h*$h)/6.0;
    }
}

可以在此处找到使用样条线的代码示例（http://jpgraph.net/download/manuals/chunkhtml/example_src/splineex1.html）

Answer 1

如果您真的想使用样条线来插值数据值，并且希望最大限度地减少实际样本（数据点）之上和之下的偏差，那么您将需要更多样本。您可以通过样本之间的线性插值生成假/人工数据点，然后重新生成样条曲线来获得它们（在这种情况下，您最终会得到更像触摸板绘图的东西）。但是如果你试图生成一个“平滑”图形，它可以最大限度地减少某种与现实的平均偏差，那么你可能需要使用样条以外的其他东西。

更新图表上的样条线类，以便最高峰是标记位置 - JPGraph

1 个答案: