Question

对于绘图应用程序，我将鼠标移动坐标保存到数组，然后使用lineTo绘制它们。生成的线条不平滑。如何在所有聚集点之间生成单条曲线？

我用谷歌搜索但我只找到3个绘制线的函数：对于2个样本点，只需使用lineTo。对于3个样本点，quadraticCurveTo，对于4个样本点，bezierCurveTo。

（我尝试在阵列中每4个点绘制一个bezierCurveTo，但这会导致每4个采样点扭结，而不是连续的平滑曲线。）

如何编写一个函数来绘制一个包含5个样本点的平滑曲线？

Answer 1

将后续采样点与不相交的“curveTo”类型函数连接在一起的问题是曲线相遇的位置不平滑。这是因为两条曲线共享一个终点，但受完全不相交的控制点的影响。一种解决方案是“曲线化”接下来的两个后续采样点之间的中点。使用这些新的插值点连接曲线可以在端点处进行平滑过渡（一次迭代的终点是下一次迭代的控制点。）换句话说，两条离散曲线具有现在更多的共同点。

此解决方案摘自“Foundation ActionScript 3.0动画：让事情发生变化”一书。第95页 - 渲染技术：创建多条曲线。

注意：这个解决方案并没有实际绘制每个点，这是我的问题的标题（相反，它通过采样点逼近曲线，但从不经过采样点），但是出于我的目的（绘图）应用程序），它对我来说足够好，在视觉上你无法分辨出来。是一个解决所有样本点的解决方案，但它要复杂得多（参见http://www.cartogrammar.com/blog/actionscript-curves-update/）

以下是近似方法的绘图代码：

// move to the first point
   ctx.moveTo(points[0].x, points[0].y);


   for (i = 1; i < points.length - 2; i ++)
   {
      var xc = (points[i].x + points[i + 1].x) / 2;
      var yc = (points[i].y + points[i + 1].y) / 2;
      ctx.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
   }
 // curve through the last two points
 ctx.quadraticCurveTo(points[i].x, points[i].y, points[i+1].x,points[i+1].y);

Answer 2

有点晚了，但记录了。

使用cardinal splines（也称为规范样条曲线）绘制穿过点的平滑曲线，可以获得平滑的线条。

我为画布制作了这个功能 - 它分为三个功能，以增加多功能性。主包装函数如下所示：

function drawCurve(ctx, ptsa, tension, isClosed, numOfSegments, showPoints) {

    showPoints  = showPoints ? showPoints : false;

    ctx.beginPath();

    drawLines(ctx, getCurvePoints(ptsa, tension, isClosed, numOfSegments));

    if (showPoints) {
        ctx.stroke();
        ctx.beginPath();
        for(var i=0;i<ptsa.length-1;i+=2) 
                ctx.rect(ptsa[i] - 2, ptsa[i+1] - 2, 4, 4);
    }
}

绘制曲线时，数组的x，y点顺序为：x1,y1, x2,y2, ...xn,yn。

像这样使用：

var myPoints = [10,10, 40,30, 100,10]; //minimum two points
var tension = 1;

drawCurve(ctx, myPoints); //default tension=0.5
drawCurve(ctx, myPoints, tension);

上述函数调用两个子函数，一个用于计算平滑点。这将返回一个带有新点的数组 - 这是计算平滑点的核心函数：

function getCurvePoints(pts, tension, isClosed, numOfSegments) {

    // use input value if provided, or use a default value   
    tension = (typeof tension != 'undefined') ? tension : 0.5;
    isClosed = isClosed ? isClosed : false;
    numOfSegments = numOfSegments ? numOfSegments : 16;

    var _pts = [], res = [],    // clone array
        x, y,           // our x,y coords
        t1x, t2x, t1y, t2y, // tension vectors
        c1, c2, c3, c4,     // cardinal points
        st, t, i;       // steps based on num. of segments

    // clone array so we don't change the original
    //
    _pts = pts.slice(0);

    // The algorithm require a previous and next point to the actual point array.
    // Check if we will draw closed or open curve.
    // If closed, copy end points to beginning and first points to end
    // If open, duplicate first points to befinning, end points to end
    if (isClosed) {
        _pts.unshift(pts[pts.length - 1]);
        _pts.unshift(pts[pts.length - 2]);
        _pts.unshift(pts[pts.length - 1]);
        _pts.unshift(pts[pts.length - 2]);
        _pts.push(pts[0]);
        _pts.push(pts[1]);
    }
    else {
        _pts.unshift(pts[1]);   //copy 1. point and insert at beginning
        _pts.unshift(pts[0]);
        _pts.push(pts[pts.length - 2]); //copy last point and append
        _pts.push(pts[pts.length - 1]);
    }

    // ok, lets start..

    // 1. loop goes through point array
    // 2. loop goes through each segment between the 2 pts + 1e point before and after
    for (i=2; i < (_pts.length - 4); i+=2) {
        for (t=0; t <= numOfSegments; t++) {

            // calc tension vectors
            t1x = (_pts[i+2] - _pts[i-2]) * tension;
            t2x = (_pts[i+4] - _pts[i]) * tension;

            t1y = (_pts[i+3] - _pts[i-1]) * tension;
            t2y = (_pts[i+5] - _pts[i+1]) * tension;

            // calc step
            st = t / numOfSegments;

            // calc cardinals
            c1 =   2 * Math.pow(st, 3)  - 3 * Math.pow(st, 2) + 1; 
            c2 = -(2 * Math.pow(st, 3)) + 3 * Math.pow(st, 2); 
            c3 =       Math.pow(st, 3)  - 2 * Math.pow(st, 2) + st; 
            c4 =       Math.pow(st, 3)  -     Math.pow(st, 2);

            // calc x and y cords with common control vectors
            x = c1 * _pts[i]    + c2 * _pts[i+2] + c3 * t1x + c4 * t2x;
            y = c1 * _pts[i+1]  + c2 * _pts[i+3] + c3 * t1y + c4 * t2y;

            //store points in array
            res.push(x);
            res.push(y);

        }
    }

    return res;
}

实际绘制点作为平滑曲线（或任何其他分段线，只要你有一个x，y数组）：

function drawLines(ctx, pts) {
    ctx.moveTo(pts[0], pts[1]);
    for(i=2;i<pts.length-1;i+=2) ctx.lineTo(pts[i], pts[i+1]);
}

var ctx = document.getElementById("c").getContext("2d");


function drawCurve(ctx, ptsa, tension, isClosed, numOfSegments, showPoints) {

  ctx.beginPath();

  drawLines(ctx, getCurvePoints(ptsa, tension, isClosed, numOfSegments));
  
  if (showPoints) {
    ctx.beginPath();
    for(var i=0;i<ptsa.length-1;i+=2) 
      ctx.rect(ptsa[i] - 2, ptsa[i+1] - 2, 4, 4);
  }

  ctx.stroke();
}


var myPoints = [10,10, 40,30, 100,10, 200, 100, 200, 50, 250, 120]; //minimum two points
var tension = 1;

drawCurve(ctx, myPoints); //default tension=0.5
drawCurve(ctx, myPoints, tension);


function getCurvePoints(pts, tension, isClosed, numOfSegments) {

  // use input value if provided, or use a default value	 
  tension = (typeof tension != 'undefined') ? tension : 0.5;
  isClosed = isClosed ? isClosed : false;
  numOfSegments = numOfSegments ? numOfSegments : 16;

  var _pts = [], res = [],	// clone array
      x, y,			// our x,y coords
      t1x, t2x, t1y, t2y,	// tension vectors
      c1, c2, c3, c4,		// cardinal points
      st, t, i;		// steps based on num. of segments

  // clone array so we don't change the original
  //
  _pts = pts.slice(0);

  // The algorithm require a previous and next point to the actual point array.
  // Check if we will draw closed or open curve.
  // If closed, copy end points to beginning and first points to end
  // If open, duplicate first points to befinning, end points to end
  if (isClosed) {
    _pts.unshift(pts[pts.length - 1]);
    _pts.unshift(pts[pts.length - 2]);
    _pts.unshift(pts[pts.length - 1]);
    _pts.unshift(pts[pts.length - 2]);
    _pts.push(pts[0]);
    _pts.push(pts[1]);
  }
  else {
    _pts.unshift(pts[1]);	//copy 1. point and insert at beginning
    _pts.unshift(pts[0]);
    _pts.push(pts[pts.length - 2]);	//copy last point and append
    _pts.push(pts[pts.length - 1]);
  }

  // ok, lets start..

  // 1. loop goes through point array
  // 2. loop goes through each segment between the 2 pts + 1e point before and after
  for (i=2; i < (_pts.length - 4); i+=2) {
    for (t=0; t <= numOfSegments; t++) {

      // calc tension vectors
      t1x = (_pts[i+2] - _pts[i-2]) * tension;
      t2x = (_pts[i+4] - _pts[i]) * tension;

      t1y = (_pts[i+3] - _pts[i-1]) * tension;
      t2y = (_pts[i+5] - _pts[i+1]) * tension;

      // calc step
      st = t / numOfSegments;

      // calc cardinals
      c1 =   2 * Math.pow(st, 3) 	- 3 * Math.pow(st, 2) + 1; 
      c2 = -(2 * Math.pow(st, 3)) + 3 * Math.pow(st, 2); 
      c3 = 	   Math.pow(st, 3)	- 2 * Math.pow(st, 2) + st; 
      c4 = 	   Math.pow(st, 3)	- 	  Math.pow(st, 2);

      // calc x and y cords with common control vectors
      x = c1 * _pts[i]	+ c2 * _pts[i+2] + c3 * t1x + c4 * t2x;
      y = c1 * _pts[i+1]	+ c2 * _pts[i+3] + c3 * t1y + c4 * t2y;

      //store points in array
      res.push(x);
      res.push(y);

    }
  }

  return res;
}

function drawLines(ctx, pts) {
  ctx.moveTo(pts[0], pts[1]);
  for(i=2;i<pts.length-1;i+=2) ctx.lineTo(pts[i], pts[i+1]);
}

canvas { border: 1px solid red; }

<canvas id="c"><canvas>

结果如下：

Example pix

您可以轻松扩展画布，以便您可以像这样调用它：

ctx.drawCurve(myPoints);

将以下内容添加到javascript：

if (CanvasRenderingContext2D != 'undefined') {
    CanvasRenderingContext2D.prototype.drawCurve = 
        function(pts, tension, isClosed, numOfSegments, showPoints) {
       drawCurve(this, pts, tension, isClosed, numOfSegments, showPoints)}
}

您可以在NPM（npm i cardinal-spline-js）或GitLab上找到更优化的版本。

Answer 3

作为Daniel Howard points out，Rob Spencer在http://scaledinnovation.com/analytics/splines/aboutSplines.html描述了您想要的内容。

这是一个互动演示：http://jsbin.com/ApitIxo/2/

这是jsbin关闭时的一个片段。

<!DOCTYPE html>
    <html>
      <head>
        <meta charset=utf-8 />
        <title>Demo smooth connection</title>
      </head>
      <body>
        <div id="display">
          Click to build a smooth path. 
          (See Rob Spencer's <a href="http://scaledinnovation.com/analytics/splines/aboutSplines.html">article</a>)
          <br><label><input type="checkbox" id="showPoints" checked> Show points</label>
          <br><label><input type="checkbox" id="showControlLines" checked> Show control lines</label>
          <br>
          <label>
            <input type="range" id="tension" min="-1" max="2" step=".1" value=".5" > Tension <span id="tensionvalue">(0.5)</span>
          </label>
        <div id="mouse"></div>
        </div>
        <canvas id="canvas"></canvas>
        <style>
          html { position: relative; height: 100%; width: 100%; }
          body { position: absolute; left: 0; right: 0; top: 0; bottom: 0; } 
          canvas { outline: 1px solid red; }
          #display { position: fixed; margin: 8px; background: white; z-index: 1; }
        </style>
        <script>
          function update() {
            $("tensionvalue").innerHTML="("+$("tension").value+")";
            drawSplines();
          }
          $("showPoints").onchange = $("showControlLines").onchange = $("tension").onchange = update;
      
          // utility function
          function $(id){ return document.getElementById(id); }
          var canvas=$("canvas"), ctx=canvas.getContext("2d");

          function setCanvasSize() {
            canvas.width = parseInt(window.getComputedStyle(document.body).width);
            canvas.height = parseInt(window.getComputedStyle(document.body).height);
          }
          window.onload = window.onresize = setCanvasSize();
      
          function mousePositionOnCanvas(e) {
            var el=e.target, c=el;
            var scaleX = c.width/c.offsetWidth || 1;
            var scaleY = c.height/c.offsetHeight || 1;
          
            if (!isNaN(e.offsetX)) 
              return { x:e.offsetX*scaleX, y:e.offsetY*scaleY };
          
            var x=e.pageX, y=e.pageY;
            do {
              x -= el.offsetLeft;
              y -= el.offsetTop;
              el = el.offsetParent;
            } while (el);
            return { x: x*scaleX, y: y*scaleY };
          }
      
          canvas.onclick = function(e){
            var p = mousePositionOnCanvas(e);
            addSplinePoint(p.x, p.y);
          };
      
          function drawPoint(x,y,color){
            ctx.save();
            ctx.fillStyle=color;
            ctx.beginPath();
            ctx.arc(x,y,3,0,2*Math.PI);
            ctx.fill()
            ctx.restore();
          }
          canvas.onmousemove = function(e) {
            var p = mousePositionOnCanvas(e);
            $("mouse").innerHTML = p.x+","+p.y;
          };
      
          var pts=[]; // a list of x and ys

          // given an array of x,y's, return distance between any two,
          // note that i and j are indexes to the points, not directly into the array.
          function dista(arr, i, j) {
            return Math.sqrt(Math.pow(arr[2*i]-arr[2*j], 2) + Math.pow(arr[2*i+1]-arr[2*j+1], 2));
          }

          // return vector from i to j where i and j are indexes pointing into an array of points.
          function va(arr, i, j){
            return [arr[2*j]-arr[2*i], arr[2*j+1]-arr[2*i+1]]
          }
      
          function ctlpts(x1,y1,x2,y2,x3,y3) {
            var t = $("tension").value;
            var v = va(arguments, 0, 2);
            var d01 = dista(arguments, 0, 1);
            var d12 = dista(arguments, 1, 2);
            var d012 = d01 + d12;
            return [x2 - v[0] * t * d01 / d012, y2 - v[1] * t * d01 / d012,
                    x2 + v[0] * t * d12 / d012, y2 + v[1] * t * d12 / d012 ];
          }

          function addSplinePoint(x, y){
            pts.push(x); pts.push(y);
            drawSplines();
          }
          function drawSplines() {
            clear();
            cps = []; // There will be two control points for each "middle" point, 1 ... len-2e
            for (var i = 0; i < pts.length - 2; i += 1) {
              cps = cps.concat(ctlpts(pts[2*i], pts[2*i+1], 
                                      pts[2*i+2], pts[2*i+3], 
                                      pts[2*i+4], pts[2*i+5]));
            }
            if ($("showControlLines").checked) drawControlPoints(cps);
            if ($("showPoints").checked) drawPoints(pts);
    
            drawCurvedPath(cps, pts);
 
          }
          function drawControlPoints(cps) {
            for (var i = 0; i < cps.length; i += 4) {
              showPt(cps[i], cps[i+1], "pink");
              showPt(cps[i+2], cps[i+3], "pink");
              drawLine(cps[i], cps[i+1], cps[i+2], cps[i+3], "pink");
            } 
          }
      
          function drawPoints(pts) {
            for (var i = 0; i < pts.length; i += 2) {
              showPt(pts[i], pts[i+1], "black");
            } 
          }
      
          function drawCurvedPath(cps, pts){
            var len = pts.length / 2; // number of points
            if (len < 2) return;
            if (len == 2) {
              ctx.beginPath();
              ctx.moveTo(pts[0], pts[1]);
              ctx.lineTo(pts[2], pts[3]);
              ctx.stroke();
            }
            else {
              ctx.beginPath();
              ctx.moveTo(pts[0], pts[1]);
              // from point 0 to point 1 is a quadratic
              ctx.quadraticCurveTo(cps[0], cps[1], pts[2], pts[3]);
              // for all middle points, connect with bezier
              for (var i = 2; i < len-1; i += 1) {
                // console.log("to", pts[2*i], pts[2*i+1]);
                ctx.bezierCurveTo(
                  cps[(2*(i-1)-1)*2], cps[(2*(i-1)-1)*2+1],
                  cps[(2*(i-1))*2], cps[(2*(i-1))*2+1],
                  pts[i*2], pts[i*2+1]);
              }
              ctx.quadraticCurveTo(
                cps[(2*(i-1)-1)*2], cps[(2*(i-1)-1)*2+1],
                pts[i*2], pts[i*2+1]);
              ctx.stroke();
            }
          }
          function clear() {
            ctx.save();
            // use alpha to fade out
            ctx.fillStyle = "rgba(255,255,255,.7)"; // clear screen
            ctx.fillRect(0,0,canvas.width,canvas.height);
            ctx.restore();
          }
      
          function showPt(x,y,fillStyle) {
            ctx.save();
            ctx.beginPath();
            if (fillStyle) {
              ctx.fillStyle = fillStyle;
            }
            ctx.arc(x, y, 5, 0, 2*Math.PI);
            ctx.fill();
            ctx.restore();
          }

          function drawLine(x1, y1, x2, y2, strokeStyle){
            ctx.beginPath();
            ctx.moveTo(x1, y1);
            ctx.lineTo(x2, y2);
            if (strokeStyle) {
              ctx.save();
              ctx.strokeStyle = strokeStyle;
              ctx.stroke();
              ctx.restore();
            }
            else {
              ctx.save();
              ctx.strokeStyle = "pink";
              ctx.stroke();
              ctx.restore();
            }
          }

        </script>


      </body>
    </html>

Answer 4

尝试使用KineticJS - 您可以使用一系列点定义样条曲线。这是一个例子：

旧网址：http://www.html5canvastutorials.com/kineticjs/html5-canvas-kineticjs-spline-tutorial/

请参阅存档网址：https://web.archive.org/web/20141204030628/http://www.html5canvastutorials.com/kineticjs/html5-canvas-kineticjs-spline-tutorial/

Answer 5

我决定添加，而不是将我的解决方案发布到另一篇文章。以下是我构建的解决方案，可能不完美，但到目前为止输出都很好。

重要提示：它将通过所有点！

如果您有任何想法，让它变得更好，请与我分享。感谢。

以下是之前的比较：

将此代码保存为HTML以进行测试。

<!DOCTYPE html>
<html>
<body>
    <canvas id="myCanvas" width="1200" height="700" style="border:1px solid #d3d3d3;">Your browser does not support the HTML5 canvas tag.</canvas>
    <script>
        var cv = document.getElementById("myCanvas");
        var ctx = cv.getContext("2d");

        function gradient(a, b) {
            return (b.y-a.y)/(b.x-a.x);
        }

        function bzCurve(points, f, t) {
            //f = 0, will be straight line
            //t suppose to be 1, but changing the value can control the smoothness too
            if (typeof(f) == 'undefined') f = 0.3;
            if (typeof(t) == 'undefined') t = 0.6;

            ctx.beginPath();
            ctx.moveTo(points[0].x, points[0].y);

            var m = 0;
            var dx1 = 0;
            var dy1 = 0;

            var preP = points[0];
            for (var i = 1; i < points.length; i++) {
                var curP = points[i];
                nexP = points[i + 1];
                if (nexP) {
                    m = gradient(preP, nexP);
                    dx2 = (nexP.x - curP.x) * -f;
                    dy2 = dx2 * m * t;
                } else {
                    dx2 = 0;
                    dy2 = 0;
                }
                ctx.bezierCurveTo(preP.x - dx1, preP.y - dy1, curP.x + dx2, curP.y + dy2, curP.x, curP.y);
                dx1 = dx2;
                dy1 = dy2;
                preP = curP;
            }
            ctx.stroke();
        }

        // Generate random data
        var lines = [];
        var X = 10;
        var t = 40; //to control width of X
        for (var i = 0; i < 100; i++ ) {
            Y = Math.floor((Math.random() * 300) + 50);
            p = { x: X, y: Y };
            lines.push(p);
            X = X + t;
        }

        //draw straight line
        ctx.beginPath();
        ctx.setLineDash([5]);
        ctx.lineWidth = 1;
        bzCurve(lines, 0, 1);

        //draw smooth line
        ctx.setLineDash([0]);
        ctx.lineWidth = 2;
        ctx.strokeStyle = "blue";
        bzCurve(lines, 0.3, 1);
    </script>
</body>
</html>

Answer 6

我发现这很好用

function drawCurve(points, tension) {
    ctx.beginPath();
    ctx.moveTo(points[0].x, points[0].y);

    var t = (tension != null) ? tension : 1;
    for (var i = 0; i < points.length - 1; i++) {
        var p0 = (i > 0) ? points[i - 1] : points[0];
        var p1 = points[i];
        var p2 = points[i + 1];
        var p3 = (i != points.length - 2) ? points[i + 2] : p2;

        var cp1x = p1.x + (p2.x - p0.x) / 6 * t;
        var cp1y = p1.y + (p2.y - p0.y) / 6 * t;

        var cp2x = p2.x - (p3.x - p1.x) / 6 * t;
        var cp2y = p2.y - (p3.y - p1.y) / 6 * t;

        ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, p2.x, p2.y);
    }
    ctx.stroke();
}

Answer 7

此代码最适合我：

this.context.beginPath();
this.context.moveTo(data[0].x, data[0].y);
for (let i = 1; i < data.length; i++) {
  this.context.bezierCurveTo(
    data[i - 1].x + (data[i].x - data[i - 1].x) / 2,
    data[i - 1].y,
    data[i - 1].x + (data[i].x - data[i - 1].x) / 2,
    data[i].y,
    data[i].x,
    data[i].y);
}

您具有正确的平滑线和正确的端点注意！（y =“画布高度”-y）;

Answer 8

要添加到K3N的基数样条方法，并且可能解决T. J. Crowder对误导性位置中曲线'浸渍'的担忧，我在getCurvePoints()函数中插入了以下代码，就在res.push(x);之前

if ((y < _pts[i+1] && y < _pts[i+3]) || (y > _pts[i+1] && y > _pts[i+3])) {
    y = (_pts[i+1] + _pts[i+3]) / 2;
}
if ((x < _pts[i] && x < _pts[i+2]) || (x > _pts[i] && x > _pts[i+2])) {
    x = (_pts[i] + _pts[i+2]) / 2;
}

这有效地在每对连续点之间创建一个（不可见的）边界框，并确保曲线保持在此边界框内 - 即。如果曲线上的一个点位于两个点的上方/下方/左侧/右侧，则它将其位置改变为在框内。这里使用了中点，但可以使用线性插值来改进。

Answer 9

令人难以置信的迟到但受到Homan出色简单答案的启发，请允许我发布一个更通用的解决方案（一般意义上Homan的解决方案在少于3个顶点的点阵列上崩溃）：

function smooth(ctx, points)
{
    if(points == undefined || points.length == 0)
    {
        return true;
    }
    if(points.length == 1)
    {
        ctx.moveTo(points[0].x, points[0].y);
        ctx.lineTo(points[0].x, points[0].y);
        return true;
    }
    if(points.length == 2)
    {
        ctx.moveTo(points[0].x, points[0].y);
        ctx.lineTo(points[1].x, points[1].y);
        return true;
    }
    ctx.moveTo(points[0].x, points[0].y);
    for (var i = 1; i < points.length - 2; i ++)
    {
        var xc = (points[i].x + points[i + 1].x) / 2;
        var yc = (points[i].y + points[i + 1].y) / 2;
        ctx.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
    }
    ctx.quadraticCurveTo(points[i].x, points[i].y, points[i+1].x, points[i+1].y);
}

Answer 10

如果要通过n个点确定曲线方程，则以下代码将为您提供n-1阶多项式的系数，并将这些系数保存到coefficients[]数组中（从常数开始术语）。 x坐标不必按顺序排列。这是Lagrange polynomial的示例。

var xPoints=[2,4,3,6,7,10]; //example coordinates
var yPoints=[2,5,-2,0,2,8];
var coefficients=[];
for (var m=0; m<xPoints.length; m++) coefficients[m]=0;
    for (var m=0; m<xPoints.length; m++) {
        var newCoefficients=[];
        for (var nc=0; nc<xPoints.length; nc++) newCoefficients[nc]=0;
        if (m>0) {
            newCoefficients[0]=-xPoints[0]/(xPoints[m]-xPoints[0]);
            newCoefficients[1]=1/(xPoints[m]-xPoints[0]);
    } else {
        newCoefficients[0]=-xPoints[1]/(xPoints[m]-xPoints[1]);
        newCoefficients[1]=1/(xPoints[m]-xPoints[1]);
    }
    var startIndex=1; 
    if (m==0) startIndex=2; 
    for (var n=startIndex; n<xPoints.length; n++) {
        if (m==n) continue;
        for (var nc=xPoints.length-1; nc>=1; nc--) {
        newCoefficients[nc]=newCoefficients[nc]*(-xPoints[n]/(xPoints[m]-xPoints[n]))+newCoefficients[nc-1]/(xPoints[m]-xPoints[n]);
        }
        newCoefficients[0]=newCoefficients[0]*(-xPoints[n]/(xPoints[m]-xPoints[n]));
    }    
    for (var nc=0; nc<xPoints.length; nc++) coefficients[nc]+=yPoints[m]*newCoefficients[nc];
}

Answer 11

对原始问题的回答略有不同；

如果有人想画一个形状：

由一系列点描述
线在点处有一条小曲线
这条线不一定要通过这些点（即稍微通过点“内部”）

那么希望我的以下功能可以帮助

<!DOCTYPE html>
<html>

<body>
<canvas id="myCanvas" width="1200" height="700" style="border: 1px solid #d3d3d3">Your browser does not support the
    HTML5 canvas tag.</canvas>
<script>
    var cv = document.getElementById("myCanvas");
    var ctx = cv.getContext("2d");

    const drawPointsWithCurvedCorners = (points, ctx) => {
        for (let n = 0; n <= points.length - 1; n++) {
            let pointA = points[n];
            let pointB = points[(n + 1) % points.length];
            let pointC = points[(n + 2) % points.length];

            const midPointAB = {
                x: pointA.x + (pointB.x - pointA.x) / 2,
                y: pointA.y + (pointB.y - pointA.y) / 2,
            };
            const midPointBC = {
                x: pointB.x + (pointC.x - pointB.x) / 2,
                y: pointB.y + (pointC.y - pointB.y) / 2,
            };
            ctx.moveTo(midPointAB.x, midPointAB.y);
            ctx.arcTo(
                pointB.x,
                pointB.y,
                midPointBC.x,
                midPointBC.y,
                radii[pointB.r]
            );
            ctx.lineTo(midPointBC.x, midPointBC.y);
        }
    };

    const shapeWidth = 200;
    const shapeHeight = 150;

    const topInsetDepth = 35;
    const topInsetSideWidth = 20;
    const topInsetHorizOffset = shapeWidth * 0.25;

    const radii = {
        small: 15,
        large: 30,
    };

    const points = [
        {
            // TOP-LEFT
            x: 0,
            y: 0,
            r: "large",
        },
        {
            x: topInsetHorizOffset,
            y: 0,
            r: "small",
        },
        {
            x: topInsetHorizOffset + topInsetSideWidth,
            y: topInsetDepth,
            r: "small",
        },
        {
            x: shapeWidth - (topInsetHorizOffset + topInsetSideWidth),
            y: topInsetDepth,
            r: "small",
        },
        {
            x: shapeWidth - topInsetHorizOffset,
            y: 0,
            r: "small",
        },
        {
            // TOP-RIGHT
            x: shapeWidth,
            y: 0,
            r: "large",
        },
        {
            // BOTTOM-RIGHT
            x: shapeWidth,
            y: shapeHeight,
            r: "large",
        },
        {
            // BOTTOM-LEFT
            x: 0,
            y: shapeHeight,
            r: "large",
        },
    ];

    // ACTUAL DRAWING OF POINTS
    ctx.beginPath();
    drawPointsWithCurvedCorners(points, ctx);
    ctx.stroke();
</script>
</body>

</html>

Answer 12

卓悦

我很欣赏 user1693593 的解决方案：Hermite 多项式似乎是控制绘制内容的最佳方式，从数学的角度来看也是最令人满意的。这个话题好像关闭了很长时间，但可能有一些像我这样的后来者仍然对它感兴趣。我一直在寻找一个免费的交互式绘图生成器，它可以让我存储曲线并在其他任何地方重复使用它，但在网络上没有找到这种东西：所以我用自己的方式制作了它，来自维基百科由用户 1693593 提及。在这里很难解释它是如何工作的，了解它是否值得的最好方法是查看 https://sites.google.com/view/divertissements/accueil/splines。

如何使用javascript HTML5画布通过N个点绘制平滑曲线？

12 个答案: