从二维距离数组在浏览器中绘制地图

Question

我正在接收二维坐标系中随机点之间的所有距离。

如何在我的浏览器中将其可视化为地图上的坐标？如果有很多解决方案，我只想查看我的算法可以提出的第一个可能的解决方案。

所以这是一个非常简单的例子：

PointCount = 3  
Distances:  
0-1 = 2  
0-2 = 4  
1-2 = 2

有没有人知道一个简单的方法（可能 solution/framework 已经存在）来使用现有的任何东西来使它更容易实施？
我在想也许可以使用 html canvas 元素进行绘图，但我不知道如何创建一个算法来为这些点提供可能的坐标。

上面的例子是简化的-
实际距离值可能如下所示：

       (0)  (1)     (2)     (3)
   (0)  0   2344    3333    10000   
   (1)      0       3566    10333   
   (2)              0       12520

Answer 1

最小工作示例。请记住，在 canvas 坐标中，y 值是倒置的，但您可以这样做：

y = canvasHeight - y

如果您也有负分，那么需要一些额外的工作。在这种情况下，绘制线条和刻度线以可视化轴也可能会有所帮助。

let canvas = document.getElementById("canvas");
let ctx = canvas.getContext("2d");

let scale = 10;
let radius = 10;

function point(x, y) {
    ctx.fillRect(x*scale, y*scale, radius, radius);
}

// test
point(10, 15);
point(20, 8);

<html>
<body>
<canvas id="canvas" width=1000 height=1000></canvas>
</body>
</html>

Answer 2

我不确定这是否与 SO 相关，但无论如何...

执行此操作的方法非常简单，使用数据将点一一放置：

为第一个点选择一个随机位置（假设它是 0,0）。
第二个点在以第一个点为圆心的半径为d(0,1)的圆上，所以可以在圆上任意取一点。让我们选择 (d(0,1),0).
第三个点在半径为d(0,2)圆心为1的圆和半径为d(1,2)圆心为2的圆的交点处。你将得到 0、1、2 或无穷多个解决方案。如果数据来自真实点，则不应出现 0。 1 和无穷大是边缘情况，但您仍应处理它们。选择任何解决方案。
第四个点在3个圆的交点处。除非你很倒霉（但你应该考虑到这一点），否则应该只有一种解决方案。
以此类推，直到所有点都被放置。

请注意，这并不意味着您将获得原始点的确切位置：您可以任意组合平移（选择第一个点）、旋转（选择第二个点）和对称性（第三点的选择）产生了差异。

一个快速而肮脏的实现（没有处理很多情况，并且测试得很少）：

function distance(p1, p2) {
  return Math.sqrt(Math.pow(p2[0] - p1[0], 2) + Math.pow(p2[1] - p1[1], 2));
}

// adapted from 
function intersection(x0, y0, r0, x1, y1, r1) {
  var a, dx, dy, d, h, rx, ry;
  var x2, y2;

  /* dx and dy are the vertical and horizontal distances between
   * the circle centers.
   */
  dx = x1 - x0;
  dy = y1 - y0;

  /* Determine the straight-line distance between the centers. */
  d = Math.sqrt((dy * dy) + (dx * dx));

  /* Check for solvability. */
  if (d > (r0 + r1)) {
/* no solution. circles do not intersect. */
return false;
  }
  if (d < Math.abs(r0 - r1)) {
/* no solution. one circle is contained in the other */
return false;
  }

  /* 'point 2' is the point where the line through the circle
   * intersection points crosses the line between the circle
   * centers.  
   */

  /* Determine the distance from point 0 to point 2. */
  a = ((r0 * r0) - (r1 * r1) + (d * d)) / (2.0 * d);

  /* Determine the coordinates of point 2. */
  x2 = x0 + (dx * a / d);
  y2 = y0 + (dy * a / d);

  /* Determine the distance from point 2 to either of the
   * intersection points.
   */
  h = Math.sqrt((r0 * r0) - (a * a));

  /* Now determine the offsets of the intersection points from
   * point 2.
   */
  rx = -dy * (h / d);
  ry = dx * (h / d);

  /* Determine the absolute intersection points. */
  var xi = x2 + rx;
  var xi_prime = x2 - rx;
  var yi = y2 + ry;
  var yi_prime = y2 - ry;

  return [
[xi, yi],
[xi_prime, yi_prime]
  ];
}

function generateData(nbPoints) {
  var i, j, k;
  var originalPoints = [];

  for (i = 0; i < nbPoints; i++) {
originalPoints.push([Math.random() * 20000 - 10000, Math.random() * 20000 - 10000]);
  }
  var data = [];
  var distances;
  for (i = 0; i < nbPoints; i++) {
distances = [];
for (j = 0; j < i; j++) {
  distances.push(distance(originalPoints[i], originalPoints[j]));
}
data.push(distances);
  }
  //console.log("original points", originalPoints);
  //console.log("distance data", data);
  return data;
}

function findPointsForDistances(data, threshold) {
  var points = [];
  var solutions;
  var solutions1, solutions2;
  var point;
  var i, j, k;

  if (!threshold)
threshold = 0.01;

  // First point, arbitrarily set at 0,0
  points.push([0, 0]);

  // Second point, arbitrarily set at d(0,1),0
  points.push([data[1][0], 0]);

  // Third point, intersection of two circles, pick any solution
  solutions = intersection(
points[0][0], points[0][1], data[2][0],
points[1][0], points[1][1], data[2][1]);
  //console.log("possible solutions for point 3", solutions);
  points.push(solutions[0]);

  //console.log("solution for points 1, 2 and 3", points);
  found = true;

  // Subsequent points, intersections of n-1 circles, use first two to find 2 solutions,
  // the 3rd to pick one of the two
  // then use others to check it's valid
  for (i = 3; i < data.length; i++) {
// distances to points 1 and 2 give two circles and two possible solutions
solutions = intersection(
  points[0][0], points[0][1], data[i][0],
  points[1][0], points[1][1], data[i][1]);
//console.log("possible solutions for point " + (i + 1), solutions);
// try to find which solution is compatible with distance to point 3
found = false;
for (j = 0; j < 2; j++) {
  if (Math.abs(distance(solutions[j], points[2]) - data[i][2]) <= threshold) {
    point = solutions[j];
    found = true;
    break;
  }
}
if (!found) {
  console.log("could not find solution for point " + (i + 1));
  console.log("distance data", data);
  console.log("solution for points 1, 2 and 3", points);
  console.log("possible solutions for point " + (i + 1), solutions);
  console.log("distances to point 3",
   distance(solutions[0], points[2]),
    distance(solutions[1], points[2]),
    data[i][2]
    );

  break;
}
// We have found a solution, we need to check it's valid
for (j = 3; j < i; j++) {
  if (Math.abs(distance(point, points[j]) - data[i][j]) > threshold) {
    console.log("Could not verify solution", point, "for point " + (i + 1) + " against distance to point " + (j + 1));
    found = false;
    break;
  }
}
if (!found) {
  console.log("stopping");
  break;
}
points.push(point);
  }
  if (found) {
//console.log("complete solution", points);
return points;
  }
}

console.log(findPointsForDistances([
  [],
  [2344],
  [3333, 3566],
  [10000, 10333, 12520],
]));
console.log(findPointsForDistances([
  [],
  [2],
  [4, 2],
]));
console.log(findPointsForDistances([
  [],
  [4000],
  [5000, 3000],
  [3000, 5000, 4000]
]));
console.log(findPointsForDistances([
  [],
  [2928],
  [4938, 3437],
  [10557, 10726, 13535]
]));

var nbPoints, i;
for (nbPoints = 4; nbPoints < 8; nbPoints++) {
  for (i = 0; i < 10; i++) {
console.log(findPointsForDistances(generateData(nbPoints)));
  }
}

Fiddle 这里：https://jsfiddle.net/jacquesc/82aqmpnb/15/

Answer 3

那里有很多图书馆。

chartist.js 是易于使用且响应迅速的 JavaS 脚本库。在尝试了许多其他图表后，我去年将它用于基本图表，但它是唯一一个可以在不同屏幕尺寸下轻松缩放的图表。

chartJS 是另一个更好看的库。

您可以使用 html5 canvas 它既简单又有趣，但需要时间，尤其是在缩放方面。

要缩放和定位，您应该使用 x 和 y 的最小值和最大值。

祝你好运

从二维距离数组在浏览器中绘制地图

Draw Map in Browser out of 2 Dimensional Array of Distances

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