如何在Javascript中获取围绕特定坐标（x，y）的坐标

Question

与标题一样，我正在努力获取围绕特定坐标的坐标。在此示例中，我将说24、7。

Answer 1

在这种情况下，我通常使用两个循环：

 const results = [];

  for(const dx of [-1, 0, 1]) {
   for(const dy of [-1, 0, 1]) {
     if(dx === 0 && dy === 0) continue;
     results.push({ x: x + dx, y: y + dy });
  }
}

或者只是：

 const neighbours = [
    [-1, -1], [-1,  0], [-1,  1],
    [ 0, -1], /*0, 0*/  [ 0,  1], 
    [ 1, -1], [ 1,  0], [ 1,  1]
 ];

 const results = neighbours.map(([dx, dy]) => ({ x: x + dx, y: y + dy }));

Answer 2

您不仅要实现这一目标，而且要很好地实现它是一件好事！在我个人看来，可以通过使用以下内容中明确的“邻接”定义来使之“好”。代码。

// This code makes "adjacency" explicit by working with "adjacency offset generators"
// These are functions which return all offsets that move a point to all of its adjacent points

let manhattanAdjacencyOffsets = dist => {
  // Returns adjacency offsets corresponding to all points within a manhattan distance of `dist`
  let ret = [];
  for (let x = -dist; x <= dist; x++) { for (let y = -dist; y <= dist; y++) {
    if (x === 0 && y === 0) continue; // Don't include the 0,0 point
    ret.push({ x, y });
  }}
  return ret;
};

// Now `nCoordinates` becomes very short:
let nCoordinates = (x0, y0, adjacencyOffsets) => {
  return adjacencyOffsets.map(({ x, y }) => ({ x: x + x0, y: y + y0 }));
};

// Here's manhattan adjacency offsets with a distance of 1
let adjacencyOffsets = manhattanAdjacencyOffsets(1);

// And here's all neighbours of the point (10,20), with regard to those offsets
let neighbours = nCoordinates(10, 20, adjacencyOffsets);
console.log(neighbours);

我喜欢这种方法，因为它使“邻接”变得明确。您甚至可以定义新的邻接函数来更改代码对“邻接”的定义。

例如，您可以使用笛卡尔距离而不是曼哈顿距离（源图块的圆形范围内的所有点）：

let cartesianAdjacencyOffsets = dist => {
  let sqrDist = dist * dist;
  let ret = [];
  let min = Math.ceil(-dist);
  let max = Math.floor(dist);
  for (let x = min; x <= max; x++) { for (let y = min; y <= max; y++) {
    if (x === 0 && y === 0) continue; // Don't include the 0,0 point
    if (sqrDist < x * x + y * y) continue; // Don't include points too far away
    ret.push({ x, y });
  }}
  return ret;
};