javascript - 将同一线性方程的所有直线分组

Question

let input = [
  [[1, 4], [40, 4]],
  [[1, 5], [40, 5]],
  [[4, 7], [4, 24]],
  [[1, 9], [4, 1]],
  [[1, 2], [6, 4]],
  [[80, 4], [90, 4]],
  [[4, 1], [4, 40]],
  [[4, 35], [4, 29]],
  [[4, 28], [4, 35]],
  [[5, 3.6], [9, 5.2]],
]; // Input

Output = [
  [[[1, 4], [40, 4]], [[80, 4], [90, 4]]],
  [[[1, 5], [40, 5]]],
  [[[4, 7], [4, 24]], [[4, 1], [4, 40]]],
  [[[4, 35], [4, 29]], [[4, 28], [4, 35]]],
  [[[1, 9], [4, 1]]],
  [[[1, 2], [6, 4]], [[5, 3.6], [9, 5.2]]],
];

如果给定一条线的每个起始和结束坐标的系列输入，例如 [[1,4],[40,4]] 意味着它有 2 个点连接 [1,4] 和 [40 ,4] 形成一条直线。我现在的目标是将共享相同方程 y=mx+c 的所有线组合在一起，形成一个嵌套数组，如上所示。例如，

[[1,4],[40,4]] and [[80,4],[90,4]] share the same linear equation y=4

[[4,7],[4,24]],[[4,1],[4,40]]      share the same linear equation x=4

 [[1,2],[6,4]] and [[5,3.6],[9,5.2]]  share the same linear equation y=0.4x+1.6

[[1,9],[4,1]]   is alone and it has the linear equation of -2.67x+11.67

Here is my working codepen demo

我知道如何编码以找到 y=mx+c 中的 m 和 c，但问题是例如 [[4,7],[4,24]] 和 [[4,1] ,[4,40]]，m梯度变为无穷大，无解。

任何人都可以指导我如何获得正确的输出吗？

Answer 1

您可以计算每组点的斜率方程并将其分配给每个数组项，然后分组:

const input=[[[1,4],[40,4]],[[1,5],[40,5]],[[4,7],[4,24]],[[1,9],[4,1]],[[1,2],[6,4]],[[80,4],[90,4]],[[4,1],[4,40]],[[4,35],[4,29]],[[4,28],[4,35]],[[5,3.6],[9,5.2]]];

const inputsWithSlope = input.map((points) => {
  const [[x, y], [x1, y1]] = points;
  const slope = (y1 - y) / (x1 - x)
  const b = y1 - slope * x1
  return {
    points,
    line: x1 == x ? `x = ${x}` : `y = ${slope}x + ${b}`
  }
})

const res = inputsWithSlope.reduce((acc, curr) => {
  const accProp = acc[curr.line]
  acc[curr.line] = !accProp ? [curr.points] : [...accProp, curr.points]
  return acc
}, {})
const result = Object.values(res)
result.forEach(e => console.log(JSON.stringify(e)))

要处理舍入问题，您必须对其进行舍入:

const input=[[[1,4],[40,4]],[[1,5],[40,5]],[[4,7],[4,24]],[[1,9],[4,1]],[[1,2],[6,4]],[[80,4],[90,4]],[[4,1],[4,40]],[[4,35],[4,29]],[[4,28],[4,35]],[[5,3.6],[9,5.2]]];

const inputsWithSlope = input.map((points) => {
  const [[x, y], [x1, y1]] = points;
  const slope = (y1 - y) / (x1 - x)
  const b = y1 - slope * x1
  return {
    points,
    line: x1 == x ? `x = ${x}` : `y = ${slope.toFixed(2)}x + ${b.toFixed(2)}`
  }
})

const res = inputsWithSlope.reduce((acc, curr) => {
  const accProp = acc[curr.line]
  acc[curr.line] = !accProp ? [curr.points] : [...accProp, curr.points]
  return acc
}, {})
const result = Object.values(res)
result.forEach(e => console.log(JSON.stringify(e)))