使用 javascript 数组减少给定的 n 个输入以产生 m 个输出

Use javascript array reduce given n inputs to yield m outputs

我有一系列的点,我想把它们变成一系列的线。

这是我希望代码执行的操作的示例

[p1, p2, p3] -> [line1, line2]

每个循环:

(p1, p2) -> line
(p2, p3) -> line

执行此操作的标准方法是:

const triangle = [[0,0], [0,1], [1,2]]

const lines = []
for (let i = 1; i < triangle.length; ++i) {
  const slope = findSlopeFromPoints(...triangle[i - 1], ...triangle[i])
  const yIntercept = findYIntercept(...triangle[i], slope)
  lines.push({
    slope,
    yIntercept
  })
}

这是我可以使用 Array.prototype.reduce 获得的收盘价。但是感觉更难推理

const initial = {
  array: [], // actual returned array we care about
  lastPoint: null // "triangle[i - 1]"
}
const linesR = triangle.reduce( (lines, point) => {
  if (lines.lastPoint === null)
    return { 
      ...lines, 
      lastPoint: point 
    }
  else {
    const slope = findSlopeFromPoints(...lines.lastPoint, ...point)
    const yIntercept = findYIntercept(...point, slope)
    lines.array.push({
      slope,
      yIntercept
    })
    lines.lastPoint = point
    return lines

  }
}, initial )

简而言之,有没有更好的方法使用reduce将N个输入组合成N - 1个输出?

当然可以,使用 currentIndex 参数应用偏移量。您的回调函数接收的参数1 比您使用的参数多:

[{x:0, y:0}, {x:0, y:1}, {x:1, y:2}].reduce((lines, point, currentIndex, source) => {
  currentIndex < source.length -1 && lines.push({
    from: point,
    to: source[currentIndex + 1]
  }); 
  return lines;     
}, []);

1有关详细信息,请参阅 Array.prototype。<a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce" rel="nofollow noreferrer">reduce()</a>