如何使資料點適合與主曲線隨機變化的曲線？

所以基本上，假設你有一條曲線，如y = log(x).

const y = x => Math.log(x)
console.log(generate(10))
// [0, 0.6931471805599453, 1.0986122886681096, 1.3862943611198906, 1.6094379124341003, 1.791759469228055, 1.9459101490553132, 2.0794415416798357, 2.1972245773362196]

function generate(max) {
  let i = 1
  let ypoints = []
  while (i < max) {
    ypoints.push(y(i  ))
  }
  return ypoints
}

現在有了這些點，我們也想將它們隨機發散，也就是在變化或隨機發散量方面遵循某種曲線。因此，例如，隨著曲線向上和向右變平，曲線y會略微向上或略微向下移動，離log(x)標記更遠的概率會降低y。所以說在 處消失1/x。

這意味著，它更有可能直接位于log(x)或靠近它，而不是遠離它。但是我們可以用任何類似的等式交換。

How can you make a simple JavaScript function give you the final set of coordinates (x, y), an array of 2-tuples containing the x and y coordinate? My attempt is not getting the part right about the "variational-decay according to a secondary curve" that I've been trying to describe.

// calling `u` the thing that converts x into a y point.
const u = x => Math.log(x)
// here, `v` is the equation for variation from the main curve, given a y.
const v = y => 1 / y
// `r` is the random variation generator
const r = y => y   (Math.random() * v(y))

const coordinates = generate(10, u, v, r)

console.log(coordinates)

function generate(max, u, v, r) {
  let i = 1
  let coordinates = []
  while (i < max) {
    const x = i  
    const y = r(u(x))
    coordinates.push([ x, y ])
  }
  return coordinates
}

How can this be made to take the two curves and generate the randomish variation away from the curve that has probability/decay-rate to be away from the main curve according to the second curve?

The expected output is to be - some small amount away from that first array in the comments (if we have 10 points along the x axis).

// [0, 0.6931471805599453, 1.0986122886681096, 1.3862943611198906, 1.6094379124341003, 1.791759469228055, 1.9459101490553132, 2.0794415416798357, 2.1972245773362196]

So it might be like (just making these up, they would be more randomly determined):

// [0, 0.69, 1.1, 1.3, 1.43, 1.71, 1.93, 2, 2.21]

Notice how they are more likely to be close to the log(x) curve (because 1/x, and randomness), than further away, and it goes -.

The main reason for asking (which is tangential to this abstracted question), is for generating dummy data during development, to test out UI data visualization features, simulating somewhat realistic looking data. I would pick a much more complicated equation for the main curve, and a similar decay equation for variation, and generate potentially millions of points, so this is just a simplification of that problem.

I am talking like given a curve equation like the one in the next picture, generate random points that are like the points in this next picture too.

uj5u.com熱心網友回復：

一種方法：

• 隨機取一個x。
• 計算y = f(x)。
• 通過想要的分布獲得該點的隨機偏移。
• 回到這一點。

const
    f = x => 10 * Math.log(x),
    offset = () => (1 / Math.random() - 1) * (Math.random() < 0.5 || -1),
    canvas = document.getElementById('canvas'),
    ctx = canvas.getContext('2d');

for (x = 0; x < 100; x  ) {
    const
        y = f(x),
        dx = offset(),
        dy = offset();
        
    console.log(x.toFixed(2), y.toFixed(2), dx.toFixed(2), dy.toFixed(2));
    ctx.beginPath();
    ctx.strokeStyle = '#000000';
    ctx.arc(x * 4 , (100 - y) * 4, 0.5, 0, Math.PI * 2, true);
    ctx.stroke();        

    ctx.beginPath();
    ctx.strokeStyle = '#ff0000';
    ctx.arc((x   dx) * 4 , (100 - y   dy) * 4, 2, 0, Math.PI * 2, true);
    ctx.stroke();        
}

<canvas id="canvas" style="border-width: 0; display: block; padding: 0; margin: 0;" width="400" height="400"></canvas>

標籤：javascript algorithm math statistics

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