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我一直在尝试找到一个我不久前在 LPC 中使用（未编写）的方便方法的 Javascript 版本，它被称为 dimval()，它采用以下形式：

NAME
     dimval() - returns values with a reduced increase in output
                as input values grow larger.

SYNOPSIS
     float dimval(float input, float max_input, float max_output,
                  float min_input, float min_output, float rate);

DESCRIPTION
     Returns (as a float) a value between min_output and max_output,
     with values increasing at a reduced rate as they move from
     min_input toward max_output.

     Input is the input value.

     Max_input is the maximum acceptable input. Any higher input
     value will be capped to this.

     Max_output is the maximum value returned.

     Min_input is the (optional) minimum input. Default is zero.

     Min_output is the (optional) minimum output. Default is zero.

     Rate determines how quickly the cost increases to achieve
     greater return values. Higher numbers are faster, lower numbers
     are slower.

我读了这篇文章，但它似乎并没有捕捉到我想要的东西（开始看起来要简单得多）。我还阅读了这个 SO 问题，并且......好吧，我认为这可以工作......但说实话，数学超出了我的范围。我了解上面的描述以及参数如何协同工作以产生我想要的结果。

如果有人可以在 Javascript 中提供具有上述约束的方法，我将不胜感激。

干杯！

编辑：原始方法的示例输出。

  • 评估返回值 (5.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 22.360680
  • 评估返回值 (10.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 31.622776

  • 评估返回 dimval(50.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 70.710678

  • 评估返回值 (10.0, 100.0, 100.0, 0.0, 0.0, 2.0) => 15.811388

  • 评估返回暗值（10.0、100.0、100.0、0.0、0.0、10.0）=> 3.162278

  • 评估返回值 (200.0, 100.0, 100.0, 0.0, 0.0, 10.0) => 10.000000

  • 评估返回值 (200.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 100.000000

  • 评估返回值 (1.0, 100.0, 100.0, 10.0, 0.0, 10.0) => 0.000000

如果您希望我运行更多示例，请告诉我。

4

3 回答 3

2

也许是这样的？

function dimval(n, min_in, max_in, min_out, max_out, exponent) {
  // unscale input
  n -= min_in
  n /= max_in - min_in

  n = Math.pow(n, exponent)

  // scale output
  n *= max_out - min_out
  n += min_out
  return n
}

0 < exponent < 1先快速增加，再小幅增加，exponent > 1反之。

例子：

> dimval(0, 0, 1, 0, 100, 2)
0
> dimval(0.1, 0, 1, 0, 100, 2)
1.0000000000000002
> dimval(0.2, 0, 1, 0, 100, 2)
4.000000000000001


> dimval(0, 0, 1, 0, 100, 0.5)
0
> dimval(0.1, 0, 1, 0, 100, 0.5)
31.622776601683793
> dimval(0.2, 0, 1, 0, 100, 0.5)
44.721359549995796
> dimval(0.3, 0, 1, 0, 100, 0.5)
54.77225575051661
> dimval(0.4, 0, 1, 0, 100, 0.5)
63.245553203367585
> dimval(0.5, 0, 1, 0, 100, 0.5)
70.71067811865476
> dimval(0.6, 0, 1, 0, 100, 0.5)
77.45966692414834
> dimval(0.7, 0, 1, 0, 100, 0.5)
83.66600265340756
> dimval(0.8, 0, 1, 0, 100, 0.5)
89.44271909999159
> dimval(0.9, 0, 1, 0, 100, 0.5)
94.86832980505137
> dimval(1, 0, 1, 0, 100, 0.5)
100
于 2012-04-10T12:10:35.227 回答
1

我刚刚通过一个练习来创建类似的东西，并想出了：

z*(最大 - 最大-x )

其中 z 是起始值，max 是渐近线，x 是变化值。

于 2014-05-01T16:30:32.100 回答
1

功能：

function dimval( input, max_in, max_out, min_in, min_out, rate) {
    if (rate < 0.000001) {rate = 0.000001}
    if (input > max_in) {input = max_in}
    if (input < min_in) {input = min_in}
    mult = (max_out - min_out);

    input = (input - min_in) / (max_in - min_in);
    input = Math.sqrt(input) / rate;
    input = (input * mult) + min_out;
    if (input > max_out) {input = max_out}
    return input;
}

测试：

dim1 = 'dimval(5.0, 100.0, 100.0, 0.0, 0.0, 1.0)';
dim2 = 'dimval(10.0, 100.0, 100.0, 0.0, 0.0, 1.0)';
dim3 = 'dimval(50.0, 100.0, 100.0, 0.0, 0.0, 1.0)';
dim4 = 'dimval(10.0, 100.0, 100.0, 0.0, 0.0, 2.0)';
dim5 = 'dimval(10.0, 100.0, 100.0, 0.0, 0.0, 10.0)';
dim6 = 'dimval(200.0, 100.0, 100.0, 0.0, 0.0, 10.0)';
dim7 = 'dimval(200.0, 100.0, 100.0, 0.0, 0.0, 1.0)';
dim8 = 'dimval(1.0, 100.0, 100.0, 10.0, 0.0, 10.0)';


console.log(dim1 + ' => ' + eval(dim1).toFixed(6));
console.log(dim2 + ' => ' + eval(dim2).toFixed(6));
console.log(dim3 + ' => ' + eval(dim3).toFixed(6));
console.log(dim4 + ' => ' + eval(dim4).toFixed(6));
console.log(dim5 + ' => ' + eval(dim5).toFixed(6));
console.log(dim6 + ' => ' + eval(dim6).toFixed(6));
console.log(dim7 + ' => ' + eval(dim7).toFixed(6));
console.log(dim8 + ' => ' + eval(dim8).toFixed(6));

结果：

dimval(5.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 22.360680
dimval(10.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 31.622777
dimval(50.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 70.710678
dimval(10.0, 100.0, 100.0, 0.0, 0.0, 2.0) => 15.811388
dimval(10.0, 100.0, 100.0, 0.0, 0.0, 10.0) => 3.162278
dimval(200.0, 100.0, 100.0, 0.0, 0.0, 10.0) => 10.000000
dimval(200.0, 100.0, 100.0, 0.0, 0.0, 1.0) => 100.000000
dimval(1.0, 100.0, 100.0, 10.0, 0.0, 10.0) => 0.000000
于 2013-06-29T04:33:58.913 回答