Question

我正在为我正在处理的网站编写一些分析模块，我需要估算当前小时后的总观看次数。我有每分钟的数据直到当前分钟，所以如果时间是12:28，我将有一个看起来像这样的数组：

0: "21410"
1: "21886"
2: "21837"
3: "21895"
4: "21564"
5: "21714"
6: "21571"
7: "21324"
8: "21310"
9: "21390"
10: "21764"
11: "21598"
12: "21493"
13: "21352"
14: "21478"
15: "21058"
16: "20942"
17: "20825"
18: "21321"
19: "20950"
20: "21039"
21: "21117"
22: "20733"
23: "20773"
24: "20929"
25: "20900"
26: "20687"
27: "20999"

目前我正在预测小时的价值：

(60/minsSoFar)*totalSoFar

这种方法运作得相当好，但我宁愿在数学上做得更多。我想计算出迄今为止我所拥有的数据的最佳拟合线，并计算到第60分钟。这将考虑加速和减速。

根据我目前正在使用的方法，我实际上假设趋势是一条直线。我如何计算多项式或幂次数的公式？

我在NodeJS中写这个，所以JavaScript是理想的，但我也会采用伪代码！

如果需要，可以使用更简单的格式：

[21410, 21886, 21837, 21895, 21564, 21714, 21571, 21324, 21310, 21390, 21764, 21598, 21493, 21352, 21478, 21058, 20942, 20825, 21321, 20950, 21039, 21117, 20733, 20773, 20929, 20900, 20687, 20999]

感谢您的帮助！

Answer 1

你可以least-squares fit行。

function LineFitter()
{
    this.count = 0;
    this.sumX = 0;
    this.sumX2 = 0;
    this.sumXY = 0;
    this.sumY = 0;
}

LineFitter.prototype = {
    'add': function(x, y)
    {
        this.count++;
        this.sumX += x;
        this.sumX2 += x*x;
        this.sumXY += x*y;
        this.sumY += y;
    },
    'project': function(x)
    {
        var det = this.count * this.sumX2 - this.sumX * this.sumX;
        var offset = (this.sumX2 * this.sumY - this.sumX * this.sumXY) / det;
        var scale = (this.count * this.sumXY - this.sumX * this.sumY) / det;
        return offset + x * scale;
    }
};

function linearProject(data, x)
{
    var fitter = new LineFitter();
    for (var i = 0; i < data.length; i++)
    {
        fitter.add(i, data[i]);
    }
    return fitter.project(x);
}

示例：

>>> linearProject([ 21410, 21886, 21837, 21895, 21564, 21714, 21571, 21324, 21310, 21390, 21764, 21598, 21493, 21352, 21478, 21058, 20942, 20825, 21321, 20950, 21039, 21117, 20733, 20773, 20929, 20900, 20687, 20999 ], 60); 19489.614121510676

对方形多项式做类似的事情有点复杂：

function SquareFitter() { this.count = 0; this.sumX = 0; this.sumX2 = 0; this.sumX3 = 0; this.sumX4 = 0; this.sumY = 0; this.sumXY = 0; this.sumX2Y = 0; } SquareFitter.prototype = { 'add': function(x, y) { this.count++; this.sumX += x; this.sumX2 += x*x; this.sumX3 += x*x*x; this.sumX4 += x*x*x*x; this.sumY += y; this.sumXY += x*y; this.sumX2Y += x*x*y; }, 'project': function(x) { var det = this.count*this.sumX2*this.sumX4 - this.count*this.sumX3*this.sumX3 - this.sumX*this.sumX*this.sumX4 + 2*this.sumX*this.sumX2*this.sumX3 - this.sumX2*this.sumX2*this.sumX2; var offset = this.sumX*this.sumX2Y*this.sumX3 - this.sumX*this.sumX4*this.sumXY - this.sumX2*this.sumX2*this.sumX2Y + this.sumX2*this.sumX3*this.sumXY + this.sumX2*this.sumX4*this.sumY - this.sumX3*this.sumX3*this.sumY; var scale = -this.count*this.sumX2Y*this.sumX3 + this.count*this.sumX4*this.sumXY + this.sumX*this.sumX2*this.sumX2Y - this.sumX*this.sumX4*this.sumY - this.sumX2*this.sumX2*this.sumXY + this.sumX2*this.sumX3*this.sumY; var accel = this.sumY*this.sumX*this.sumX3 - this.sumY*this.sumX2*this.sumX2 - this.sumXY*this.count*this.sumX3 + this.sumXY*this.sumX2*this.sumX - this.sumX2Y*this.sumX*this.sumX + this.sumX2Y*this.count*this.sumX2; return (offset + x*scale + x*x*accel)/det; } }; function squareProject(data) { var fitter = new SquareFitter(); for (var i = 0; i < data.length; i++) { fitter.add(i, data[i]); } return fitter.project(60); }

<强>例2：

>>> squareProject([ 21410, 21886, 21837, 21895, 21564, 21714, 21571, 21324, 21310, 21390, 21764, 21598, 21493, 21352, 21478, 21058, 20942, 20825, 21321, 20950, 21039, 21117, 20733, 20773, 20929, 20900, 20687, 20999 ], 60); 19282.85862700518

我可以为更高次多项式做到这一点，但表达式会变得更长。对于任意程度，您必须查看矩阵。

计算趋势线并预测未来结果

1 个答案: