问题描述
|
我想在网络浏览器中用Java最小二乘拟合。
当前，用户使用HTML文本输入输入数据点信息，然后我使用jQuery抓取该数据并使用Flot将其绘制成图形。
用户输入他们的数据点后，我想向他们显示“最合适的行”。我想我会计算线性，多项式，指数和对数方程，然后选择具有最高“ 0”值的方程。
我似乎找不到任何可以帮助我完成此操作的库。我偶然发现了jstat，但据我所知，它完全缺少文档，在探究了源代码之后，它似乎没有内置任何线性回归功能-我纯粹是以此为基础在函数名称上。
有谁知道任何提供简单回归分析的Javascript库？
希望我可以像这样使用该库...
如果我在数组
var points = [[3,4],[15,45],...[23,78]]
lin_reg(points)
y = 7.12 x + 3
[7.12,3]
解决方法
什么样的线性回归？对于像最小二乘法这样的简单事物，我自己编写程序即可：
http://mathworld.wolfram.com/LeastSquaresFitting.html
那里的数学并不难，可以试一小时左右，如果太难了，请告诉我，我可以尝试一下。
编辑：
找到做到这一点的人：
http://dracoblue.net/dev/linear-least-squares-in-javascript/159/
, 我为手头问题找到的最简单的解决方案可以在以下文章中找到：
http://trentrichardson.com/2010/04/06/compute-linear-regressions-in-javascript/
请注意，除了线性方程式，它还会返回R2分数，这可能很有用。
**编辑**
这是实际的代码片段：
function linearRegression(y,x){
var lr = {};
var n = y.length;
var sum_x = 0;
var sum_y = 0;
var sum_xy = 0;
var sum_xx = 0;
var sum_yy = 0;
for (var i = 0; i < y.length; i++) {
sum_x += x[i];
sum_y += y[i];
sum_xy += (x[i]*y[i]);
sum_xx += (x[i]*x[i]);
sum_yy += (y[i]*y[i]);
}
lr[\'slope\'] = (n * sum_xy - sum_x * sum_y) / (n*sum_xx - sum_x * sum_x);
lr[\'intercept\'] = (sum_y - lr.slope * sum_x)/n;
lr[\'r2\'] = Math.pow((n*sum_xy - sum_x*sum_y)/Math.sqrt((n*sum_xx-sum_x*sum_x)*(n*sum_yy-sum_y*sum_y)),2);
return lr;
}
var known_y = [1,2,3,4];
var known_x = [5.2,5.7,5.0,4.2];
var lr = linearRegression(known_y,known_x);
// now you have:
// lr.slope
// lr.intercept
// lr.r2
// return (a,b) that minimize
// sum_i r_i * (a*x_i+b - y_i)^2
function linear_regression( xyr )
{
var i,x,y,r,sumx=0,sumy=0,sumx2=0,sumy2=0,sumxy=0,sumr=0,a,b;
for(i=0;i<xyr.length;i++)
{
// this is our data pair
x = xyr[i][0]; y = xyr[i][1];
// this is the weight for that pair
// set to 1 (and simplify code accordingly,ie,sumr becomes xy.length) if weighting is not needed
r = xyr[i][2];
// consider checking for NaN in the x,y and r variables here
// (add a continue statement in that case)
sumr += r;
sumx += r*x;
sumx2 += r*(x*x);
sumy += r*y;
sumy2 += r*(y*y);
sumxy += r*(x*y);
}
// note: the denominator is the variance of the random variable X
// the only case when it is 0 is the degenerate case X==constant
b = (sumy*sumx2 - sumx*sumxy)/(sumr*sumx2-sumx*sumx);
a = (sumr*sumxy - sumx*sumy)/(sumr*sumx2-sumx*sumx);
return [a,b];
}
const regress = (x,y) => {
const n = y.length;
let sx = 0;
let sy = 0;
let sxy = 0;
let sxx = 0;
let syy = 0;
for (let i = 0; i < n; i++) {
sx += x[i];
sy += y[i];
sxy += x[i] * y[i];
sxx += x[i] * x[i];
syy += y[i] * y[i];
}
const mx = sx / n;
const my = sy / n;
const yy = n * syy - sy * sy;
const xx = n * sxx - sx * sx;
const xy = n * sxy - sx * sy;
const slope = xy / xx;
const intercept = my - slope * mx;
const r = xy / Math.sqrt(xx * yy);
const r2 = Math.pow(r,2);
let sst = 0;
for (let i = 0; i < n; i++) {
sst += Math.pow((y[i] - my),2);
}
const sse = sst - r2 * sst;
const see = Math.sqrt(sse / (n - 2));
const ssr = sst - sse;
return {slope,intercept,r2,sse,ssr,sst,sy,sx,see};
}
regress([1,4,5],[1,3]);
function linearRegression(x,y)
{
var xs = 0; // sum(x)
var ys = 0; // sum(y)
var xxs = 0; // sum(x*x)
var xys = 0; // sum(x*y)
var yys = 0; // sum(y*y)
var n = 0;
for (; n < x.length && n < y.length; n++)
{
xs += x[n];
ys += y[n];
xxs += x[n] * x[n];
xys += x[n] * y[n];
yys += y[n] * y[n];
}
var div = n * xxs - xs * xs;
var gain = (n * xys - xs * ys) / div;
var offset = (ys * xxs - xs * xys) / div;
var correlation = Math.abs((xys * n - xs * ys) / Math.sqrt((xxs * n - xs * xs) * (yys * n - ys * ys)));
return { gain: gain,offset: offset,correlation: correlation };
}