线性拟合

[dragonwong]

公式

相关系数计算：

$$r=\frac{\sum\left[\left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)\right]}{\sqrt{\mathrm{\Sigma}\left(x_i-\overline{x}\right)^2\ \ast\ \mathrm{\Sigma}(y_i\ -\overline{y})^2}}$$

当 ∣r∣≥0.8时，可视为高度相关； 当 0.5≤∣r∣<0.8时，可视为中度相关； 当 0.3≤∣r∣<0.5时，可视为低度相关； 当 ∣r∣<0.3时，可视为相关性极弱，可视为不相关。

一元线性回归方程 $y=\beta{0}+\beta{1}x$，计算：

$$\left{\begin{matrix} \beta{1} = \frac{n\sum x{i}y{i}-\sum x{i}\sum y{i}}{n\sum x{i}^{2}-(\sum x{i})^2} \ \beta{0} = \bar{y} - \beta_1\bar{x} \end{matrix}\right.$$

代码

const variable1 = [3471994, 3522934, 3435623, 3368437, 3412836, 3590946, 3485158, 3181179];
const variable2 = [27.7, 27.8, 28.27, 27.58, 28.75, 34.3, 28.55, 26.74];

const sum = arr => arr.reduce((a, b) => a + b, 0);

const mean1 = sum(variable1) / variable1.length;
const mean2 = sum(variable2) / variable2.length;

const differences1 = variable1.map((value, index) => (value - mean1));
const differences2 = variable2.map((value, index) => (value - mean2));

const differences = differences1.map((value, index) => (differences1[index] * differences2[index]));

const squaredDifferences1 = differences1.map(value => Math.pow(value, 2));
const squaredDifferences2 = differences2.map(value => Math.pow(value, 2));

const variance1 = sum(squaredDifferences1);
const variance2 = sum(squaredDifferences2);

const covariance = differences.reduce((a, b) => a + b, 0);

const correlationCoefficient = covariance / Math.sqrt(variance1 * variance2);
console.log('相关系数为：', correlationCoefficient);

const b1 = ((variable1.length * sum(variable1.map((value, index) => variable1[index] * variable2[index]))) - (sum(variable1) * sum(variable2))) / ((variable1.length * sum(variable1.map((value, index) => Math.pow(value, 2)))) - Math.pow(sum(variable1), 2));

const b0 = mean2 - (b1 * mean1);

console.log('b0为：', b0);
console.log('b1为：', b1);

工具

• 散点拟合在线制图：Scatter Plot | XY Graph Maker
• 在线LaTeX公式编辑器-编辑器

参考资料

• 公式介绍：相关系数 | 统计学简介 | JMP
• javascript算相关系数
• 简单相关性分析(两个连续型变量)
• 【统计学笔记】如何判断变量间相关关系，并建立一元线性回归模型？