JavaScript: fitting a line with least squares

Compute the slope, intercept, and R² of the best-fit line in closed form.

JavaScript

const x = [1, 2, 3, 4, 5];
const y = [2.2, 4.1, 6.2, 7.9, 10.1];

const mean = (values) => values.reduce((a, b) => a + b, 0) / values.length;
const mx = mean(x);
const my = mean(y);

let sxy = 0;
let sxx = 0;
for (let i = 0; i < x.length; i++) {
  sxy += (x[i] - mx) * (y[i] - my);
  sxx += (x[i] - mx) ** 2;
}
const slope = sxy / sxx;
const intercept = my - slope * mx;

let residual = 0;
let total = 0;
for (let i = 0; i < x.length; i++) {
  residual += (y[i] - (slope * x[i] + intercept)) ** 2;
  total += (y[i] - my) ** 2;
}

console.log("slope " + slope.toFixed(3) + ", intercept " + intercept.toFixed(3));
console.log("R squared " + (1 - residual / total).toFixed(4));
console.log("prediction for x=6: " + (slope * 6 + intercept).toFixed(2));

Output

slope 1.960, intercept 0.220
R squared 0.9989
prediction for x=6: 11.98

Explanation

Simple linear regression finds the line y = m·x + b that minimizes the sum of squared vertical errors. For one input variable there is a closed-form solution: the slope is the covariance of x and y divided by the variance of x, and the intercept makes the line pass through the point of means.

R² measures how much of the variation in y the line explains. It is 1 minus the ratio of the residual sum of squares to the total sum of squares, and it is 1 for a perfect fit and near 0 for a useless one. The data here are roughly y = 2x with a little noise, so the slope should be close to 2.

Explore the dry run

Guided walkthroughReady

1const x = [1, 2, 3, 4, 5];

2const y = [2.2, 4.1, 6.2, 7.9, 10.1];

4const mean = (values) => values.reduce((a, b) => a + b, 0) / values.length;

5const mx = mean(x);

6const my = mean(y);

8let sxy = 0;

9let sxx = 0;

10for (let i = 0; i < x.length; i++) {

11 sxy += (x[i] - mx) * (y[i] - my);

12 sxx += (x[i] - mx) ** 2;

13}

14const slope = sxy / sxx;

15const intercept = my - slope * mx;

17let residual = 0;

18let total = 0;

19for (let i = 0; i < x.length; i++) {

20 residual += (y[i] - (slope * x[i] + intercept)) ** 2;

21 total += (y[i] - my) ** 2;

22}

24console.log("slope " + slope.toFixed(3) + ", intercept " + intercept.toFixed(3));

25console.log("R squared " + (1 - residual / total).toFixed(4));

26console.log("prediction for x=6: " + (slope * 6 + intercept).toFixed(2));

Follow the example

Start to inspect the source, values, and output. These prepared steps explain the published sample; grouped loop steps describe multiple iterations.

View the complete dry-run table

Prepared trace of the published program. Some rows group repeated iterations.
What happens	State after step	Output so far
Covariance divided by variance: 19.6 / 10 = 1.96.	`slope = 1.96`	—
The line passes through the means (3, 6.1): 6.1 − 1.96·3 = 0.22.	`intercept = 0.22`	—
The fitted line is close to y = 2x.	`intercept = 0.22`	slope 1.960, intercept 0.220
Almost all of the variation is explained by the line.	`intercept = 0.22`	slope 1.960, intercept 0.220 R squared 0.9989
1.96·6 + 0.22.	`intercept = 0.22`	slope 1.960, intercept 0.220 R squared 0.9989 prediction for x=6: 11.98

Cost and common mistakes

O(n) time and O(1) extra space.

A high R² does not prove that the relationship is causal, or that the model will predict well outside the observed range.
Least squares is sensitive to outliers, because errors are squared.
toFixed returns a string and rounds for display only. Keep full precision for calculation.

Try it yourself

What does the model predict for x = 0?

Reveal answer and reasoning

The intercept, which is 0.22.

JavaScript: fitting a line with least squares

Explanation

Follow the example

Cost and common mistakes

Try it yourself

Further reading