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Simple Linear Regression (Least Squares)

Fit the best straight line through data points with the least squares formulas.

배우게 될 내용

  • Fit a line with closed-form formulas
  • Compute slope and intercept
  • Understand what best fit means

문제

번역본 보기

Linear regression finds the line y = slope × x + intercept that best fits the data, meaning the sum of squared errors is as small as possible. For one input variable there is an exact formula:

  • slope = Σ(x - x̄)(y - ȳ) / Σ(x - x̄)²
  • intercept = ȳ - slope × x̄

where x̄ and ȳ are the means of xs and ys. If all xs are equal, return slope 0 and intercept ȳ.

Implement linear_regression(xs, ys).

Input

xs, ys: The x and y values of the data points (same length).

Output

Return [slope, intercept].

제약 조건

1 ≤ number of points ≤ 10,000

-10,000 ≤ each value ≤ 10,000

예시

예시 1
입력:
xs = [1.0, 2.0, 3.0, 4.0]
ys = [3.0, 5.0, 7.0, 9.0]

예상 출력:
expected = [2.0, 1.0]
입력:
std::vector<double> xs = {1.0, 2.0, 3.0, 4.0};
std::vector<double> ys = {3.0, 5.0, 7.0, 9.0};

예상 출력:
std::vector<double> expected = {2.0, 1.0};
입력:
double[] xs = new double[] {1.0, 2.0, 3.0, 4.0};
double[] ys = new double[] {3.0, 5.0, 7.0, 9.0};

예상 출력:
double[] expected = new double[] {2.0, 1.0};

설명

The points lie exactly on y = 2x + 1. x̄ = 2.5, ȳ = 6, Σ(x - x̄)(y - ȳ) = 10 and Σ(x - x̄)² = 5, so slope = 2 and intercept = 6 - 2 × 2.5 = 1.

예시 2
입력:
xs = [0.0, 1.0, 2.0]
ys = [1.0, 3.0, 2.0]

예상 출력:
expected = [0.5, 1.5]
입력:
std::vector<double> xs = {0.0, 1.0, 2.0};
std::vector<double> ys = {1.0, 3.0, 2.0};

예상 출력:
std::vector<double> expected = {0.5, 1.5};
입력:
double[] xs = new double[] {0.0, 1.0, 2.0};
double[] ys = new double[] {1.0, 3.0, 2.0};

예상 출력:
double[] expected = new double[] {0.5, 1.5};

설명

x̄ = 1 and ȳ = 2. Σ(x - x̄)(y - ȳ) = (-1)(-1) + 0 + (1)(0) = 1 and Σ(x - x̄)² = 2, so slope = 0.5 and intercept = 2 - 0.5 × 1 = 1.5.

참고 자료

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