Math.NET Regression：如何获取回归的斜率、截距及系数？

Hey there, let's clarify exactly how to get the intercept from a simple linear regression using Math.NET Numerics, since you already know how to work with the xdata and ydata arrays.

Using the SimpleRegression class instance #

The easiest way is to use the built-in SimpleRegression class, which calculates and exposes the intercept directly as a property once you've loaded your data:

using MathNet.Numerics.Statistics;

// Your input data arrays
double[] xdata = new double[] { 1, 2, 3, 4, 5 };
double[] ydata = new double[] { 2, 4, 5, 4, 6 };

// Initialize the regression and add your data
var regression = new SimpleRegression();
regression.AddRange(xdata, ydata);

// Grab the intercept directly
double intercept = regression.Intercept;
// You can also get the slope here for reference
double slope = regression.Slope;

Using the static Fit method (shorter syntax) #

If you prefer a more concise approach, the static SimpleRegression.Fit method returns a tuple containing both the slope and intercept in one go:

// Directly get slope and intercept from your data arrays
var (slope, intercept) = SimpleRegression.Fit(xdata, ydata);

Manual verification (optional) #

If you want to cross-check the result using the linear regression formula, the intercept is calculated as:

intercept = mean(ydata) - slope * mean(xdata)

You can compute this manually with Math.NET's built-in statistics methods too:

double meanX = xdata.Mean();
double meanY = ydata.Mean();
double manualIntercept = meanY - slope * meanX;

All these approaches will give you the exact intercept value you're looking for—no extra hoops to jump through, since Math.NET handles the heavy lifting for you.

内容的提问来源于stack exchange，提问作者Jon