/// <summary>
/// Calculate line of best fit.
/// </summary>
/// <returns>The regression.</returns>
/// <param name="x">The x coordinate.</param>
/// <param name="y">The y coordinate.</param>
/// I'm not going to try and explain linear regression in comments.
/// Check out this link for more info: http://mathworld.wolfram.com/LeastSquaresFitting.html
public static BestFit LinRegression(float[] x, float[] y)
{
if (x.Length != y.Length)
{
throw new Exception("Error: x-values and y-values must be of equal length!");
}
//m = (n*sum(products) - sum(x's)sum(y's))/(n*sum(x squares) - sum(x's)^2)
//Calculate:
//1: The sum of the squares of the inputs.
//2: The sum of the products of the inputs and outputs.
float productSum = 0f;
float productSquares = 0f;
for (int a = 0; a < x.Length; a++)
{
productSum += x[a] * y[a];
productSquares += x[a] * x[a];
}
//Calculus to solve for the line of best fit's slope.
float dividend = (x.Length * productSum) - (Sum(x) * Sum(y));
float divisor = (x.Length * productSquares) - (float)Math.Pow(Sum(x), 2.0);
float m = dividend / divisor;
//Calculus to solve for the line of best fit's y-intercept.
float b = (Sum(y) - m * (Sum(x))) / x.Length;
//Return the result.
LinearEquation eq = new LinearEquation(m, b);
return(new BestFit(eq, x, y));
}
/// <summary>
/// Calculate the exponential function of best fit and return it.
/// </summary>
/// <returns>The exponential function of best fit.</returns>
/// <param name="x">The x inputs.</param>
/// <param name="y">The y values.</param>
/// For a clearer understanding of exponential regression, check out this
/// link: http://mathworld.wolfram.com/LeastSquaresFittingExponential.html
public static BestFit ExpRegression(float[] x, float[] y)
{
//Convert the line into a "linear" form. So it changes from
//y = a * 10^bx to log(y) = log(a) + bx. Solve for the best fit of this "line".
BestFit bestLin = LinRegression(x, Log(y));
//Convert to a linear equation. This will allow us to use the slope and
//y-intercept in calculations.
LinearEquation eq = (LinearEquation)bestLin.equation;
//Find line of best fit for actual x-inputs and y-inputs.
bestLin = LinRegression(x, y);
//Looking at the equation from earlier, a is the y-intercept of the line.
float A = ((LinearEquation)bestLin.equation).b;
//The exponential equation should be in the form y = a * r^x.
//From earlier, that means r = 10^b, which in this case makes r = 10^m
float r = (float)Math.Pow(10, eq.m);
//Return exponential equation.
ExponentialEquation exp = new ExponentialEquation(A, r);
return(new BestFit(exp, x, y));
}