Created
March 17, 2017 13:43
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Linear regression implementation in pure C# with example of Bulgarian population prediction
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namespace LinearRegression | |
{ | |
using System; | |
using System.Diagnostics; | |
public static class Program | |
{ | |
public static void Main() | |
{ | |
var xValues = new double[] | |
{ | |
1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, | |
2005, 2006, 2007, 2008, 2009 | |
}; | |
var yValues = new double[] | |
{ | |
8669269, 8595500, 8484900, 8459800, 8427400, 8384700, 8340900, 8283200, 8230400, 8190900, | |
8149468, 7932984, 7845841, 7801273, 7761049, 7720000, 7679290, 7640238, 7606551, | |
7563710 | |
}; | |
double rSquared, intercept, slope; | |
LinearRegression(xValues, yValues, out rSquared, out intercept, out slope); | |
Console.WriteLine($"R-squared = {rSquared}"); | |
Console.WriteLine($"Intercept = {intercept}"); | |
Console.WriteLine($"Slope = {slope}"); | |
var predictedValue = (slope * 2017) + intercept; | |
Console.WriteLine($"Prediction for 2017: {predictedValue}"); | |
} | |
/// <summary> | |
/// Fits a line to a collection of (x,y) points. | |
/// </summary> | |
/// <param name="xVals">The x-axis values.</param> | |
/// <param name="yVals">The y-axis values.</param> | |
/// <param name="rSquared">The r^2 value of the line.</param> | |
/// <param name="yIntercept">The y-intercept value of the line (i.e. y = ax + b, yIntercept is b).</param> | |
/// <param name="slope">The slop of the line (i.e. y = ax + b, slope is a).</param> | |
public static void LinearRegression( | |
double[] xVals, | |
double[] yVals, | |
out double rSquared, | |
out double yIntercept, | |
out double slope) | |
{ | |
if (xVals.Length != yVals.Length) | |
{ | |
throw new Exception("Input values should be with the same length."); | |
} | |
double sumOfX = 0; | |
double sumOfY = 0; | |
double sumOfXSq = 0; | |
double sumOfYSq = 0; | |
double sumCodeviates = 0; | |
for (var i = 0; i < xVals.Length; i++) | |
{ | |
var x = xVals[i]; | |
var y = yVals[i]; | |
sumCodeviates += x * y; | |
sumOfX += x; | |
sumOfY += y; | |
sumOfXSq += x * x; | |
sumOfYSq += y * y; | |
} | |
var count = xVals.Length; | |
var ssX = sumOfXSq - ((sumOfX * sumOfX) / count); | |
var ssY = sumOfYSq - ((sumOfY * sumOfY) / count); | |
var rNumerator = (count * sumCodeviates) - (sumOfX * sumOfY); | |
var rDenom = (count * sumOfXSq - (sumOfX * sumOfX)) * (count * sumOfYSq - (sumOfY * sumOfY)); | |
var sCo = sumCodeviates - ((sumOfX * sumOfY) / count); | |
var meanX = sumOfX / count; | |
var meanY = sumOfY / count; | |
var dblR = rNumerator / Math.Sqrt(rDenom); | |
rSquared = dblR * dblR; | |
yIntercept = meanY - ((sCo / ssX) * meanX); | |
slope = sCo / ssX; | |
} | |
} | |
} |
@dittytwo Yes you are right. Its not needed
Thanks a lot! :)
If anyone wants to calculate the sample standard deviation (sigma) with this, just put this at the bottom and also add it as another out parameter: stdDev = Math.Sqrt(ssY / (count - 1));
Hi,
If we removed the last value of each of the above arrays (2009 & 7563710).
What formulae would I use to calculate the x-result (approx. 2009) when the y-input is 7563710?
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HI @NikolayIT
i noticed that the var ssY is not used is there a reason for this