Linear Regression and Standard Deviation

Category:

C#, VB.NET, ASP.NET

Type:

Classes

Difficulty:

Intermediate

Author: Anonymous

Version Compatibility: Visual Basic.NET

More information:

This class will accept an input of arrays and determine the regression correlation between them. The user may also input a predictor value to determine what the likely value would be.

Instructions: Copy the declarations and code below and paste directly into your VB project.

Declarations:
' (None)

Code:

Public Class Stats
        'This class was created to hold various mathimatical functions

Public Class RegressInfo

Public samplecount As Integer = 0
            Public sigmaX As Double
            Public sigmaY As Double
            Public sigmaXX As Double
            Public sigmaXY As Double
            Public sigmaYY As Double
            Public regSlope As Double
            Public regYInt As Double
            Public regValue As Double
            Public ResidualErr As Double
            Public PearsonsR As Double

Public Overrides Function ToString() As String
                'This function builds the RegressInfo class into a string
                Dim ret As String = _
                "SampleSize       = " & Me.samplecount & vbCrLf & _
                "Formula used     =   y=" & Me.regSlope & "x + " & Me.regYInt & vbCrLf & _
                "Residual Error   = " & Me.ResidualErr & vbCrLf & _
                "Pearsons R value = " & Me.PearsonsR
                Return ret
            End Function

End Class

Public Shared Function _
        Regression(ByVal predictor As Double, ByVal xvalues() As Double, ByVal yvalues() As Double)
            'Caution: This function is dependent on three things
            '   1. A valid input value for the prediction
            '   2. Both input arrays MUST be the same length
            '   3. Both input arrays MUST contain numeric data
            '   These dependencies could be verified in this function
            '   with a couple of error traps but my application includes
            '   the traps in the calling procedures.
            '   Once computations have finished the function will return
            '   the "data" object

Dim i As Integer
            Dim reserrX As Double
            Dim reserrY As Double
            Dim resultY As Double
            Dim resErr As Double
            Dim sumsigX As Double
            Dim sumsigY As Double
            Dim sumsigXY As Double

Dim data As RegressInfo = New RegressInfo

'First we need to find the sums of X, Y, X*Y, and X^2 for the input arrays
            For i = LBound(xvalues) To UBound(xvalues)
                data.sigmaXY += xvalues(i) * yvalues(i)
                data.sigmaXX += xvalues(i) ^ 2
                data.sigmaYY += yvalues(i) ^ 2
                data.sigmaX += xvalues(i)
                data.sigmaY += yvalues(i)
            Next

'Now we need to determine what the slope of the regression line will be
            'Slope(b) = NÓXY - (ÓX)(ÓY) / (NÓX2 - (ÓX)2)
            data.regSlope = ((xvalues.Length * data.sigmaXY) - (data.sigmaX * data.sigmaY)) / _
                       ((xvalues.Length * data.sigmaXX) - (data.sigmaX ^ 2))

'Next we need to determine what the point of Y intercept is
            'Intercept(a) = (ÓY - b(ÓX)) / N 
            data.regYInt = (data.sigmaY - (data.regSlope * data.sigmaX)) / xvalues.Length

'Now use slope and intercept and predictor value in regression equation
            'Regression Equation(y) = a + bx
            data.regValue = data.regYInt + (data.regSlope * predictor)

'Now we will use the regression formula to find the residual error
            For i = LBound(xvalues) To UBound(xvalues)
                reserrX = xvalues(i)
                reserrY = yvalues(i)
                'Use regression formula
                resultY = data.regYInt + (data.regSlope * xvalues(i))
                'Now figure error value
                resErr = yvalues(i) - resultY
                data.ResidualErr += resErr ^ 2
            Next

'Now we will calculate Pearsons R value
            sumsigX = data.sigmaXX - ((data.sigmaX * data.sigmaX) / xvalues.Length)
            sumsigY = data.sigmaYY - ((data.sigmaY * data.sigmaY) / xvalues.Length)
            sumsigXY = data.sigmaXY - ((data.sigmaX * data.sigmaY) / xvalues.Length)
            data.PearsonsR = sumsigXY / Math.Sqrt(sumsigX * sumsigY)

Return data

End Function

Public Shared Function _
        StandardDeviation(ByVal Values() As Double) As Double

Dim Mean As Double = 0
            Dim Sum As Double = 0
            Dim SquaredDeviations(Values.Length - 1) As Double
            Dim i As Integer

'Sum the values in the array
            For i = 0 To Values.Length - 2
                Sum += Values(i)
            Next

'Calculate the mean of the array
            Mean = Sum / (Values.Length - 1)
            Sum = 0

'Find the squared deviation of each array value
            '   and calculate the sum
            For i = 0 To Values.Length - 2
                SquaredDeviations(i) = (Values(i) - Mean) ^ 2
                Sum += SquaredDeviations(i)
            Next

'Return the standard deviation value for the array
            Return _
            System.Math.Sqrt(Sum / (SquaredDeviations.Length - 1))

End Function
    End Class