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# Slope Function for VB 6.0

Posted on 2006-11-30
Medium Priority
1,560 Views
Last Modified: 2008-01-09
Anyone know where I can see code for the Slope funtion written in VB?

I suppose I could write one but why reinvent the wheel.

I actually don't need it for two arrays necessarily.  The Y data will just be ascending dates
so I figure i can just use a count of the data in the X array.  I would think that would be valid?
Y       X
1       5
2       9
3       2
4       6
5       5
6       7

0
Question by:mayfieldjr
• 3
• 2
5 Comments

LVL 70

Accepted Solution

Éric Moreau earned 500 total points
ID: 18048242
Hi mayfieldjr,

I did that in VB.Net a couple of months ago. You should be able to convert it easily.

Here is my code:
Public Shared Function Slope(ByVal pArrayY() As Double, ByVal pArrayX() As Double) As Double
Dim dblEX As Double
Dim dblEX2 As Double
Dim dblEXY As Double
Dim dblEY As Double
Dim intN As Integer
For intCtr As Integer = 0 To pArrayX.Length - 1
If (pArrayX(intCtr) <> Double.MinValue) AndAlso (pArrayY(intCtr) <> Double.MinValue) Then
dblEX += pArrayX(intCtr)
dblEX2 += (pArrayX(intCtr) * pArrayX(intCtr))
dblEXY += (pArrayX(intCtr) * pArrayY(intCtr))
dblEY += pArrayY(intCtr)
intN += 1
End If
Next

Dim dblA As Double
Dim dblB As Double
Dim dblAvgX As Double
Dim dblAvgY As Double

dblAvgX = dblEX / intN
dblAvgY = dblEY / intN
dblA = dblEXY - ((dblEX * dblEY) / intN)
dblB = dblEX2 - ((dblEX * dblEX) / intN)
Return (dblA / dblB)
End Function

Cheers!
0

Author Comment

ID: 18048677
I'm not sure what Double.MinValue is????

I believe I have everything else.  Is Double.MinValue mean the min value in the array?

Thanks
0

LVL 70

Expert Comment

ID: 18048861
it is the lowest value a double can hold. You can replace this with a value you are sure you won't find into your array.
0

Author Comment

ID: 18049043
I have tested your function but am getting a different answer than Excel gives me.

Example:
Y     X
1     7
2     1
3     4
4     8
5     8
6     4
7     9
8     8
9     1

I get .038   Excel gets .05

I either did not translate to VB properly (probably the case) or made some other error.

What do you get using the above data with your function.

I'll keep trying.

Thanks
0

Author Comment

ID: 18049312
I see MY error now.  Works very well.

Thank you.  You saved me allot of time.
0

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