Find min and max value from array data

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September 22nd, 2009, 08:43 AM

mien

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Find min and max value from array data

i try to get min and max value from array data but i seem not work.and it give error subscript out of range.here is my coding.

[CODE]
Private Sub getminval_Click()

'Get Minimum and maximum Value
Dim strFileName As String
Dim myFile As Integer
Dim strTextLine As String

Dim dblX(10000) As Long
Dim dblY(10000) As Long
Dim dblZ(10000) As Double

Dim arrText As Variant
Dim strName As String

Dim minva, maxva As Integer
Dim i As Integer
Dim min(10000), max(10000) As Double

strFileName = mstrinpfile
myFile = FreeFile
If strFileName = "" Then
MsgBox "No Data ", vbOKOnly, "HydroLab File Error"
Me.Show
End If

i = 1
min(0) = 1000
max(0) = -1000

Open strFileName For Input As #myFile
Line Input #myFile, strTextLine
Do While Not EOF(myFile)
arrText = Split(strTextLine, ",")

dblZ(i) = Str(arrText(2))

min(i) = dblZ(i)

If min(i) < min(i - 1) Then
min(i) = Val(arrText(2))
minval.Text = min(i)
End If

max(i) = dblZ(i)

If max(i) > max(i - 1) Then
max(i) = Val(arrText(2))
maxval.Text = max(i)
End If

i = i + 1
Loop
End Sub
[CODE]
thanks...

September 22nd, 2009, 01:36 PM

June7

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Arrays index numbering start with 0 unless you use Option Base 1 in the procedure header or declare the array variable with lower bound starting at 1. Your code doesn't show either so suspect the 'subscript out of range' error is because you have your index variable start with 1 and the array lowerbound is 0. Check link http://msdn.microsoft.com/en-us/lib...179(VS.60).aspx

Use '[/code]' in the second code tag.

Last edited by June7 : September 22nd, 2009 at 01:40 PM.

September 24th, 2009, 09:55 AM

mien

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i already declare i=1...how can i make this code work?

September 24th, 2009, 01:25 PM

June7

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If you want i to start with 1 and you use i to refer to the array index then the array lower bound must start with 1 not 0. The default is 0 unless you use the Option Base 1 in the Declarations section or you declare the array with specific bounds, example: Dim dblArray(1 to 10) As Double

Check out this tutorial:http://patorjk.com/programming/tutorials/vbarrays.htm

September 29th, 2009, 12:43 PM

mien

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i tried this code.but i get new error..type mismatch

Code:

Dim strFileName, strTextLine As String
Dim arrText(0 To 10000) As Variant
Dim dblZ(0 To 10000), min(0 To 10000), max(0 To 10000) As Double
Dim strName As String
Dim minva, maxva, i, myFile As Integer

    strFileName = mstrinpfile
    myFile = FreeFile
    If strFileName = "" Then
    MsgBox "No Data ", vbOKOnly, "Geo-Data Plotter File Error"
    Me.Show
    End If
    

min(0) = 1000
max(0) = -1000
i = 1

    Open strFileName For Input As #myFile
    Line Input #myFile, strTextLine
    Do While Not EOF(myFile)
    arrText(i) = Split(strTextLine, ",")
        
    dblZ(i) = Str(arrText(2))
     
    min(i) = dblZ(i)
       
    If min(i) < min(i - 1) Then
    min(i) = dblZ(i)
    minval.Text = min(i)
    Else
    minval.Text = min(i - 1)
    End If
    
    max(i) = dblZ(i)
    
    If max(i) > max(i - 1) Then
    max(i) = dblZ(i)
    maxval.Text = max(i)
    Else
    maxval.Text = max(i - 1)
    End If

                     
i = i + 1
Loop
End Sub

September 29th, 2009, 04:31 PM

June7

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The error happens on what line?

Might also help to show example of the input data.

Now I see how you are using the 0 index in the array so my prior post about the array lower bound really wasn't relevant, sorry.

In case you don't know:
When declaring more than one variable on a single line, you have to explicitely declare each one with a datatype or else will default to variant. So unless you want these to be variant need to do:

Code:

Dim strFileName As String, strTextLine As String
Dim minva As Integer, maxva As Integer, i As Integer, myFile As Integer

Same for the array variables.

Last edited by June7 : September 30th, 2009 at 02:54 PM.

September 30th, 2009, 08:18 AM

mien

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thanks for ur guide...my code still have problem.it seem not give correct answer..it should compare the value to get min and max.

Code:

Private Sub getminval_Click()

'Get Minimum and maximum Value
Dim strFileName As String, strTextLine As String
Dim arrText As Variant
Dim min As Double, max As Double
Dim strName As String
Dim i As Integer, myFile As Integer
Dim x(100000), y(100000), z(100000), diff, minva, maxva As Double

    strFileName = mstrinpfile
    myFile = FreeFile
    If strFileName = "" Then
    MsgBox "No Data ", vbOKOnly, "File Error"
    GoTo 10
    End If
    

i = 1
   
    Open strFileName For Input As #myFile
    
    Do While Not EOF(myFile)
    Input #myFile, x(i), y(i), z(i)
       
        min = z(0)
        max = z(0)
       
        If i > 1 Then
                          
        If z(i) < z(i - 1) Then
            min = z(i)
            minval.Text = min
        Else
            minval.Text = z(i - 1)
        End If
        
        If z(i) > z(i - 1) Then
            max = z(i)
            maxval.Text = maxva
        Else
            maxval.Text = max
        End If
       
    End If

i = i + 1
Loop
10
End Sub

here is the example of the data

Code:

4501.87, -35959.47, -6.3
4501.87, -35959.47, -6.3
4513.21, -35952.61, -6.6
4522.91, -35944.57, -6.5
4532.12, -35936.75, -6.7
4541.03, -35929.66, -6.7
4548.59, -35920.53, -7.0
4556.43, -35910.96, -7.0
4566.46, -35903.94, -6.7

September 30th, 2009, 03:18 PM

June7

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Here is what I did with your procedure. It works with the example data.

Code:

Private Sub getminval_Click()

'Get Minimum and maximum Value
Dim strFileName As String, strTextLine As String
Dim arrText As Variant
Dim min As Double, max As Double
Dim strName As String
Dim i As Integer, myFile As Integer
Dim x(100000), y(100000), z(100000), diff, minva, maxva As Double

'example of data
'4501.87, -35959.47, -6.3
'4501.87, -35959.47, -6.3
'4513.21, -35952.61, -6.6
'4522.91, -35944.57, -6.5
'4532.12, -35936.75, -6.7
'4541.03, -35929.66, -6.7
'4548.59, -35920.53, -7.0
'4556.43, -35910.96, -7.0
'4566.46, -35903.94, -6.7

z(1) = -6.3
z(2) = -6.3
z(3) = -6.6
z(4) = -6.5
z(5) = -6.7
z(6) = -6.7
z(7) = -7
z(8) = -7
z(9) = -6.7

'    strFileName = mstrinpfile
'    myFile = FreeFile
'    If strFileName = "" Then
'    MsgBox "No Data ", vbOKOnly, "File Error"
'    GoTo 10
'    End If
  
'    Open strFileName For Input As #myFile
    
'    Do While Not EOF(myFile)
'    Input #myFile, x(i), y(i), z(i)
min = z(1)
max = z(1)

For i = 1 To 9
    If z(i) < min Then
        min = z(i)
    End If
    If z(i) > max Then
        max = z(i)
    End If
       
'    If i > 1 Then
'
'        If z(i) < z(i - 1) Then
'            min = z(i)
'            minval.Text = min
'        Else
'            minval.Text = z(i - 1)
'        End If
'
'        If z(i) > z(i - 1) Then
'            max = z(i)
'            maxval.Text = maxva
'        Else
'            maxval.Text = max
'        End If
'
'    End If
Next
Debug.Print min
Debug.Print max

'Loop
'10
End Sub

Last edited by June7 : September 30th, 2009 at 03:26 PM.

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