Function MyPV(CF As Variant, PositiveR As Double, NegativeR As Double)
Dim n
Dim i, soma
soma = 0
For i = 1 To n
If CF(i) > 0 Then
soma = soma + CF(i) / (1 + PositiveR) ^ i
ElseIf CF(i) < 0 Then
soma = soma + CF(i) / (1 + NegativeR) ^ i
Else
MyPV = "ERRO"
End If
Next i
MyPV = soma
End Function
In this code, I have to select the Cashflows and then return the present value. The book I'm using suggests doing CF as a Variant, but I can't get the value of its length. How can I do it?
I know that excel in English "," is used to separate the parameters of a function, but in Portuguese is ";"
Related
I have a set which has an unknown number of objects. I want to associate a label to each one of these objects. Instead of labeling each object with a number I want to label them with letters.
For example the first object would be labeled A the second B and so on.
When I get to Z, the next object would be labeled AA
AZ? then BA, BB, BC.
ZZ? then AAA, AAB, AAC and so on.
I'm working using Mapbasic (similar to VBA), but I can't seem to wrap my head around a dynamic solution. My solution assumes that there will be a max number of objects that the set may or may not exceed.
label = pos1 & pos2
Once pos2 reaches ASCII "Z" then pos1 will be "A" and pos2 will be "A". However, if there is another object after "ZZ" this will fail.
How do I overcome this static solution?
Basically what I needed was a Base 26 Counter. The function takes a parameter like "A" or "AAA" and determines the next letter in the sequence.
Function IncrementAlpha(ByVal alpha As String) As String
Dim N As Integer
Dim num As Integer
Dim str As String
Do While Len(alpha)
num = num * 26 + (Asc(alpha) - Asc("A") + 1)
alpha = Mid$(alpha, 2,1)
Loop
N = num + 1
Do While N > 0
str = Chr$(Asc("A") + (N - 1) Mod 26) & str
N = (N - 1) \ 26
Loop
IncrementAlpha = str
End Function
If we need to convert numbers to a "letter format" where:
1 = A
26 = Z
27 = AA
702 = ZZ
703 = AAA etc
...and it needs to be in Excel VBA, then we're in luck. Excel's columns are "numbered" the same way!
Function numToLetters(num As Integer) As String
numToLetters = Split(Cells(1, num).Address(, 0), "$")(0)
End Function
Pass this function a number between 1 and 16384 and it will return a string between A and XFD.
Edit:
I guess I misread; you're not using Excel. If you're using VBA you should still be able to do this will the help of an reference to an Excel Object Library.
This should get you going in terms of the logic. Haven't tested it completely, but you should be able to work from here.
Public Function GenerateLabel(ByVal Number As Long) As String
Const TOKENS As String = "ZABCDEFGHIJKLMNOPQRSTUVWXY"
Dim i As Long
Dim j As Long
Dim Prev As String
j = 1
Prev = ""
Do While Number > 0
i = (Number Mod 26) + 1
GenerateLabel = Prev & Mid(TOKENS, i, 1)
Number = Number - 26
If j > 0 Then Prev = Mid(TOKENS, j + 1, 1)
j = j + Abs(Number Mod 26 = 0)
Loop
End Function
I'm am using the code example below to represent an integer as an alphabetic string
Private Function GetExcelColumnName(columnNumber As Integer) As String
Dim dividend As Integer = columnNumber
Dim columnName As String = String.Empty
Dim modulo As Integer
While dividend > 0
modulo = (dividend - 1) Mod 26
columnName = Convert.ToChar(65 + modulo).ToString() & columnName
dividend = CInt((dividend - modulo) / 26)
End While
Return columnName
End Function
I found the above example here:
Converting Numbers to Excel Letter Column vb.net
How do I get the reverse, for example:
123 = DS -- Reverse -- DS = 123
35623789 = BYXUWS -- Reverse -- BYXUWS = 35623789
Is it possible to get the number from the alphabetic string without importing Excel?
I found an answer from another post. This function below will work to get the reverse
Public Function GetCol(c As String) As Long
Dim i As Long, t As Long
c = UCase(c)
For i = Len(c) To 1 Step -1
t = t + ((Asc(Mid(c, i, 1)) - 64) * (26 ^ (Len(c) - i)))
Next i
GetCol = t
End Function
I am getting the "next without for" error. I checked other questions on this and looked for any open if statements or loops in my code, but could find none. I'm need an extra set of eyes to catch my error here.
I am trying to loop through this code and advance the torque value 3 times each times it gets to the 30th i.
'This is Holzer's method for finding the torsional natural frequency
Option Explicit
Sub TorsionalVibrationAnalysis_()
Dim n As Integer 'position along stucture
Dim m As Integer
Dim i As Long 'frequency to be used
Dim j As Variant 'moment of inertia
Dim k As Variant 'stiffness
Dim theta As Long 'angular displacement
Dim torque As ListRow 'torque
Dim lambda As Long 'ListRow 'omega^2
Dim w As Variant
Dim s As Long
'equations relating the displacement and torque
n = 1
Set j = Range("d2:f2").Value 'Range("d2:f2").Value
Set k = Range("d3:f3").Value
'initial value
Set w = Range("B1:B30").Value
For i = 1 To 30
'start at 40 and increment frequency by 20
w = 40 + (i - 1) * 20
lambda = w ^ 2
theta = 1
s = 1
Do While i = 30 & s <= 3
torque = lambda * j(1, s)
s = s + 1
End
m = n + 1
theta = theta - torque(i, n) / k(n)
torque(i, m) = torque(i, n) + lambda * j(m) * theta
If m = 4 & i < 30 Then
w(i) = 40 + (i - 1) * 20
lambda = w(i) ^ 2
ElseIf m = 4 & i >= 30 Then
Cells([d], [5+i]).display (i)
Cells([e], [5+i]).display (theta)
Cells([f], [5+i]).display (torque)
Else
End If
If m <> 4 Then
n = n + 1
End If
Next i
End Sub
You are trying to terminate your While with an End instead of Loop
Try changing your End to Loop in your Do While loop. I think you are terming the loop when you hit that End
Proper indentation makes the problem rather apparent.
You have:
For i = 1 To 30
'...
Do While i = 30 & s <= 3
'...
End
'...
If m = 4 & i < 30 Then
'...
ElseIf m = 4 & i >= 30 Then
'...
Else
End If
If m <> 4 Then
'...
End If
Next i
But run it through Rubberduck's Smart Indenter and you get:
For i = 1 To 30
'...
Do While i = 30 & s <= 3
'...
End
'...
If m = 4 & i < 30 Then
'...
ElseIf m = 4 & i >= 30 Then
'...
Else
End If
If m <> 4 Then
'...
End If
Next i
End Sub
Notice how the End other answers are pointing out, is clearly not delimiting the Do While loop.
The Next i is inside the Do While block, which isn't terminated - when the VBA compiler encounters that Next i, it doesn't know how it could possibly relate to any previously encountered For statement, and thus issues a "Next without For" compile error.
Use an indenter.
I have this code below, and I'm getting an overflow error at the line:
s = s + (x Mod 10) [first line in the Do Loop]
Why? I declared x and s to be of type Double. Adding two doubles, why is this not working?
Thanks for your help.
Public Sub bidon1()
Dim i As Double, x As Double, s As Double, k As Byte, h As Byte
Dim y(1 To 6) As Double
For i = 1 To 1000000
x = i ^ 3
Do
s = s + (x Mod 10)
x = x \ 10
Loop Until x = 0
If s = x Then
k = k + 1
y(k) = x
If y(6) > 0 Then
For h = 1 To 6
Debug.Print y(h)
Next
Exit Sub
End If
End If
Next
End Sub
The problem is that the VBA mod operator coerces its arguments to be integers (if they are not already so). It is this implicit coercion which is causing the overflow. See this question: Mod with Doubles
On Edit:
Based on your comments, you want to be able to add together the digits in a largish integer. The following function might help:
Function DigitSum(num As Variant) As Long
'Takes a variant which represents an integer type
'such as Integer, Long or Decimal
'and returns the sum of its digits
Dim sum As Long, i As Long, s As String
s = CStr(num)
For i = 1 To Len(s)
sum = sum + Val(Mid(s, i, 1))
Next i
DigitSum = sum
End Function
The following test sub shows how it can be used to correctly get the sum of the digits in 999999^3:
Sub test()
Dim x As Variant, y As Variant
Debug.Print "Naive approach: " & DigitSum(999999 ^ 3)
y = CDec(999999)
x = y * y * y
Debug.Print "CDec approach: " & DigitSum(x)
End Sub
Output:
Naive approach: 63
CDec approach: 108
Since 999999^3 = 999997000002999999, only the second result is accurate. The first result is only the sum of the digits in the string representation of the double 999999^3 = 9.99997000003E+17
I have read through the answers here https://stackoverflow.com/a/14332574/44080
I've also tried to produce equivalent VB.net code:
Option Strict ON
Public Function ParseHex(hexString As String) As Byte()
If (hexString.Length And 1) <> 0 Then
Throw New ArgumentException("Input must have even number of characters")
End If
Dim length As Integer = hexString.Length \ 2
Dim ret(length - 1) As Byte
Dim i As Integer = 0
Dim j As Integer = 0
Do While i < length
Dim high As Integer = ParseNybble(hexString.Chars(j))
j += 1
Dim low As Integer = ParseNybble(hexString.Chars(j))
j += 1
ret(i) = CByte((high << 4) Or low)
i += 1
Loop
Return ret
End Function
Private Function ParseNybble(c As Char) As Integer
If c >= "0"C AndAlso c <= "9"C Then
Return c - "0"C
End If
c = ChrW(c And Not &H20)
If c >= "A"C AndAlso c <= "F"C Then
Return c - ("A"C - 10)
End If
Throw New ArgumentException("Invalid nybble: " & c)
End Function
Can we remove the compile errors in ParseNybble without introducing data conversions?
Return c - "0"c Operator '-' is not defined for types 'Char' and 'Char'
c = ChrW(c And Not &H20) Operator 'And' is not defined for types 'Char' and 'Integer'
As it stands, no.
However, you could change ParseNybble to take an integer and pass AscW(hexString.Chars(j)) to it, so that the data conversion takes place outside of ParseNybble.
This solution is much much faster than all the alternative i have tried. And it avoids any ParseNybble lookup.
Function hex2byte(s As String) As Byte()
Dim l = s.Length \ 2
Dim hi, lo As Integer
Dim b(l - 1) As Byte
For i = 0 To l - 1
hi = AscW(s(i + i))
lo = AscW(s(i + i + 1))
hi = (hi And 15) + ((hi And 64) >> 6) * 9
lo = (lo And 15) + ((lo And 64) >> 6) * 9
b(i) = CByte((hi << 4) Or lo)
Next
Return b
End Function