I am stuck with this issue on VBA...
I need to create combinations but when N>22 and K=5, VBA is returning Run-time error '6': Overflow.
I've tried to fix this by saving the file with .xlxs but the problem still remains.
So I've tried to switch the variables to 32-bit but I honestly don't know how to modify the code.
Please find it here below:
Public col(100), r, n, nr As Integer
Function comb(k)
col(k) = col(k - 1)
While col(k) < n - r + k
col(k) = col(k) + 1
If k < r Then
comb (k + 1)
Else
nr = nr + 1
For i = 1 To r
Cells(nr, i) = col(i)
Next
End If
Wend
End Function
Public col(100), r, n, nr As Integer
You must be assigning a higher value than an Integer can handle. Use Single, or for debugging reasons, declare them individually inside the function that way if the error persist it will highlight the variable with the error.
Do not use double as the output may require format, double = decimals, etc (as stated by enderland in the comments).
The numbers are becoming too big for the int or long to handle. for small numbers you can use the direct formula, and for larger numbers you can use either Stirling's approximation or Ramanujan's formula for factorial and store it in a double
http://www.johndcook.com/blog/2012/09/25/ramanujans-factorial-approximation/
Related
I created a simple function in MATLAB, and am trying to convert the function into Excel VBA function. My goal is to create an Excel formula =RT('range of dB levels', 'delta-time') and output the estimated reverberation time. The math is simple, see MATLAB code below:
function rr=RT(lvl_broad, dt)
n=12; %number of samples within slope calc
slope=zeros(length(lvl_broad), 1);
for i=1:length(lvl_broad)
if i<((n/2)+1) | i>length(lvl_broad)-(n/2)-1
slope(i)=0;
else
slope(i)=(lvl_broad(i+(n/2))-lvl_broad(i-(n/2)))/n;
end
end
min_slope=min(slope);
rr=abs(dt./min_slope)*60;
end
In excel, I modified/simplified this until I no longer got errors, however, the cell that I enter my 'RT' function in returns #VALUE and I do not know why. Does anything stand out in the code below? (note I changed the input range from lvl_broad to InterruptedNZ)
Function RT(InterruptedNZ, dt)
Dim Slope As Double
Slope = Slope(InterruptedNZ.Height, 1)
For i = 1 To InterruptedNZ.Height
If i < ((6) + 1) Or i > (InterruptedNZ.Height - (6) - 1) Then
Slope(i) = 0
Else
Slope(i) = (InterruptedNZ(i + (6)) - InterruptedNZ(i - (6))) / 12
End If
Next
End
min_slope = Application.WorksheetFunction.Min(Slope)
RT = Abs((dt / min_slope) * 60)
End Function
Here are some tips to translate MATLAB code into VBA code:
length()
If you are trying to get the dimensions of a range, you'll need to use the .Rows.Count or .Columns.Count properties on the range you are working with.
PERFORMANCE NOTE:
When you have a large enough range, it's a good idea to store the values of the range inside an array since you will reduce the number of times you access data from the sheets which can comme with lot of overhead. If so, you'll have to use ubound() and lbound().
zeros()
In VBA, there is no exact equivalent to the zeros() function in MATLAB. The way we would initialize an array of zeros would simply be by initializing an array of doubles (or another numerical type). And since the default value of a double is zero, we don't need to do anything else :
Dim Slope() As Double
ReDim Slope(1 To InterruptedNZ.Rows.Count)
Note that you cannot pass the dimensions in the Dim statement since it only accepts constants as arguments, so we need to create Slope as a dynamic array of doubles and then redimension it to the desired size.
Putting these two principles together, it seems like your code would look something like this:
Function RT(ByRef InterruptedNZ As Range, ByVal dt As Double)
Dim Slope() As Double
ReDim Slope(1 To InterruptedNZ.Rows.Count)
Dim i As Long
For i = 1 To InterruptedNZ.Rows.Count
If i < ((6) + 1) Or i > (InterruptedNZ.Rows.Count - (6) - 1) Then
Slope(i) = 0
Else
Slope(i) = (InterruptedNZ(i + (6)) - InterruptedNZ(i - (6))) / 12
End If
Next
Dim min_slope As Double
min_slope = Application.WorksheetFunction.Min(Slope)
RT = Abs((dt / min_slope) * 60)
End Function
Addtionnal notes:
Refering to cells from a range like this InterruptedNZ(i) works but it is good practice to be more specific like this (assuming column range) :
InterruptedNZ.Cells(i,1)
During my tests, I had a division by zero error since min_slope was zero. You might want to account for that in your code.
Having a problem with this Error. I am creating a GA and the loop is to assign my fitness value to an array.
some of the variables
Dim Chromolength as integer
Chromolength = varchromolength * aVariables
Dim i as integer, j as integer, counter as integer
Dim Poparr() As Integer
Dim FitValarr() As Integer
the code:
ReDim Poparr(1 To PopSize, 1 To Chromolength)
For i = 1 To PopSize
For j = 1 To Chromolength
If Rnd < 0.5 Then
Poparr(i, j) = 0
Else
Poparr(i, j) = 1
End If
Next j
Next i
For i = 1 To PopSize
j = 1
counter = Chromolength
Do While counter > 0
FitValarr(i) = FitValarr(i) + Poparr(i, counter) * 2 ^ (j - 1)
j = j + 1
counter = counter - 1
Loop
Next i
I am having problems with:
FitValarr(i) = FitValarr(i) + Poparr(i, counter) * 2 ^ (j - 1)
I apologize, I am fairly new to VBA.
An overflow condition arises when you create an integer expression that evaluates to a value larger than can be expressed in a 16-bit signed integer. Given the expression, either the contents of FitValarr(i), or the expression 2^(j-1) could be overflowing. Suggest all the the variables presently declared as Int be changed to Long. Long integers are 32-bit signed values and provide a correspondingly larger range of possible values.
I had the same run time error 6. After much investigation l discovered that mine was a simple 'divide by zero' error.
I set up an integer value to hold Zip codes, and Error 6 events plagued me - until I realized that a zip code of 85338 exceeded the capacity of an int...
While I didn't think of a zip code as a "value" it was nonetheless certainly interpreted as one. I suspect the same could happen with addresses as well as other "non-numeric" numeric values. Changing the variable to a string resolved the problem.
It just didn't occur to me that a zip code was a "numeric value." Lesson learned.
I have run into an overflow error in Excel VBA and cannot find my way around it. While Microsoft's documentation indicates that the range for doubles should reach ~1.8E308, I am receiving an overflow error for numbers significantly lower than that threshold. My code is as follows:
Public Function Fixed_Sample_Nums(ByVal n As Long, seed As Long) As Double()
Dim x() As Double, y() As Double, i As Long
ReDim y(1 To n)
ReDim x(1 To n)
x(1) = (CDbl(48271) * seed) Mod CDbl(2 ^ 31 - 1)
For i = 2 To n
x(i) = (CDbl(48271) * CDbl(x(i - 1))) Mod (CDbl(2 ^ 31 - 1))
y(i) = CDbl(x(i)) / CDbl(2 ^ 31 - 1)
Next i
Fixed_Sample_Nums = y
End Function
'I receive the error in the first iteration of the for loop with
'seed equal to any value >= 1 (i.e. w/ seed = 1):
Debug.Print((CDbl(48271) * CDbl(48271)) Mod (CDbl(2 ^ 31 - 1)))
'results in an overflow error
I am attempting to create a pseudo-random number generator that can take in any 'seed' value up to and including 2 ^ 31 - 1. The for loop should be able to iterate at least 9,999 times (i.e. n = 10000). If the overflow error is not encountered within the first few iterations, it most likely will not be encountered for any subsequent iteration.
As you can see, I am converting each integer to a double before any calculation. I am aware of the fact that arrays substantially increase the byte size of the calculation, but that does not appear to be the current issue as I directly copied the example calculation above into the immediate window and still received the overflow error. My attempts to find a solution online have resulted in no avail, so I would really appreciate any input. Thanks in advance!
Try using Chip Pearson's XMod function:
x(i) = XMod((CDbl(48271) * seed), CDbl(2 ^ 31 - 1))
As he notes:
You can also get overflow errors in VBA using the Mod operator with
very large numbers. For example,
Dim Number As Double
Dim Divisor As Double
Dim Result As Double
Number = 2 ^ 31
Divisor = 7
Result = Number Mod Divisor ' Overflow error here.
Code for the function:
Function XMod(ByVal Number As Double, ByVal Divisor As Double) As Double
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
' XMod
' Performs the same function as Mod but will not overflow
' with very large numbers. Both Mod and integer division ( \ )
' will overflow with very large numbers. XMod will not.
' Existing code like:
' Result = Number Mod Divisor
' should be changed to:
' Result = XMod(Number, Divisor)
' Input values that are not integers are truncated to integers. Negative
' numbers are converted to postive numbers.
' This can be used in VBA code and can be called directly from
' a worksheet cell.
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
Number = Int(Abs(Number))
Divisor = Int(Abs(Divisor))
XMod = Number - (Int(Number / Divisor) * Divisor)
End Function
Additional details:
http://www.cpearson.com/excel/ModFunction.aspx
I have an issue where I am trying to compare a values that can be alphanumeric, only numeric, or only alphabetic.
The code originally worked fine for comparing anything within the same 100s group (IE 1-99 with alphabetic components). However when I included 100+ into it, it malfunctioned.
The current part of the code reads:
For j = 1 To thislength
If lennew < j Then
enteredval = Left("100A", lennew)
ElseIf lennew >= j Then
enteredval = Left("100A", j)
End If
If lenold < j Then
cellval = Left("67", lenold)
ElseIf lenold >= j Then
cellval = Left("67", j)
End If
'issue occurs here
If enteredval >= cellval Then
newrow = newrow+1
End If
Next j
The issue occurs in the last if statement.
When cycling through the 100 is greater than the 67 but still skips over. I tried to declare them both as strings (above this part of code) to see if that would help but it didn't.
What I am trying to accomplish is to sort through a bunch of rows and find where it should go. IE the 100A should go between 100 and 100B.
Sorry lennew=len("100A") and lennold=len("67"). And thislength=4or whatever is larger of the two lengths.
The problem is that you're trying to solve the comparison problem by attacking specific values, and that's going to be a problem to maintain. I'd make the problem more generic by creating a function that supplies takes two values returns -1 if the first operand is "before" the second, 0 if they are the same, and 1 if the first operand is "after" the second per your rules.
You could then restructure your code to eliminate the specific hardcoded prefix testing and then just call the comparison function directly, eg (and this is COMPLETELY untested, off-the-cuff, and my VBA is VERRRRRY stale :) but the idea is there: (it also assumes the existence of a simple string function called StripPrefix that just takes a string and strips off any leading digits, which I suspect you can spin up fairly readily yourself)
Function CompareCell(Cell1 as String, Cell2 as String) as Integer
Dim result as integer
Dim suffix1 as string
Dim suffix2 as string
if val(cell1)< val(cell2) Then
result = -1
else if val(cell1)>val(cell2) then
result = 1
else if val(cell1)=val(cell2) then
if len(cell1)=len(cell2) then
result =0
else
' write code to strip leading numeric prefixes
' You must supply StripPrefix, but it's pretty simple
' I just omitted it here for clarity
suffix1=StripPrefix(cell1) ' eg returns "ABC" for "1000ABC"
suffix2=StripPrefix(cell2)
if suffix1 < suffix2 then
result = -1
else if suffix1 > suffix2 then
result = 1
else
result = 0
end if
end if
return result
end function
A function like this then allows you to take any two cell references and compare them directly to make whatever decision you need:
if CompareCell(enteredval,newval)>=0 then
newrow=newrow+1
end if
I'm designing a function in VBA of the form myFunction(x,y,z) where z is a table, and x can take the values of the column headings. As part of the function I need to find the number of rows in z.
I'm having problems with this, as everywhere I look suggests using length = z.Rows.Count, but when I try and output this value (as in, set myFunction = length), it produces a VALUE error. However, when I output myFunction = a which doesn't directly use length (it will eventually form part of an IF statement once I get it working), the function works fine. My code is below:
Public Function myFunction(x As String, y As Double, z As Range) As Double
Dim upper_threshold As Double
Dim lower_threshold As Double
Dim a As Double
Dim rates As Variant
Dim u As Byte
Dim l As Byte
Dim r As Byte
Dim length As Byte
a = 0
u = 2
l = 1
rates = Application.WorksheetFunction.Index(z, 1, 0)
r = Application.WorksheetFunction.Match(x, rates, 0)
length = z.rows.Count
upper_threshold = z(u, 1)
Do While y > upper_threshold
u = u + 1
l = l + 1
upper_threshold = z(u, 1)
lower_threshold = z(l, 1)
If y < upper_threshold Then
a = a + z(l, r) * (y - lower_threshold)
Else
a = a + z(l, r) * (upper_threshold - lower_threshold)
End If
Loop
myFunction = a
End Function
To test it out I also created another function:
Public Function myRows(myTable As Range) As Double
myRows = myTable.rows.Count
End Function
This one works fine on its own, but when I try to use it within the other function, I still get a VALUE error. I've tried declaring length as every type I can think of and it doesn't seem to help.
Can anyone see what's going on?
EDIT: I'm obviously not making myself very clear. The function without the two lines referring to length works as I intended. However, I need to add a bit of code to increase its functionality and this involves calculating the number of rows in the table z. When I add the two lines shown here into the function it continues to work, since it doesn't affect the output. However, if I then set the output to show length, i.e. change the penultimate line to myFunction = length it gives me a VALUE error. This leaves me with two options as far as I can see: either something else in the program is impacting on these two lines (some clashes of syntax or something), or I'm making a mistake in just assuming I can output length like that.
Your problem is with:
rates = Application.WorksheetFunction.Index(z, 1, 0)
Index only accepts a single row or column, otherwise you get a VALUE error.