I am having trouble with some loops in some code that I have inherited but that I can't get to run. I have passed what I assume are the correct values into the function (see my previous question), but now I am struggling to have the initial loops work.
The function works fine when called in a cell. However, when it runs in this macro, it is giving me a few errors, which I have chased back to the ReDim of the Lambda and v variables.
When I check it in the debugger, Lambda has the value of 0.33, but the value of the variable Splines is 1, and there is no variable with the value 0.33 that it could be picking up nearby.
Meanwhile I would expect V to pick up the value of 2, i.e. 1+1, but instead it has the value of 0.
I cannot work out why this is happening. Here is the code.
Function Spline(form As String, Splines As Integer, params As Variant, knots As Variant, coef As Variant, tstar As Range) 'form is type of spline, splines is number of knots, params is gammas, knots as knot posititions, coef as 0, tstar as cycle length in days
Dim Lambda As Variant
Dim v As Variant
Dim kmin, kmax As Double
Dim vtstar As Variant
Dim vspline As Variant
Dim a As Long
vtstar = tstar.Value2 'set vtstar as second value of the array of the cycle length
ReDim vspline(1 To UBound(vtstar), 1 To 1) 'set vspline to be cycle lengths without the 0
For a = 1 To UBound(vtstar) 'for each cycle to the max cycle
If vtstar(a, 1) = 0 Then 'for each cycle
vspline(a, 1) = 1 'recoding the days cycles into cycle number
'Exit Function
GoTo Avert 'unsure of this functionality, i think its just part of the loop syntax, by the 'exit function notation
End If
timeS = Log(vtstar(a, 1)) 'set timeS as the log of the cycle length
ReDim Lambda(Splines, 0) 'set lambda to number of knots
ReDim v(Splines + 1, 0) 'recode v to number of knots plus 1
v(0, 0) = 1
v(1, 0) = timeS 'unsure about this part
This is causing errors when the lambda and v variables are used in calculations further down in the function, first causing the lambda calculations to give "subscript out of range" error when called in this loop, I assume because 0.33 isn't a cycle number:
For i = 1 To Splines 'loop for each variable in splines (number of knots)
Lambda(i, 0) = (kmax - knots(i + 1)) / ulamb 'calculate lambda as kmax - knots+1 divided by difference between max and min knot values
Next i
(kmax, knots and ulamb appear to be calculating correctly)
And then I am getting a Error2015 error on the following loop which uses v
s = Application.SumProduct(params, v) 'set s as multiply gammas by v
Thanks in advance
Unclear what your previous question was.
Struggle a bit with your description.
the value of the variable Splines is 1
Is Splines a constant = 1? You do not show how Splines is a variable but you state it is equal to the number of knots...
ReDim Lambda(Splines, 0) 'set lambda to number of knots
If Splines = 1, you are re-dimensioning Lambda to an array of size 2x1. So your array will hold two values.
For i = 1 To Splines 'loop for each variable in splines (number of knots)
Lambda(i, 0) = (kmax - knots(i + 1)) / ulamb 'calculate lambda as kmax - knots+1 divided by difference between max and min knot values
Next i
If Splines = 1 you are starting your loop at 1 and ending at 1... So your array Lambda(0, 0) is empty. You will have a value in Lambda(1, 0).
I am trying to generate random colors (2,6) using the following code below; however, my end goal is to generate white color (2) more than the rest of the colors. Would appreciate if someone could help. Thank you.
GenerateColor = Int(Rnd() * 5) + 2
It is a probably a good idea to separate the randomization logic and the logic, which forces a given color to be created more often. E.g., this one works quite ok, giving equal chances to each number:
randomColor = CLng(rnd() * 5) + 2
However, once you obtain the randomColor, it could be changed based on some percentage, named priorityPercentage in the function:
Public Sub TestMe()
Dim counter As Long
Dim randomColor As Long
With Worksheets(1)
.Cells.Clear
For counter = 1 To 1000000
randomColor = CLng(rnd() * 5) + 2
.Cells(counter, 1) = GenerateColor(randomColor, 2, (0.4 - 0.4 * 1 / 6))
Next
.Cells(1, 2).Formula = "=COUNTIF(A:A,2)"
End With
End Sub
Public Function GenerateColor(randomColor As Long, _
priorityColor As Long, _
priorityPercentage As Double) As Long
If rnd() <= priorityPercentage Then
GenerateColor = priorityColor
Exit Function
End If
GenerateColor = CLng(rnd() * 5) + 2
End Function
This example runs 1 million times and it writes in B2 the count of the 2. The reason to pass 0.4 - 0.4 * 1.6 in the parameter, is to make sure, that the chance for 2 is exactly 40%. We have 1/6 for each of the possible 6 numbers - [2,3,4,5,6,7]. Thus, the times when we do not enter in If rnd() <= priorityPercentage are also taken into account.
So I'm new to VBA and it has been a long time since I had to program something like this. I'm trying to use six different inputs to format cells. Right now, if I could just get the inputs in x, y, z to format the corresponding cell, that would be great! I'm trying to avoid writing over 125 different if statements or cases and need a way to copy whatever color is in the B column to the appropriate x, y, z box. How can I do this iteratively?
Example
Edit: Basically what I'm trying to do is locate a cell that corresponds to an x, y, z input. Right now, the l, w, h don't matter. I'm just trying to go through all the cells in the z-planes and to find the one that matches the three input conditions (x,y,z).
Unfinished but, this should get you started in the right direction.
This will add your 3d object to an array.
AddValue will insert a value in the array at the appropriate locations
PostArray will display the array at a given location.
Note: There is no error checking in either function.
Your example Height (named depth in mine) is incorrect for sub_3. If height 0 shows on a single depth, then 3 should show on 4 depths (the starting depth extended 3 depth layers). Much like length and width values of 1 extend one past the original position.
Sub Test()
Dim oArr(1 To 5, 1 To 5, 1 To 5) As String
Call AddValue(oArr, 4, 4, 1, 0, 0, 0, "s1")
Call AddValue(oArr, 1, 3, 2, 0, 0, 0, "s2")
Call AddValue(oArr, 2, 2, 1, 1, 1, 3, "s3")
Call PostArray(oArr, Sheet1.Range("I4"))
End Sub
Private Sub AddValue(ByRef Arr() As String, ByVal iX As Integer, ByVal iY As Integer, ByVal iZ As Integer, _
ByVal iLength As Integer, ByVal iWidth As Integer, iDepth As Integer, _
ByVal Value As String)
Dim oLength As Integer
Dim oWidth As Integer
Dim oDepth As Integer
For oLength = iX To iX + iLength
For oWidth = iY To iY + iWidth
For oDepth = iZ To iZ + iDepth
Arr(oLength, oWidth, oDepth) = Value
Next
Next
Next
End Sub
Private Sub PostArray(ByRef Arr() As String, ByVal PostRange As Range)
Dim oX As Integer
Dim oY As Integer
Dim oZ As Integer
For oX = 1 To 5
For oY = 1 To 5
For oZ = 1 To 5
PostRange.Offset(oX - 1, (oY - 1) + ((oZ - 1) * 6)).Value = Arr(oX, oY, oZ)
Next
Next
Next
End Sub
Additional work
PostArray values from are going left to right, rather than your example of Right to left. If you read from right to left in your grids why arn't the depths also listed in the same direction? If you need it in that order, you'll likely want to separate the math to get the cooridinate of the 1,1 position for each depth, and a separate calculation for the offsets within that grid.
PostArray listed values instead of color, but if you enter the color as your value for each position you could set the color instead of posting the value.
Will need a loop to call AddValue using values from your worksheet rather than manually entering them into an array.
In Excel VBA, I have a working function to calculate an Exponentially Weighted Moving Average, following http://www.value-at-risk.net/exponentially-weighted-moving-average-ewma/
I want to convert this function to Access VBA to use with some data I have in Access.
I have data of the form:
BucketDate InterpRate
8/17/2015 5.56992228328638E-03
8/18/2015 5.64693660341032E-03
8/19/2015 5.72395092353427E-03
8/20/2015 5.80096524365821E-03
8/21/2015 5.87797956378215E-03
8/22/2015 5.9549938839061E-03
8/23/2015 6.03200820403004E-03
8/24/2015 6.10902252415399E-03
... ...
for 76 datapoints. The VBA subroutine is as follows:
Function EWMA(InterpRate As Range, Lambda As Double, _
MarkDate As Date, MaturityDate As Date) As Double
Dim vZeros() As Variant
Dim Price1 As Double, Price2 As Double
Dim SumWtdRtn As Double
Dim I As Long
Dim m As Double
Dim LogRtn As Double, RtnSQ As Double, WT As Double, WtdRtn As Double
vZeros = InterpRate
m = Month(MaturityDate) - Month(MarkAsOfDate)
For I = 2 To UBound(vZeros, 1)
Price1 = Exp(-vZeros(I - 1, 1) * (m / 12))
Price2 = Exp(-vZeros(I, 1) * (m / 12))
LogRtn = Log(Price1 / Price2)
RtnSQ = LogRtn ^ 2
WT = (1 - Lambda) * Lambda ^ (I - 2)
WtdRtn = WT * RtnSQ
SumWtdRtn = SumWtdRtn + WtdRtn
Next I
EWMA = SumWtdRtn ^ (1 / 2)
End Function
However, I cannot pass InterpRate into an array to use to calculate the exponentially weighted moving average. How can I change this in order to calculate the exponential moving average?
Whats the best way to round in VBA Access?
My current method utilizes the Excel method
Excel.WorksheetFunction.Round(...
But I am looking for a means that does not rely on Excel.
Be careful, the VBA Round function uses Banker's rounding, where it rounds .5 to an even number, like so:
Round (12.55, 1) would return 12.6 (rounds up)
Round (12.65, 1) would return 12.6 (rounds down)
Round (12.75, 1) would return 12.8 (rounds up)
Whereas the Excel Worksheet Function Round, always rounds .5 up.
I've done some tests and it looks like .5 up rounding (symmetric rounding) is also used by cell formatting, and also for Column Width rounding (when using the General Number format). The 'Precision as displayed' flag doesn't appear to do any rounding itself, it just uses the rounded result of the cell format.
I tried to implement the SymArith function from Microsoft in VBA for my rounding, but found that Fix has an error when you try to give it a number like 58.55; the function giving a result of 58.5 instead of 58.6. I then finally discovered that you can use the Excel Worksheet Round function, like so:
Application.Round(58.55, 1)
This will allow you to do normal rounding in VBA, though it may not be as quick as some custom function. I realize that this has come full circle from the question, but wanted to include it for completeness.
To expand a little on the accepted answer:
"The Round function performs round to even, which is different from round to larger."--Microsoft
Format always rounds up.
Debug.Print Round(19.955, 2)
'Answer: 19.95
Debug.Print Format(19.955, "#.00")
'Answer: 19.96
ACC2000: Rounding Errors When You Use Floating-Point Numbers: http://support.microsoft.com/kb/210423
ACC2000: How to Round a Number Up or Down by a Desired Increment: http://support.microsoft.com/kb/209996
Round Function: http://msdn2.microsoft.com/en-us/library/se6f2zfx.aspx
How To Implement Custom Rounding Procedures: http://support.microsoft.com/kb/196652
In Switzerland and in particulat in the insurance industry, we have to use several rounding rules, depending if it chash out, a benefit etc.
I currently use the function
Function roundit(value As Double, precision As Double) As Double
roundit = Int(value / precision + 0.5) * precision
End Function
which seems to work fine
Int and Fix are both useful rounding functions, which give you the integer part of a number.
Int always rounds down - Int(3.5) = 3, Int(-3.5) = -4
Fix always rounds towards zero - Fix(3.5) = 3, Fix(-3.5) = -3
There's also the coercion functions, in particular CInt and CLng, which try to coerce a number to an integer type or a long type (integers are between -32,768 and 32,767, longs are between-2,147,483,648 and 2,147,483,647). These will both round towards the nearest whole number, rounding away from zero from .5 - CInt(3.5) = 4, Cint(3.49) = 3, CInt(-3.5) = -4, etc.
1 place = INT(number x 10 + .5)/10
3 places = INT(number x 1000 + .5)/1000
and so on.You'll often find that apparently kludgy solutions like this are much faster than using Excel functions, because VBA seems to operate in a different memory space.
eg If A > B Then MaxAB = A Else MaxAB = B is about 40 x faster than using ExcelWorksheetFunction.Max
Unfortunately, the native functions of VBA that can perform rounding are either missing, limited, inaccurate, or buggy, and each addresses only a single rounding method. The upside is that they are fast, and that may in some situations be important.
However, often precision is mandatory, and with the speed of computers today, a little slower processing will hardly be noticed, indeed not for processing of single values. All the functions at the links below run at about 1 µs.
The complete set of functions - for all common rounding methods, all data types of VBA, for any value, and not returning unexpected values - can be found here:
Rounding values up, down, by 4/5, or to significant figures (EE)
or here:
Rounding values up, down, by 4/5, or to significant figures (CodePlex)
Code only at GitHub:
VBA.Round
They cover the normal rounding methods:
Round down, with the option to round negative values towards zero
Round up, with the option to round negative values away from zero
Round by 4/5, either away from zero or to even (Banker's Rounding)
Round to a count of significant figures
The first three functions accept all the numeric data types, while the last exists in three varieties - for Currency, Decimal, and Double respectively.
They all accept a specified count of decimals - including a negative count which will round to tens, hundreds, etc. Those with Variant as return type will return Null for incomprehensible input
A test module for test and validating is included as well.
An example is here - for the common 4/5 rounding. Please study the in-line comments for the subtle details and the way CDec is used to avoid bit errors.
' Common constants.
'
Public Const Base10 As Double = 10
' Rounds Value by 4/5 with count of decimals as specified with parameter NumDigitsAfterDecimals.
'
' Rounds to integer if NumDigitsAfterDecimals is zero.
'
' Rounds correctly Value until max/min value limited by a Scaling of 10
' raised to the power of (the number of decimals).
'
' Uses CDec() for correcting bit errors of reals.
'
' Execution time is about 1µs.
'
Public Function RoundMid( _
ByVal Value As Variant, _
Optional ByVal NumDigitsAfterDecimals As Long, _
Optional ByVal MidwayRoundingToEven As Boolean) _
As Variant
Dim Scaling As Variant
Dim Half As Variant
Dim ScaledValue As Variant
Dim ReturnValue As Variant
' Only round if Value is numeric and ReturnValue can be different from zero.
If Not IsNumeric(Value) Then
' Nothing to do.
ReturnValue = Null
ElseIf Value = 0 Then
' Nothing to round.
' Return Value as is.
ReturnValue = Value
Else
Scaling = CDec(Base10 ^ NumDigitsAfterDecimals)
If Scaling = 0 Then
' A very large value for Digits has minimized scaling.
' Return Value as is.
ReturnValue = Value
ElseIf MidwayRoundingToEven Then
' Banker's rounding.
If Scaling = 1 Then
ReturnValue = Round(Value)
Else
' First try with conversion to Decimal to avoid bit errors for some reals like 32.675.
' Very large values for NumDigitsAfterDecimals can cause an out-of-range error
' when dividing.
On Error Resume Next
ScaledValue = Round(CDec(Value) * Scaling)
ReturnValue = ScaledValue / Scaling
If Err.Number <> 0 Then
' Decimal overflow.
' Round Value without conversion to Decimal.
ReturnValue = Round(Value * Scaling) / Scaling
End If
End If
Else
' Standard 4/5 rounding.
' Very large values for NumDigitsAfterDecimals can cause an out-of-range error
' when dividing.
On Error Resume Next
Half = CDec(0.5)
If Value > 0 Then
ScaledValue = Int(CDec(Value) * Scaling + Half)
Else
ScaledValue = -Int(-CDec(Value) * Scaling + Half)
End If
ReturnValue = ScaledValue / Scaling
If Err.Number <> 0 Then
' Decimal overflow.
' Round Value without conversion to Decimal.
Half = CDbl(0.5)
If Value > 0 Then
ScaledValue = Int(Value * Scaling + Half)
Else
ScaledValue = -Int(-Value * Scaling + Half)
End If
ReturnValue = ScaledValue / Scaling
End If
End If
If Err.Number <> 0 Then
' Rounding failed because values are near one of the boundaries of type Double.
' Return value as is.
ReturnValue = Value
End If
End If
RoundMid = ReturnValue
End Function
If you're talking about rounding to an integer value (and not rounding to n decimal places), there's always the old school way:
return int(var + 0.5)
(You can make this work for n decimal places too, but it starts to get a bit messy)
Lance already mentioned the inherit rounding bug in VBA's implementation.
So I need a real rounding function in a VB6 app.
Here is one that I'm using. It is based on one I found on the web as is indicated in the comments.
' -----------------------------------------------------------------------------
' RoundPenny
'
' Description:
' rounds currency amount to nearest penny
'
' Arguments:
' strCurrency - string representation of currency value
'
' Dependencies:
'
' Notes:
' based on RoundNear found here:
' http://advisor.com/doc/08884
'
' History:
' 04/14/2005 - WSR : created
'
Function RoundPenny(ByVal strCurrency As String) As Currency
Dim mnyDollars As Variant
Dim decCents As Variant
Dim decRight As Variant
Dim lngDecPos As Long
1 On Error GoTo RoundPenny_Error
' find decimal point
2 lngDecPos = InStr(1, strCurrency, ".")
' if there is a decimal point
3 If lngDecPos > 0 Then
' take everything before decimal as dollars
4 mnyDollars = CCur(Mid(strCurrency, 1, lngDecPos - 1))
' get amount after decimal point and multiply by 100 so cents is before decimal point
5 decRight = CDec(CDec(Mid(strCurrency, lngDecPos)) / 0.01)
' get cents by getting integer portion
6 decCents = Int(decRight)
' get leftover
7 decRight = CDec(decRight - decCents)
' if leftover is equal to or above round threshold
8 If decRight >= 0.5 Then
9 RoundPenny = mnyDollars + ((decCents + 1) * 0.01)
' if leftover is less than round threshold
10 Else
11 RoundPenny = mnyDollars + (decCents * 0.01)
12 End If
' if there is no decimal point
13 Else
' return it
14 RoundPenny = CCur(strCurrency)
15 End If
16 Exit Function
RoundPenny_Error:
17 Select Case Err.Number
Case 6
18 Err.Raise vbObjectError + 334, c_strComponent & ".RoundPenny", "Number '" & strCurrency & "' is too big to represent as a currency value."
19 Case Else
20 DisplayError c_strComponent, "RoundPenny"
21 End Select
End Function
' -----------------------------------------------------------------------------
VBA.Round(1.23342, 2) // will return 1.23
To solve the problem of penny splits not adding up to the amount that they were originally split from, I created a user defined function.
Function PennySplitR(amount As Double, Optional splitRange As Variant, Optional index As Integer = 0, Optional n As Integer = 0, Optional flip As Boolean = False) As Double
' This Excel function takes either a range or an index to calculate how to "evenly" split up dollar amounts
' when each split amount must be in pennies. The amounts might vary by a penny but the total of all the
' splits will add up to the input amount.
' Splits a dollar amount up either over a range or by index
' Example for passing a range: set range $I$18:$K$21 to =PennySplitR($E$15,$I$18:$K$21) where $E$15 is the amount and $I$18:$K$21 is the range
' it is intended that the element calling this function will be in the range
' or to use an index and total items instead of a range: =PennySplitR($E$15,,index,N)
' The flip argument is to swap rows and columns in calculating the index for the element in the range.
' Thanks to: http://stackoverflow.com/questions/5559279/excel-cell-from-which-a-function-is-called for the application.caller.row hint.
Dim evenSplit As Double, spCols As Integer, spRows As Integer
If (index = 0 Or n = 0) Then
spRows = splitRange.Rows.count
spCols = splitRange.Columns.count
n = spCols * spRows
If (flip = False) Then
index = (Application.Caller.Row - splitRange.Cells.Row) * spCols + Application.Caller.Column - splitRange.Cells.Column + 1
Else
index = (Application.Caller.Column - splitRange.Cells.Column) * spRows + Application.Caller.Row - splitRange.Cells.Row + 1
End If
End If
If (n < 1) Then
PennySplitR = 0
Return
Else
evenSplit = amount / n
If (index = 1) Then
PennySplitR = Round(evenSplit, 2)
Else
PennySplitR = Round(evenSplit * index, 2) - Round(evenSplit * (index - 1), 2)
End If
End If
End Function
I used the following simple function to round my currencies as in our company we always round up.
Function RoundUp(Number As Variant)
RoundUp = Int(-100 * Number) / -100
If Round(Number, 2) = Number Then RoundUp = Number
End Function
but this will ALWAYS round up to 2 decimals and may also error.
even if it is negative it will round up (-1.011 will be -1.01 and 1.011 will be 1.02)
so to provide more options for rounding up (or down for negative) you could use this function:
Function RoundUp(Number As Variant, Optional RoundDownIfNegative As Boolean = False)
On Error GoTo err
If Number = 0 Then
err:
RoundUp = 0
ElseIf RoundDownIfNegative And Number < 0 Then
RoundUp = -1 * Int(-100 * (-1 * Number)) / -100
Else
RoundUp = Int(-100 * Number) / -100
End If
If Round(Number, 2) = Number Then RoundUp = Number
End Function
(used in a module, if it isn't obvious)
Here is easy way to always round up to next whole number in Access 2003:
BillWt = IIf([Weight]-Int([Weight])=0,[Weight],Int([Weight])+1)
For example:
[Weight] = 5.33 ; Int([Weight]) = 5 ; so 5.33-5 = 0.33 (<>0), so answer is BillWt = 5+1 = 6.
[Weight] = 6.000, Int([Weight]) = 6 , so 6.000-6 = 0, so answer is BillWt = 6.
Public Function RoundUpDown(value, decimals, updown)
If IsNumeric(value) Then
rValue = Round(value, decimals)
rDec = 10 ^ (-(decimals))
rDif = rValue - value
If updown = "down" Then 'rounding for "down" explicitly.
If rDif > 0 Then ' if the difference is more than 0, it rounded up.
RoundUpDown = rValue - rDec
ElseIf rDif < 0 Then ' if the difference is less than 0, it rounded down.
RoundUpDown = rValue
Else
RoundUpDown = rValue
End If
Else 'rounding for anything thats not "down"
If rDif > 0 Then ' if the difference is more than 0, it rounded up.
RoundUpDown = rValue
ElseIf rDif < 0 Then ' if the difference is less than 0, it rounded down.
RoundUpDown = rValue + rDec
Else
RoundUpDown = rValue
End If
End If
End If
'RoundUpDown(value, decimals, updown) 'where updown is "down" if down. else rounds up. put this in your program.
End Function