How can I convert string to long in excel macro. CLng is giving me type mismatch error
Dim wStr As String
Dim w As Long
wStr = "=RAND() * 0.3 + 0.35"
w = CLng(wStr)
The root cause of your error is that CDbl expects a numeric value or a string that looks like a number. the string "=RAND() * 0.3 + 0.35" itself does not look like a number, even though it will evaluate to a number.
What are you actually trying to achieve here?
If its to get a long integer result from the formula =RAND() * 0.3 + 0.35, use
Dim w as Long
w = Rnd() * 0.3 + 0.35
If its to emulate a cell formula use
Dim w as Long
w = Application.Evaluate("=RAND() * 0.3 + 0.35")
As to the formula itself, why this construct? It will return Single in the range [0.35, 0.65) which when rounded to a Long will return 0 or 1 at 50% probability of each.
Why not use
w = Rnd()
or
w = Application.Evaluate("=RAND()")
or
w = Application.WorksheetFunction.RandBetween(0, 1)
or is there some other reason I've missed?
Try the formula for w below.
w = CLng(Evaluate(wStr))
Or forget trying to use an "Excel formula", and go straight to VBA with with its random function counterpart
w = CLng(Rnd() * 0.3 + 0.35)
Related
Calling divsig(h) results in changes to h. But divsig is a function that takes a matrix by value and not by reference. How can this happen?
I'm trying to use the sig/divsig functions on matrices of data, for example:
DenseMatrix 4x4-Double ' h before divsig(h)
0.5 0.5 0.5 0.5
0.568811 0.995811 0.418727 0.987232
0.65701 0.269138 0.990942 0.99298
0.716466 0.988705 0.98747 0.999909
divsig(h)
DenseMatrix 4x4-Double ' h after divsig
0.25 0.25 0.25 0.25
0.245265 0.00417185 0.243395 0.0126045
0.225348 0.196703 0.00897602 0.00697036
0.203142 0.0111678 0.0123732 9.14075E-05
Makes no sense to me what so ever, i'm even setting up a new variable called matrix in the function instead of editing 'mat' it's self.
Function divsig(ByVal mat As LinearAlgebra.Double.Matrix)
Dim matrix = mat
For _x = 0 To matrix.RowCount() - 1
For _y = 0 To matrix.ColumnCount() - 1
matrix(_x, _y) = derivsigmoid(matrix(_x, _y))
Next
Next
Return matrix
End Function
Function sigmoid(ByVal x As Double) As Double
Return 1 / (1 + Math.Exp(-x))
End Function
Function derivsigmoid(ByVal x As Double) As Double
Return x * (1 - x)
End Function
Function sig(ByVal mat As LinearAlgebra.Double.Matrix)
Dim matrix = mat
For _x = 0 To matrix.RowCount() - 1
For _y = 0 To matrix.ColumnCount() - 1
matrix(_x, _y) = sigmoid(matrix(_x, _y))
Next
Next
Return matrix
End Function
I have fixed it, turns out matrixes are classes which means passing them ByVal passes the reference anyway. I fixed it by replacing the Dim matrix = mat with
Dim matrix As LinearAlgebra.Double.Matrix = LinearAlgebra.Double.Matrix.Build.DenseOfMatrix(mat)
So matrix becomes a copy of mat instead of just giving the same reference a different identifier.
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.
I am trying to implement Haversine formula into excel function. Its looks like this:
Public Function Haversine(Lat1 As Variant, Lon1 As Variant, Lat2 As Variant, Lon2 As Variant)
Dim R As Integer, dlon As Variant, dlat As Variant, Rad1 As Variant
Dim a As Variant, c As Variant, d As Variant, Rad2 As Variant
R = 6371
dlon = Excel.WorksheetFunction.Radians(Lon2 - Lon1)
dlat = Excel.WorksheetFunction.Radians(Lat2 - Lat1)
Rad1 = Excel.WorksheetFunction.Radians(Lat1)
Rad2 = Excel.WorksheetFunction.Radians(Lat2)
a = Sin(dlat / 2) * Sin(dlat / 2) + Cos(Rad1) * Cos(Rad2) * Sin(dlon / 2) * Sin(dlon / 2)
c = 2 * Excel.WorksheetFunction.Atan2(Sqr(a), Sqr(1 - a))
d = R * c
Haversine = d
End Function
But when im testing it I am getting wrong distance... I dont understand why. For coordinates used in this topic : Function to calculate distance between two coordinates shows wrong
I am getting 20013,44 as output. Anyone knows what is wrong here? Cant find my mistake...
Atan2 is defined back to front in Excel compared to JavaScript i.e. Atan2(x,y) rather than Atan2(y,x).
You need to reverse the order of the two arguments:-
c = 2 * Excel.WorksheetFunction.Atan2(Sqr(1 - a), Sqr(a))
See this
So
=haversine(59.3293371,13.4877472,59.3225525,13.4619422)
gives
1.65 km
which is the correct distance as the crow flies.
Great tool! Just underscoring that the result will be in kilometers, so if you want miles multiply the result by 0.62137.
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?
Public Function foo()
Dim x As Double, y As Double, z As Double
x = 1.26
y = 3.175
z = Round(x + y, 2)
foo = z
End Function
Running Excel 2007 on Windows 7. This function returns 4.43 into a cell with =foo() which is annoying as I want 4.44. I found some documentation claiming VBA ROUND uses even rounding but here the last digit is odd. What is wrong here?
You can always use the Worksheet Round Function instead of VBA's built-in one
Public Function foo2()
Dim x As Double, y As Double, z As Double
x = 1.26
y = 3.175
z = Application.WorksheetFunction.Round(x + y, 2)
foo2 = z
End Function
foo2 will result in 4.44 (tested on my machine). I don't know if this would affect performance at all.
You will need to use decimal types to accomplish this which uses integer based arithmetic as opposed to floating point based.
Excel doesn't have a native data type for this, so you have to use a Variant and then convert to a decimal using the CDec function.
Public Function foo()
Dim x As Variant, y As Variant, z As Variant
x = CDec(1.26)
y = CDec(3.175)
z = Round(x + y, 2)
foo = z
End Function
While Excel.WorksheetFunction.Round does perform correct 4/5 rounding, it is terribly slow and, of course, requires Excel to be installed.
For fast and precise rounding of any value - up, down, 4/5, Banker's, significant digits, Base 2 or Base 10, and more - go to my project VBA.Round.