VBA check if whole row of multidimensional variant is empty without loops - arrays

Is there a quick way to check whether a whole row of a variant is empty?
My multi-dimensional array / variant has n-rows and m-columns.
The only way I can think of is to loop through the columns (of a specific row) and use the IsEmpty() function to determine if a cell is empty.
The variant only consists strings.
Do you know a faster way? Maybe something like this pseudo-code: IsEmpty(myarr(1,*))
this pseudocode would mean to check the all columns of the first row if they are empty.

You could try something like:
Sub Test()
Dim myarr() As Variant, indx As Long
myarr = Range("A8:C20").Value 'Or however you initialize your array.
indx = 1 'Or whichever row you would want to check.
With Application
Debug.Print Join(.Index(myarr, indx, 0), "") <> ""
End With
End Sub
Not sure if it will be faster than a loop though, since we call a worksheet application.

No, there isn't a faster way especially considering that arrays in VBA are stored column-wise in memory. The values on a single row are not stored adjacent in memory as it's the case with column values - you could easily test this by running a For Each loop on an array.
That being said, you should probably consider having a Function that checks if a specific row is empty so that you can call it repeatedly and maybe also check for null strings if needed. For example a range of formulas returning "" will not be empty but you might want to have the ability to consider them empty.
For example, you could use something like this:
Public Function Is2DArrayRowEmpty(ByRef arr As Variant _
, ByVal rowIndex As Long _
, Optional ByVal ignoreEmptyStrings As Boolean = False _
) As Boolean
Const methodName As String = "Is2DArrayRowEmpty"
'
If GetArrayDimsCount(arr) <> 2 Then
Err.Raise 5, methodName, "Array is not two-dimensional"
ElseIf rowIndex < LBound(arr, 1) Or rowIndex > UBound(arr, 1) Then
Err.Raise 5, methodName, "Row Index out of bounds"
End If
'
Dim j As Long
Dim v As Variant
'
For j = LBound(arr, 2) To UBound(arr, 2)
v = arr(rowIndex, j)
Select Case VBA.VarType(v)
Case VbVarType.vbEmpty
'Continue to next element
Case VbVarType.vbString
If Not ignoreEmptyStrings Then Exit Function
If LenB(v) > 0 Then Exit Function
Case Else
Exit Function
End Select
Next j
'
Is2DArrayRowEmpty = True 'If code reached this line then row is Empty
End Function
Public Function GetArrayDimsCount(ByRef arr As Variant) As Long
If Not IsArray(arr) Then Exit Function
'
Const MAX_DIMENSION As Long = 60
Dim dimension As Long
Dim tempBound As Long
'
'A zero-length array has 1 dimension! Ex. Array() returns (0 to -1)
On Error GoTo FinalDimension
For dimension = 1 To MAX_DIMENSION
tempBound = LBound(arr, dimension)
Next dimension
Exit Function
FinalDimension:
GetArrayDimsCount = dimension - 1
End Function
Notice that I haven't checked for IsObject as your values are coming from a range in Excel but you would normally check for that in a general case.
Your pseudocode IsEmpty(myarr(1,*)) could be translated to:
Is2DArrayRowEmpty(myarr, 1, False) 'Empty strings would not be considered Empty
or
Is2DArrayRowEmpty(myarr, 1, True) 'Empty strings would be considered Empty

Related

Store data by using FILTER Function within VBA [duplicate]

I'm trying to make a function MonstersInLevel() that filters the second column of my "LevelMonsters" named range based on the value of the first column. The range's first column represents a game level ID and the second column represents a monster ID that appears in that level. Here's what my range looks like.
If I call MonstersInLevel(2) I expect the function to return a range consisting of "2", "3" and "4".
Function MonstersInLevel(level As Integer) As Range
MonstersInLevel = Application.WorksheetFunction.Filter(Range("LevelMonsters").Columns(2), Range("LevelMonsters").Columns(1) = level)
End Function
I get:
A value used in the formula is of the wrong data type
I'm using the FILTER function as I would as an Excel formula. I assume there's some difference in the Excel and VBA syntax for FILTER's criteria.
Just encountered this problem myself and wanted to post my workaround.
We need to return an array of True/False to the worksheet function. To do this I created a Function that takes a 2D array, the column wanted and the value to compare. It then returns a 2d single column array of the necessary True/False.
Function myeval(arr() As Variant, clm As Long, vl As Variant) As Variant()
Dim temp() As Variant
ReDim temp(1 To UBound(arr, 1), 1 To 1)
Dim i As Long
For i = 1 To UBound(arr, 1)
temp(i, 1) = arr(i, clm) = vl
Next i
myeval = temp
End Function
So in this particular case it would be called:
Function MonstersInLevel(level As Integer) As Variant
MonstersInLevel = Application.WorksheetFunction.Filter(Range("LevelMonsters").Columns(2), myeval(Range("LevelMonsters").Value, 1, level),"""")
End Function
Avoid type mismatch in Worksheetfunction via VBA
Keeping in mind that the 2nd argument reflects a dynamic matrix condition
based entirely on ►worksheet logic (returning an array of 0 or 1 cell values /False or True])
it seems that you have
to execute an evaluation at least within this argument and
declare the function type (explicitly or implicitly) as Variant
Function MonstersInLevel(level As Integer) As Variant
'' Failing assignment:
' MonstersInLevel = Application.WorksheetFunction.Filter(Range("LevelMonsters").Columns(2), _
' Range("LevelMonsters").Columns(1) = level _
' )
MonstersInLevel = Application.WorksheetFunction.Filter( _
Range("LevelMonsters").Columns(2), _
Evaluate(Range("LevelMonsters").Columns(1).Address & "=" & level) _
)
End Function
...or to evaluate the complete function
Function MonstersInLevel(level As Integer) As Variant
Dim expr As String
expr = "=Filter(" & _
Range("LevelMonsters").Columns(2).Address & "," & _
Range("LevelMonsters").Columns(1).Address & "=" & level & _
")"
'Debug.Print expr
MonstersInLevel = Evaluate(expr)
End Function
Example call writing results to any target
Dim v
v = MonstersInLevel(2)
Sheet1.Range("D2").Resize(UBound(v), UBound(v, 2)) = v
Of course it would be possible as well to write .Formula2 expressions programmatically, even splitting into spill range references.
Addendum ........... //as of Jan 10th
Backwards compatible workaround via VBA.Filter()
"If you guys know any other VBA function that would be more appropriate
than Application.WorksheetFunction.Filter I'd be ok."
In order to provide also a backwards compatible alternative,
I demonstrate the following approach using the classic (VBA.)Filter() function (see section [3]) based upon
prior matching results (see [1]).
Note that Application.Match() comparing two (!) array inputs
delivers a whole array of possible findings (instead of a single result as most frequently executed).
Non findings are identified by IsError() values of -1; adding +1 results in a set
of zeros and ones. Section [2] enters corresponding data for positive findings.
Eventually non-findings (i.e. 0or zero) are removed by a tricky negative filtering.
Function getLevels()
Function getLevels(rng As Range, ByVal level As Long)
'Site: https://stackoverflow.com/questions/65630126/how-to-remove-only-the-duplicate-row-instead-of-removing-all-the-rows-that-follo
'[0] get datafield array
Dim v, v2
v = Application.Index(rng.Value2, 0, 1) ' 1st column
v2 = Application.Index(rng.Value2, 0, 2) ' 2nd column
'[1] check data (with Match comparing 2 arrays :-)
Dim results
results = Application.Transpose(Application.Match(v, Array(level), 0))
'[2] rebuild with False/True entries
Dim i As Long
For i = 1 To UBound(results)
results(i) = IsError(results(i)) + 1 ' 0 or 1-values
If results(i) Then results(i) = v2(i, 1) ' get current value if true
Next i
'[3] remove zeros (negative filtering)
results = Filter(results, "0", False)
'[4] return results as vertical 1-based array
getLevels = Application.Transpose(results)
End Function
Example call
Const LVL = 2 ' define level
With Sheet1 ' change to project's sheet Code(Name)
'define data range (assuming columns A:B)
Dim rng As Range
Set rng = .UsedRange.Resize(, 2)
'function call getLevels()
Dim levels
levels = getLevels(rng, level:=LVL)
'write to target
.Columns("I:I").Clear
.Range("I2").Resize(UBound(levels), 1) = levels
End With
Solution without any supporting VBA function:
Function MonstersInLevel(level As Integer) As Variant
With Application.WorksheetFunction
MonstersInLevel = .Filter(Range("LevelMonsters").Columns(2), _
.IfError(.XLookup(Range("LevelMonsters").Columns(1), level, True), False))
End With
End Function
XLookup returns an array of #N/A or True. IfError replaces errors with False. Finally, the Filter function receives an array of booleans as the second parameter.
EDIT
Removed the IfError function thanks to #ScottCraner:
Function MonstersInLevel(level As Integer) As Variant
With Application.WorksheetFunction
MonstersInLevel = .Filter(Range("LevelMonsters").Columns(2), _
.XLookup(Range("LevelMonsters").Columns(1), level, True, False))
End With
End Function
I couldn't resolve your question but as I did some testing on the subject trying to do so, I thought I'd share my findings:
Based on this Microsoft community post, or at least the answers there, it seems you will need to loop through the output in one way or another...
That question seems to want to achieve the same as what you are wanting to do (I think?).
On the other hand, I have never used the WorksheetFunction.Filter method, and the closest I could get it to working was like so:
Here is my sample data - RangeOne is Column A and RangeTwo is Column B. I have used the =FILTER() function in cell C1 evaluating the input in D1 for reference of expected results. Naturally this function is working as expected! The VBA routine is outputting to Columns E, F and G.
Sub TestFilterFunction()
Dim TestArray As Variant
Range("E1:E3") = Application.Filter(Range("RangeTwo"), Range("RangeOne"), Range("D1"))
Range("F1:F3") = Application.Filter(Range("RangeTwo"), Range("RangeOne") = Range("D1")) 'Runtime Error 13
Range("G1:G3") = Application.Filter(Range("RangeTwo"), Range("RangeOne"))
TestArray = Application.Filter(Range("RangeTwo"), Range("RangeOne"), Range("D1"))
TestArray = Application.Filter(Range("RangeTwo"), Range("RangeOne") = Range("D1")) 'Runtime Error 13
TestArray = Application.Filter(Range("RangeTwo"), Range("RangeOne"))
Range("H1:H3") = Application.Filter(Range("RangeTwo", "RangeOne"), Range("RangeOne"), Range("D1"))
TestArray = Application.Filter(Range("A1:B9"), Range("RangeOne"), "2")
End Sub
Column E returned the first 3 values from RangeTwo.
Column F has not been populated - This is because that line threw the Runtime error 13 - Type Mismatch
Column G returned the first 3 values from RangeTwo.
Column H returned the first 3 values from "A1:B9" (both ranges together) - specifically the first 3 values of column A.
I thought this was odd so I threw in an array to assign the values to rather than directly to the worksheet;
The first TestArray line and the third TestArray line both populated the array with the entire RangeTwo values;
I realised with the syntax of the first and third attempt at the WorksheetFunction.Filter, the entire range is returned (that being the first argument - Arg1 - range), but when trying to include the = Range("D1") , it returns the Type Mismatch error.
The final TestArray attempt being the same syntax as the Column H test, returned both columns in a 2D array (now TestArray(1 To 9, 1 To 2)).
I should note I couldn't find any documentation at all on WorksheetFunction.Filter so I'm assuming it does follow the same syntax as the Excel Sheet Function has.
If I find anything more on this topic I'll come back and edit it in, but for now it's looking like perhaps a solution using either loops or Index/Match functions also will need to happen to have the data returned in VBA.
I thought about perhaps writing the sheet formula to a cell and then grab that into an array or something but Excel inserts # into it now which only returns a single cell result, i.e.
Range("J1").Formula = "=FILTER(B1:B9, A1:A9 = D1)"
Would return in J1:
=#FILTER(B1:B9, A1:A9 = D1)
Which with our sample data, would only return 2 in J1 as opposed to the expected/desired 2, 3 and 4 in J1:J3.
I can't work out a way to remove the # as it is applied when the function is written to the cell unfortunately, but hopefully any of the above helps someone find a solution.
Just some comments to help you out.
If you are using the new FILTER() function from either a worksheet cell or within some VBA code, the first argument should be a range and the second argument should a a Boolean array. (if you don't enter something that can evaluate to a Boolean array, VBA may complain the the data type is wrong)
You would be best served (in VBA) if you:
explicitly declared a 2 dimensional, column-compatible, Boolean array
filled the array
used the array in the function call
Here is a super simple example. Say we want to filter the data from A1 to A6 to remove blanks. We could pick a cell and enter:
=FILTER(A1:A6,A1:A6<>"")
Looks like:
Now we want to perform the same activities with a VBA sub and put the result in a block starting with B9. The code:
Sub SingleColumn()
Dim r As Range, wf As WorksheetFunction, i As Long
Dim arr, s As String, dq As String, boo, rc As Long
Set wf = Application.WorksheetFunction
Set r = Range("A1:A6")
rc = r.Rows.Count
ReDim boo(1 To r.Rows.Count, 1 To 1) As Boolean
i = 1
For Each rr In r
If rr.Value = "" Then
boo(i, 1) = False
Else
boo(i, 1) = True
End If
i = i + 1
Next rr
arr = wf.Filter(r, boo)
MsgBox LBound(arr, 1) & "-" & UBound(arr, 1) & vbCrLf & LBound(arr, 2) & "-" & UBound(arr, 2)
Range("B9").Resize(UBound(arr, 1), UBound(arr, 2)) = arr
End Sub
Result:
On Excel version 15.0 (2013), I don't see Application.WorksheetFunction.Filter (tried with Show Hidden Members):
So maybe this is a newer function in later versions ?
My top Google search directs me to this question ;)
So, my answer is to avoid the function primarily from the point of view of backwards compatibility.
Alternate code options presented below returning e.g. a Range and a Variant.
Input:
Code:
Option Explicit
Sub Test()
Dim rngInput As Range
Dim rngFiltered As Range
Dim varFiltered As Variant
Dim varItem As Variant
Set rngInput = ThisWorkbook.Worksheets("Sheet1").Range("A2:B10")
' as range
Debug.Print "' Output as Range"
Set rngFiltered = MonstersInLevel_AsRange(rngInput, 2, 1, 2)
Debug.Print "' " & rngFiltered.Address ' expect B5, B6, B8
Debug.Print "' ---------------"
' as variant
Debug.Print "' Output as Variant"
varFiltered = MonstersInLevel_AsVariant(rngInput, 2, 1, 2)
For Each varItem In varFiltered
Debug.Print "' " & varItem ' expect 3, 4, 5
Next varItem
Debug.Print "' ---------------"
End Sub
Function MonstersInLevel_AsRange(rngToFilter As Range, _
ByVal lngLevel As Long, _
ByVal lngColIxToFilter As Long, _
ByVal lngColIxForValue As Long) As Range
Dim rngResult As Range
Dim lngRowIndex As Long
Dim lngResultIndex As Long
Set rngResult = Nothing
For lngRowIndex = 1 To rngToFilter.Rows.Count
If rngToFilter.Cells(lngRowIndex, lngColIxToFilter) = lngLevel Then
If rngResult Is Nothing Then
Set rngResult = rngToFilter.Cells(lngRowIndex, lngColIxForValue)
Else
Set rngResult = Union(rngResult, rngToFilter.Cells(lngRowIndex, lngColIxForValue))
End If
End If
Next lngRowIndex
Set MonstersInLevel_AsRange = rngResult
End Function
Function MonstersInLevel_AsVariant(rngToFilter As Range, _
ByVal lngLevel As Long, _
ByVal lngColIxToFilter As Long, _
ByVal lngColIxForValue As Long) As Variant
Dim varResult As Variant
Dim lngRowIndex As Long
Dim lngResultIndex As Long
lngResultIndex = 0
ReDim varResult(0)
For lngRowIndex = 1 To rngToFilter.Rows.Count
If rngToFilter.Cells(lngRowIndex, lngColIxToFilter) = lngLevel Then
lngResultIndex = lngResultIndex + 1
ReDim Preserve varResult(1 To lngResultIndex)
varResult(lngResultIndex) = rngToFilter.Cells(lngRowIndex, lngColIxForValue)
End If
Next lngRowIndex
MonstersInLevel_AsVariant = varResult
End Function
Test output:
' Output as Range
' $B$5:$B$6,$B$8
' ---------------
' Output as Variant
' 3
' 5
' 4
' ---------------
Based on Christian Buses answer (https://stackoverflow.com/a/65671334/16578424) I wrote a generic function to use the FILTER-function.
It returns a one-dimensional array with the filtered values.
Public Function getFILTERValuesFromRange(rgResult As Range, rgFilter As Range, varValue As Variant) As Variant
If rgResult.Columns.count > 1 Or rgFilter.Columns.count > 1 Then
Err.Raise vbObjectError + 512, , "Only ranges with one column are allowed."
ElseIf rgResult.Rows.count <> rgFilter.Rows.count Then
Err.Raise vbObjectError + 512, , "Both ranges have to be of the same size."
End If
Dim arr1 As Variant
With Application.WorksheetFunction
arr1 = .filter(rgResult, .XLookup(rgFilter, varValue, True, False))
End With
getFILTERValuesFromRange = getOneDimensionalArrayFromRangeArray(arr1)
End Function
Private Function getOneDimensionalArrayFromRangeArray(arr1 As Variant) As Variant
Dim arr2 As Variant
ReDim arr2(LBound(arr1, 1) To UBound(arr1, 1))
Dim i As Long
For i = 1 To UBound(arr1, 1)
arr2(i) = arr1(i, 1)
Next
getOneDimensionalArrayFromRangeArray = arr2
End Function

VLOOKUP() Alternative using Arrays

I’ve been experimenting with arrays to find a faster alternative to VLOOKUP(), which can take a long time to execute with very large data sets.
I searched SO and many other sites, grabbing snippets of code.
The data:
A1:A5 the list of values to lookup (1,2,3,4,5)
C1:C5 the range to ‘find’ the values (2,4,6,8,10)
D1:D5 the range of values to ‘return’ (a,b,c,d,e)
B1:B5 is where I’d like to paste the ‘looked-up’ values.
The code works up to a point, in that it does return correct values for the ‘looked-up’ value’s position in C1:C5 – and the correct values in the adjacent cells in D1:D5.
When I try to load the returned values to Arr4 (the array to be pasted back to the sheet) which is saying <Type mismatch> when I hover the mouse over it. It doesn’t stop the code from executing, but it doesn’t paste anything.
My questions are:
How do I populate the array Arr4 with the myVal2 values, and
How do I paste it back to the sheet?
Option Explicit
Sub testArray()
Dim ArrLookupValues As Variant
ArrLookupValues = Sheet1.Range("A1:A5") 'The Lookup Values
Dim ArrLookupRange As Variant
ArrLookupRange = Sheet1.Range("C1:C5") 'The Range to find the Value
Dim ArrReturnValues As Variant
ArrReturnValues = Sheet1.Range("D1:D5") 'The adjacent Range to return the Lookup Value
Dim ArrOutput As Variant 'output array
Dim UpperElement As Long
UpperElement = UBound(ArrLookupValues) 'Used purely for the ReDim statement
Dim i As Long
For i = LBound(ArrLookupValues) To UBound(ArrLookupValues)
Dim myVal As Variant
myVal = ArrLookupValues(i, 1)
Dim pos As Variant 'variant becaus it can return an error
pos = Application.Match(myVal, ArrLookupRange, 0) 'pos always returns the correct position
Dim myVal2 As Variant
If Not IsError(pos) Then
myVal2 = ArrReturnValues(pos, 1) 'myVal2 always returns the correct value
ReDim Preserve ArrOutput(1 To UpperElement, 1 To 1)
ArrOutput(i, 1) = myVal2
Else
ReDim Preserve ArrOutput(1 To UpperElement, 1 To 1)
myVal2 = "Not Found"
ArrOutput(i, 1) = myVal2
End If
Next i
Dim Destination As Range
Set Destination = Range("B1")
Destination.Resize(UBound(ArrOutput, 1), UBound(ArrOutput, 2)).Value =
ArrOutput
End Sub
According to #T.M 's answer, you can even do that without looping just by using VLookup instead of Match:
Public Sub testArraya()
With Sheet1
Dim ArrLookupValues() As Variant
ArrLookupValues = .Range("A1:A5").Value ' lookup values 1,2,3,4,5,6
Dim ArrLookupReturnRange() As Variant ' lookup range items 2,4,6,8,10
ArrLookupReturnRange = .Range("C1:D5").Value ' And return column D a,b,c,d,e
End With
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'[1] Match all values at once and return other values of column D
' (found position indices or Error 2042 if not found)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Dim ArrOutput() As Variant
ArrOutput = Application.VLookup(ArrLookupValues, ArrLookupReturnRange, 2, 0)
'[3] write results to any wanted target
Dim Destination As Range
Set Destination = Sheet1.Range("B1") ' ‹‹ change to your needs
Destination.Resize(UBound(ArrOutput, 1), UBound(ArrOutput, 2)).Value = ArrOutput
End Sub
Or even extremly shortened and almost a one liner:
Public Sub testArrayShort()
Const nRows As Long = 5 'amount of rows
With Sheet1
.Range("B1").Resize(nRows).Value = Application.VLookup(.Range("A1").Resize(nRows).Value, .Range("C1:D1").Resize(nRows).Value, 2, 0)
End With
End Sub
Use proper error handling and an If statement instead of On Error Resume Next.
Also your Arr4 needs to be 2 dimensional like your other arrays. Even if it is only one column it needs no be Arr4(1 To UpperElement, 1 To 1) and Arr4(i, 1) = myVal2. Ranges are always 2 dimensional (row, column) even if there is only one column.
And I highly recommend to rename your array variables. When ever you feel like you have to give your variable numbers, you can be sure you are doing it wrong.
Rename them like following for example:
Arr1 --› ArrLookupValues
Arr2 --› ArrLookupRange
Arr3 --› ArrReturnValues
Arr4 --› ArrOutput
This is only a simple modification but your code will extremely gain in human readability and maintainability. You even don't need comments to describe the arrays because their names are self descriptive now.
Finally your output array can be declared the same size as the input arrays. Using ReDim Preserve makes your code slow, so avoid using it.
So something like this should work:
Option Explicit
Public Sub testArray()
Dim ArrLookupValues() As Variant
ArrLookupValues = Sheet1.Range("A1:A5").Value
Dim ArrLookupRange() As Variant
ArrLookupRange = Sheet1.Range("C1:C5").Value
Dim ArrReturnValues() As Variant
ArrReturnValues = Sheet1.Range("D1:D5").Value
Dim UpperElement As Long
UpperElement = UBound(ArrLookupValues, 1)
'create an empty array (same row count as ArrLookupValues)
ReDim ArrOutput(1 To UpperElement, 1 To 1)
Dim i As Long
For i = LBound(ArrLookupValues, 1) To UBound(ArrLookupValues, 1)
Dim FoundAt As Variant 'variant because it can return an error
FoundAt = Application.Match(ArrLookupValues(i, 1), ArrLookupRange, 0) 'pos always returns the correct position
If Not IsError(FoundAt) Then
ArrOutput(i, 1) = ArrReturnValues(FoundAt, 1)
Else
ArrOutput(i, 1) = "Not Found"
End If
Next i
Dim Destination As Range
Set Destination = Range("B1") 'make sure to specify a sheet for that range!
Destination.Resize(UBound(ArrOutput, 1), UBound(ArrOutput, 2)).Value = ArrOutput
End Sub
Just for fun a slight modification of #PEH 's valid approach demonstrating a rather unknown way to excecute a single Match checking both arrays instead of repeated matches:
Public Sub testArray()
With Sheet1
Dim ArrLookupValues As Variant
ArrLookupValues = .Range("A1:A5").Value ' lookup values 1,2,3,4,5,6
Dim ArrLookupRange As Variant ' lookup range items 2,4,6,8,10
ArrLookupRange = .Range("C1:C5").Value
Dim ArrReturnValues As Variant ' return column D a,b,c,d,e
ArrReturnValues = .Range("D1:D5").Value
End With
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'[1] Match all item indices within ArrLookupRange at once
' (found position indices or Error 2042 if not found)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Dim ArrOutput
ArrOutput = Application.Match(ArrLookupValues, ArrLookupRange, 0)
'[2] change indices by return values
Dim i As Long
For i = 1 To UBound(ArrOutput)
If Not IsError(ArrOutput(i, 1)) Then
ArrOutput(i, 1) = ArrReturnValues(ArrOutput(i, 1), 1)
' Else
' ArrOutput(i, 1) = "Not Found" ' optional Not Found statement instead of #NV
End If
Next i
'[3] write results to any wanted target
Dim Destination As Range
Set Destination = Sheet1.Range("B1") '<< change to your needs
Destination.Resize(UBound(ArrOutput, 1), UBound(ArrOutput, 2)).Value = ArrOutput
End Sub

Can't get rid of unwanted items from an array

I'm trying to find out any way to kick out unwanted items from a list. For example, I wish to get rid of 47 and 90 from the list as they do not meet the condition. I've used Delete within the script which is definitely not the right keyword. However, consider it a placeholder.
I've tried with:
Sub DeleteItemConditionally()
Dim numList As Variant, elem As Variant
numList = Array("12", "47", "90", "15", "37")
Debug.Print UBound(numList) - LBound(numList) + 1
For Each elem In numList
If elem >= 40 Then
Delete elem
End If
Next elem
Debug.Print UBound(numList) - LBound(numList) + 1
End Sub
Expected result:
First print : 5 (already getting it)
Second print: 3 (want to achieve it)
Adding and removing additional elements to arrays is rather slow. And changing dimensions of arrays with Redim is one of the slowest operations in VBA. Anyway, if we are talking about up to a decent number of cases, then the speed is ok:
Option Explicit
Sub DeleteItemConditionally()
Dim numList As Variant
numList = Array(12, 47, 90, 15, 3)
Dim newElements() As Variant
Dim firstElement As Boolean: firstElement = True
Dim i As Long
For i = LBound(numList) To UBound(numList)
If numList(i) <= 40 Then
If firstElement Then
ReDim Preserve newElements(0)
firstElement = False
Else
ReDim Preserve newElements(UBound(newElements) + 1)
End If
newElements(UBound(newElements)) = numList(i)
End If
Next
Dim element As Variant
For Each element In newElements
Debug.Print element
Next
End Sub
With a Collection or with System.Collections.ArrayList as in the case below, the optimization and the speed would be way faster (but still slightly invisible, if the data is not more than a few hundred items). Additionally, a collection could be sorted rather quickly and then the speed of the task will be even better:
Sub TestMyCollection()
Dim myList As Object
Set myList = CreateObject("System.Collections.ArrayList")
With myList
.Add 12
.Add 47
.Add 90
.Add 15
.Add 3
End With
myList.Sort
Dim i As Long
For i = myList.Count - 1 To 0 Step -1
If Not myList.Item(i) <= 40 Then
myList.RemoveAt i
End If
Next i
Dim element As Variant
For Each element In myList
Debug.Print element
Next
End Sub
Additionally, to increase performance and to get some good usage of the .Sort() after the first number, bigger than 40 the For i = myList.Count - 1 To 0 Step -1 could exit.
If you are using a single dimension array to represent a list then you will be much better served by replacing your array with a collection (or if you wish to be more advanced a Scripting.Dictionary).
If you replace your array with a collection then essentially you don't need to make any significant changes to your code. Just a few minor tweaks to compensate for the fact that you can't query a collection to get the index of an item so you have to iterate by index rather than by item in your particular case.
I've updated your code to add a function that replaces the Array method by returning a populated Collection, and updates the loop to use indexing. You should also note that the indexing loop counts down. This is because if we remove an item from a collection, the size will no longer be the count we obtained at the start of the loop.
Sub DeleteItemConditionally()
Dim my_num_list As Collection, my_item_index As Long
Set my_num_list = FilledCollection("12", "47", "90", "15", "37")
Debug.Print my_num_list.Count
For my_item_index = my_num_list.Count To 1 Step -1
If my_num_list(my_item_index) >= 40 Then
my_num_list.Remove my_item_index
End If
Next
Debug.Print my_num_list.Count
End Sub
Public Function FilledCollection(ParamArray args() As Variant) As Collection
Dim my_return As Collection
Dim my_item As Variant
Set my_return = New Collection
For Each my_item In args
my_return.Add my_item
Next
Set FilledCollection = my_return
End Function
Note: This answer focuses on the question asked: how to conditionally delete items from an Array. Other answers deal with some of the many alternatives.
Your data. You've created an array of Strings then compared them to a Number. That won't work (well, it will give an answer, but it won't be what you expect). I've changed your data to Numbers
I've created the Delete functionality as a Function that returns a possibly reduced array. It only accepts 1D arrays (if anythig else is passed, the passed parameter is returned)
I've borrowed a couple of Utility Functions from CPearson.Com - BTW thats a great resource for all things VBA
I've included some flexibility for the test type, (>= or <) - you could add more if you want.
Speed. Whether or not this is fast enough depends on your use case. I've tested it as follows - Array Size of 5 run 1000 time in 3.9 mS. Array size of 10,000 runs 1000 times 586 mS
Included is an alternate version that can apply several of multiple conditions, >, >= <, <= a value must pass all tests to be kept (obviously, only certain conbinations make sence)
Sub Test()
Dim numList As Variant
numList = Array(12, 47, 90, 15, 37)
Debug.Print UBound(numList) - LBound(numList) + 1
numList = DeleteItemConditionally(numList, 40) ' Delete >= 40
Debug.Print UBound(numList) - LBound(numList) + 1
End Sub
' Only 1 condition may be supplied
Function DeleteItemConditionally(Arr As Variant, Optional DeleteGEQ As Variant, Optional DeleteLES As Variant) As Variant
Dim NewArr As Variant
Dim iArr As Long, iNewArr As Long
' Check if Arr is valid
If Not IsArrayAllocated(Arr) Then GoTo AbortExit
If NumberOfArrayDimensions(Arr) <> 1 Then GoTo AbortExit
' that one and only one of Delete criteria is specified
If Not (IsMissing(DeleteGEQ) Xor IsMissing(DeleteLES)) Then GoTo AbortExit
ReDim NewArr(LBound(Arr) To UBound(Arr))
If Not IsMissing(DeleteGEQ) Then
' Delete members >= DeleteGEQ
iNewArr = LBound(Arr) - 1
For iArr = LBound(Arr) To UBound(Arr)
If Arr(iArr) < DeleteGEQ Then
iNewArr = iNewArr + 1
NewArr(iNewArr) = Arr(iArr)
End If
Next
Else
' Delete members < DeleteGEQ
iNewArr = LBound(Arr) - 1
For iArr = LBound(Arr) To UBound(Arr)
If Arr(iArr) >= DeleteGEQ Then
iNewArr = iNewArr + 1
NewArr(iNewArr) = Arr(iArr)
End If
Next
End If
' ReDim Preserve is an expensive function, do it only once
ReDim Preserve NewArr(LBound(Arr) To iNewArr)
DeleteItemConditionally = NewArr
Exit Function
AbortExit:
On Error Resume Next
DeleteItemConditionally = Arr
End Function
' Several conditions may be supplied
Function DeleteItemConditionally2(Arr As Variant, Optional KeepGEQ As Variant, Optional KeepGRT As Variant, Optional KeepLEQ As Variant, Optional KeepLES As Variant) As Variant
Dim NewArr As Variant
Dim iArr As Long, iNewArr As Long
Dim Keep As Boolean
' Check if Arr is valid
If Not IsArrayAllocated(Arr) Then GoTo AbortExit
If NumberOfArrayDimensions(Arr) <> 1 Then GoTo AbortExit
ReDim NewArr(LBound(Arr) To UBound(Arr))
iNewArr = LBound(Arr) - 1
For iArr = LBound(Arr) To UBound(Arr)
Keep = True
If Not IsMissing(KeepGEQ) Then
' Keep members >= KeepGEQ
If Arr(iArr) < KeepGEQ Then
Keep = False
End If
End If
If Keep And Not IsMissing(KeepGRT) Then
' Keep members > KeepGRT
If Arr(iArr) <= KeepGRT Then
Keep = False
End If
End If
If Keep And Not IsMissing(KeepLEQ) Then
' Keep members <= KeepLEQ
If Arr(iArr) > KeepLEQ Then
Keep = False
End If
End If
If Keep And Not IsMissing(KeepLES) Then
' Keep members < KeepLES
If Arr(iArr) >= KeepGRT Then
Keep = False
End If
End If
If Keep Then
iNewArr = iNewArr + 1
NewArr(iNewArr) = Arr(iArr)
End If
Next
' ReDim Preserve is an expensive function, do it only once
ReDim Preserve NewArr(LBound(Arr) To iNewArr)
DeleteItemConditionally2 = NewArr
Exit Function
AbortExit:
On Error Resume Next
DeleteItemConditionally2 = Arr
End Function
Public Function IsArrayAllocated(Arr As Variant) As Boolean
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
' IsArrayAllocated
' Returns TRUE if the array is allocated (either a static array or a dynamic array that has been
' sized with Redim) or FALSE if the array is not allocated (a dynamic that has not yet
' been sized with Redim, or a dynamic array that has been Erased). Static arrays are always
' allocated.
'
' The VBA IsArray function indicates whether a variable is an array, but it does not
' distinguish between allocated and unallocated arrays. It will return TRUE for both
' allocated and unallocated arrays. This function tests whether the array has actually
' been allocated.
'
' This function is just the reverse of IsArrayEmpty.
'
' From http://www.cpearson.com/Excel/VBAArrays.htm
'
'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
Dim N As Long
On Error Resume Next
' if Arr is not an array, return FALSE and get out.
If IsArray(Arr) = False Then
IsArrayAllocated = False
Exit Function
End If
' Attempt to get the UBound of the array. If the array has not been allocated,
' an error will occur. Test Err.Number to see if an error occurred.
N = UBound(Arr, 1)
If (Err.Number = 0) Then
''''''''''''''''''''''''''''''''''''''
' Under some circumstances, if an array
' is not allocated, Err.Number will be
' 0. To acccomodate this case, we test
' whether LBound <= Ubound. If this
' is True, the array is allocated. Otherwise,
' the array is not allocated.
'''''''''''''''''''''''''''''''''''''''
If LBound(Arr) <= UBound(Arr) Then
' no error. array has been allocated.
IsArrayAllocated = True
Else
IsArrayAllocated = False
End If
Else
' error. unallocated array
IsArrayAllocated = False
End If
End Function
Public Function NumberOfArrayDimensions(Arr As Variant) As Integer
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
' NumberOfArrayDimensions
' This function returns the number of dimensions of an array. An unallocated dynamic array
' has 0 dimensions. This condition can also be tested with IsArrayEmpty.
'
' From http://www.cpearson.com/Excel/VBAArrays.htm
'
'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
Dim Ndx As Integer
Dim Res As Integer
On Error Resume Next
' Loop, increasing the dimension index Ndx, until an error occurs.
' An error will occur when Ndx exceeds the number of dimension
' in the array. Return Ndx - 1.
Do
Ndx = Ndx + 1
Res = UBound(Arr, Ndx)
Loop Until Err.Number <> 0
NumberOfArrayDimensions = Ndx - 1
End Function

Finding if a string is in a 2 dimensional VBA Excel array

I have a great function that I use all of the time for a 1 dimensional Excel VBA array that checks if a string is in an array:
Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
IsInArray = (UBound(Filter(arr(), stringToBeFound)) > -1)
End Function
Unfortunately it does not work when using it to check for a 2 dimensional array, like I have here:
Sub new_idea_filter()
home_sheet = ActiveSheet.Name
c = 1
Dim myfilters(1 To 4, 1 To 5000)
myfilters(1, 4) = "Test"
If IsInArray("Test", myfilters()) = True Then
killer = True
End If
End Sub
Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
IsInArray = (UBound(Filter(arr(), stringToBeFound)) > -1)
End Function
It keeps erroring out in the function saying subscript out of range, anyone have a thought how I can check if a string is in the 2 dimensional array?
Something from my code collection
You can use Application.Match. This will work for both 1D and 2D array :)
See this
Sub Sample()
Dim myfilters(1 To 4, 1 To 5000)
myfilters(1, 4) = "Test"
If IsInArray("Test", myfilters()) = True Then MsgBox "Found"
End Sub
Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
Dim bDimen As Byte, i As Long
On Error Resume Next
If IsError(UBound(arr, 2)) Then bDimen = 1 Else bDimen = 2
On Error GoTo 0
Select Case bDimen
Case 1
On Error Resume Next
IsInArray = Application.Match(stringToBeFound, arr, 0)
On Error GoTo 0
Case 2
For i = 1 To UBound(arr, 2)
On Error Resume Next
IsInArray = Application.Match(stringToBeFound, Application.Index(arr, , i), 0)
On Error GoTo 0
If IsInArray = True Then Exit For
Next
End Select
End Function
As long as you're in Excel (or have a reference to it), you can use the Index function to slice your array into rows or columns.
Public Function IsInArray(ByVal vToFind As Variant, vArr As Variant) As Boolean
Dim i As Long
Dim bReturn As Boolean
Dim vLine As Variant
For i = LBound(vArr, 1) To UBound(vArr, 1)
vLine = Application.WorksheetFunction.Index(vArr, i) 'slice off one line
If IsArray(vLine) Then 'if it's an array, use the filter
bReturn = UBound(Filter(vLine, vToFind)) > -1
Else 'if it's not an array, it was 1d so check the value
bReturn = vLine = vToFind
End If
If bReturn Then Exit For 'stop looking if one found
Next i
IsInArray = bReturn
End Function
Public Sub test()
Dim arr() As Variant
ReDim arr(1 To 2, 1 To 2)
arr(1, 2) = "Test"
Debug.Assert IsInArray("Test", arr)
arr(1, 2) = "Wrong"
Debug.Assert Not IsInArray("Test", arr)
ReDim arr(1 To 3)
arr(2) = "Test"
Debug.Assert IsInArray("Test", arr)
arr(2) = "Wrong"
Debug.Assert Not IsInArray("Test", arr)
Debug.Print "Passed"
End Sub
If you get the data from a recordset i use this method; first i use GetString for the recordset, second use Split to convert the string in a array unidimensional where each item is a string with all the information. After that you con use the function IsInArray.
The code is:
RecSet.Open strSQL, Cn
RecSet.MoveFirst
RecString = RecSet.GetString(, , ";", vbCr) 'genera una cadena con los datos. Campos separados por ; y registros por vbCR
RecSplit = Split(RecString, vbCr) 'Genera un array unidimensional con la cadena
you can test the code, but remember only works if you get the data from a recordset
You can try converting your original Function to be able to work with arrays. Please try the following, though note that I have not tested if it works.
Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
Dim cell As Variant
For Each cell In arr
IsInArray = IsInArray Or (UBound(Filter(cell(), stringToBeFound)) > -1)
Next
End Function
Regards
#Siddharth-Rout answer above is working perfectly with Application.Match in addition to the Filter function :-). - My solution tries to use the OP Filter function only: As the filter function needs a 1dim array, the array is splitted into portions.
A) Alternative solution using the original FILTER function instead of Match plus error handling
Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
Dim i As Long
If nDim(arr) = 1 Then
IsInArray = (UBound(Filter(arr(), stringToBeFound)) > -1)
Else ' allows using filter function in portions
For i = 1 To UBound(arr, 2)
If (UBound(Filter(Application.Transpose(Application.Index(arr, 0, i)), stringToBeFound)) > -1) Then IsInArray = True: Exit For
Next i
End If
End Function
Helper function to get array Dimension
Function nDim(ByVal vArray As Variant) As Long
' Purp: get number of array dimensions
' Site: http://support.microsoft.com/kb/152288
Dim dimnum As Long
Dim ErrorCheck As Variant
On Error GoTo FinalDimension
For dimnum = 1 To 60000
ErrorCheck = LBound(vArray, dimnum)
Next
FinalDimension:
nDim = dimnum - 1
End Function
B) Recursive solution using the original FILTER function instead of Match plus error handling
Function IsInArray(stringToBeFound As String, arr As Variant, Optional i As Long = 0) As Boolean
Select Case i
Case -1: ' stop 2dim calls
Case 0: IsInArray = IsInArray(stringToBeFound, arr, nDim(arr)) ' start recursive call
Case 1: IsInArray = (UBound(Filter(arr(), stringToBeFound)) > -1) ' 1dim array
Case Else ' allows using filter function in portions
If (UBound(Filter(Application.Transpose(Application.Index(arr, 0, i)), stringToBeFound)) > -1) Then
IsInArray = True
Else ' recursive calls (2dim array)
IsInArray = IsInArray(stringToBeFound, arr, IIf(i + 1 > UBound(arr), -1, i + 1))
End If
End Select
End Function
I have an Excel users version solution for this as well.
Cant you just split concatenate the array into a single column (1-d array)? you got x columns. who cares about the # of rows for now.
I would do : col 1 & "/// unique character delimiter"& col#1 & col 2 & "/// unique character delimiter"& col#2 & col 3 & "/// unique character delimiter"& col#2 & ... & & col (n-1) & "/// unique character delimiter"& col#(n-1) & & "/// unique character delimiter"& col#n
turning the 2-d array into a 1-d array.
and index match this joined-up array/column, to find the multiple occurances of the string located in the original array.
And whats good about this, because of the unique way you joined it (any unique delimator charavter + col# ) it can and will also tell you the original column each found return value of the string your looking for resided in. SO you dont loose any information.
(you can do that implementing =match ("/"&string&"/")) the position of the looked-for text in the lookup output and the next occurrence of the special unique delimiter & the next (subsequent) col # that's to the right of it.
Doesn't this do the same thing , as the macros above or the question asks for ?
and in an (almost) non-macro*/non-vba way?
*see below for why it can be done with out without macros.
So in the end, you can just turn any 2-d N.M array into an 1-d X array, while keeping all the information (of which column the text was originally belonging to) and still do a simple lookup, index-match or a LoopALL function (which is great) :
Lookupall macro to use to find and return multiple found occurrences of string:
Function LookupAll(vVal, rTable As Range, ColumnI As Long) As Variant
Dim rFound As Range, lLoop As Long
Dim strResults As String
With rTable.Columns(1)
Set rFound = .Cells(1, 1)
For lLoop = 1 To WorksheetFunction.CountIf(.Cells, vVal)
Set rFound = .Find(what:=vVal, After:=rFound, LookIn:=xlFormulas, lookAt _
:=xlWhole, SearchOrder:=xlByRows, SearchDirection:=xlNext, MatchCase:= _
False, SearchFormat:=False)
strResults = strResults & "," & rFound(1, ColumnI)
Next lLoop
End With
LookupAll = Trim(Right(strResults, Len(strResults) - 1))
End Function
Up to you whether you use VBA lookup all function above or an index-match formula in excel which can find and return multiple occurrences of a search find.
Delimation and join of separate columns of an array strips a need for an array search (which I've never been able to do as I wanted - ie. get the results all into 1 cell), and turns it into a single and simpler 1-d array without any information loss.
I believe the speed would be as fast (and accurate) as anything else. Particularly as you've reduced/condensed the array into a single array - 1 column.
Any thoughts?

ReDim Preserve to a multi-dimensional array in VB6

I'm using VB6 and I need to do a ReDim Preserve to a Multi-Dimensional Array:
Dim n, m As Integer
n = 1
m = 0
Dim arrCity() As String
ReDim arrCity(n, m)
n = n + 1
m = m + 1
ReDim Preserve arrCity(n, m)
Whenever I do it as I have written it, I get the following error:
runtime error 9: subscript out of range
Because I can only change the last array dimension, well in my task I have to change the whole array (2 dimensions in my example) !
Is there any workaround or another solution for this?
As you correctly point out, one can ReDim Preserve only the last dimension of an array (ReDim Statement on MSDN):
If you use the Preserve keyword, you can resize only the last array
dimension and you can't change the number of dimensions at all. For
example, if your array has only one dimension, you can resize that
dimension because it is the last and only dimension. However, if your
array has two or more dimensions, you can change the size of only the
last dimension and still preserve the contents of the array
Hence, the first issue to decide is whether 2-dimensional array is the best data structure for the job. Maybe, 1-dimensional array is a better fit as you need to do ReDim Preserve?
Another way is to use jagged array as per Pieter Geerkens's suggestion. There is no direct support for jagged arrays in VB6. One way to code "array of arrays" in VB6 is to declare an array of Variant and make each element an array of desired type (String in your case). Demo code is below.
Yet another option is to implement Preserve part on your own. For that you'll need to create a copy of data to be preserved and then fill redimensioned array with it.
Option Explicit
Public Sub TestMatrixResize()
Const MAX_D1 As Long = 2
Const MAX_D2 As Long = 3
Dim arr() As Variant
InitMatrix arr, MAX_D1, MAX_D2
PrintMatrix "Original array:", arr
ResizeMatrix arr, MAX_D1 + 1, MAX_D2 + 1
PrintMatrix "Resized array:", arr
End Sub
Private Sub InitMatrix(a() As Variant, n As Long, m As Long)
Dim i As Long, j As Long
Dim StringArray() As String
ReDim a(n)
For i = 0 To n
ReDim StringArray(m)
For j = 0 To m
StringArray(j) = i * (m + 1) + j
Next j
a(i) = StringArray
Next i
End Sub
Private Sub PrintMatrix(heading As String, a() As Variant)
Dim i As Long, j As Long
Dim s As String
Debug.Print heading
For i = 0 To UBound(a)
s = ""
For j = 0 To UBound(a(i))
s = s & a(i)(j) & "; "
Next j
Debug.Print s
Next i
End Sub
Private Sub ResizeMatrix(a() As Variant, n As Long, m As Long)
Dim i As Long
Dim StringArray() As String
ReDim Preserve a(n)
For i = 0 To n - 1
StringArray = a(i)
ReDim Preserve StringArray(m)
a(i) = StringArray
Next i
ReDim StringArray(m)
a(n) = StringArray
End Sub
Since VB6 is very similar to VBA, I think I might have a solution which does not require this much code to ReDim a 2-dimensional array - using Transpose, if you are working in Excel.
The solution (Excel VBA):
Dim n, m As Integer
n = 2
m = 1
Dim arrCity() As Variant
ReDim arrCity(1 To n, 1 To m)
m = m + 1
ReDim Preserve arrCity(1 To n, 1 To m)
arrCity = Application.Transpose(arrCity)
n = n + 1
ReDim Preserve arrCity(1 To m, 1 To n)
arrCity = Application.Transpose(arrCity)
What is different from OP's question: the lower bound of arrCity array is not 0, but 1. This is in order to let Application.Transpose do it's job.
Note that Transpose is a method of the Excel Application object (which in actuality is a shortcut to Application.WorksheetFunction.Transpose). And in VBA, one must take care when using Transpose as it has two significant limitations: If the array has more than 65536 elements, it will fail. If ANY element's length exceed 256 characters, it will fail. If neither of these is an issue, then Transpose will nicely convert the rank of an array form 1D to 2D or vice-versa.
Unfortunately there is nothing like 'Transpose' build into VB6.
In regards to this:
"in my task I have to change the whole array (2 dimensions"
Just use a "jagged" array (ie an array of arrays of values). Then you can change the dimensions as you wish. You can have a 1-D array of variants, and the variants can contain arrays.
A bit more work perhaps, but a solution.
I haven't tested every single one of these answers but you don't need to use complicated functions to accomplish this. It's so much easier than that! My code below will work in any office VBA application (Word, Access, Excel, Outlook, etc.) and is very simple. Hope this helps:
''Dimension 2 Arrays
Dim InnerArray(1 To 3) As Variant ''The inner is for storing each column value of the current row
Dim OuterArray() As Variant ''The outer is for storing each row in
Dim i As Byte
i = 1
Do While i <= 5
''Enlarging our outer array to store a/another row
ReDim Preserve OuterArray(1 To i)
''Loading the current row column data in
InnerArray(1) = "My First Column in Row " & i
InnerArray(2) = "My Second Column in Row " & i
InnerArray(3) = "My Third Column in Row " & i
''Loading the entire row into our array
OuterArray(i) = InnerArray
i = i + 1
Loop
''Example print out of the array to the Intermediate Window
Debug.Print OuterArray(1)(1)
Debug.Print OuterArray(1)(2)
Debug.Print OuterArray(2)(1)
Debug.Print OuterArray(2)(2)
I know this is a bit old but I think there might be a much simpler solution that requires no additional coding:
Instead of transposing, redimming and transposing again, and if we talk about a two dimensional array, why not just store the values transposed to begin with. In that case redim preserve actually increases the right (second) dimension from the start. Or in other words, to visualise it, why not store in two rows instead of two columns if only the nr of columns can be increased with redim preserve.
the indexes would than be 00-01, 01-11, 02-12, 03-13, 04-14, 05-15 ... 0 25-1 25 etcetera instead of 00-01, 10-11, 20-21, 30-31, 40-41 etcetera.
As long as there is only one dimension that needs to be redimmed-preserved the approach would still work: just put that dimension last.
As only the second (or last) dimension can be preserved while redimming, one could maybe argue that this is how arrays are supposed to be used to begin with.
I have not seen this solution anywhere so maybe I'm overlooking something?
(Posted earlier on similar question regarding two dimensions, extended answer here for more dimensions)
You can use a user defined type containing an array of strings which will be the inner array. Then you can use an array of this user defined type as your outer array.
Have a look at the following test project:
'1 form with:
' command button: name=Command1
' command button: name=Command2
Option Explicit
Private Type MyArray
strInner() As String
End Type
Private mudtOuter() As MyArray
Private Sub Command1_Click()
'change the dimensens of the outer array, and fill the extra elements with "1"
Dim intOuter As Integer
Dim intInner As Integer
Dim intOldOuter As Integer
intOldOuter = UBound(mudtOuter)
ReDim Preserve mudtOuter(intOldOuter + 2) As MyArray
For intOuter = intOldOuter + 1 To UBound(mudtOuter)
ReDim mudtOuter(intOuter).strInner(intOuter) As String
For intInner = 0 To UBound(mudtOuter(intOuter).strInner)
mudtOuter(intOuter).strInner(intInner) = "1"
Next intInner
Next intOuter
End Sub
Private Sub Command2_Click()
'change the dimensions of the middle inner array, and fill the extra elements with "2"
Dim intOuter As Integer
Dim intInner As Integer
Dim intOldInner As Integer
intOuter = UBound(mudtOuter) / 2
intOldInner = UBound(mudtOuter(intOuter).strInner)
ReDim Preserve mudtOuter(intOuter).strInner(intOldInner + 5) As String
For intInner = intOldInner + 1 To UBound(mudtOuter(intOuter).strInner)
mudtOuter(intOuter).strInner(intInner) = "2"
Next intInner
End Sub
Private Sub Form_Click()
'clear the form and print the outer,inner arrays
Dim intOuter As Integer
Dim intInner As Integer
Cls
For intOuter = 0 To UBound(mudtOuter)
For intInner = 0 To UBound(mudtOuter(intOuter).strInner)
Print CStr(intOuter) & "," & CStr(intInner) & " = " & mudtOuter(intOuter).strInner(intInner)
Next intInner
Print "" 'add an empty line between the outer array elements
Next intOuter
End Sub
Private Sub Form_Load()
'init the arrays
Dim intOuter As Integer
Dim intInner As Integer
ReDim mudtOuter(5) As MyArray
For intOuter = 0 To UBound(mudtOuter)
ReDim mudtOuter(intOuter).strInner(intOuter) As String
For intInner = 0 To UBound(mudtOuter(intOuter).strInner)
mudtOuter(intOuter).strInner(intInner) = CStr((intOuter + 1) * (intInner + 1))
Next intInner
Next intOuter
WindowState = vbMaximized
End Sub
Run the project, and click on the form to display the contents of the arrays.
Click on Command1 to enlarge the outer array, and click on the form again to show the results.
Click on Command2 to enlarge an inner array, and click on the form again to show the results.
Be careful though: when you redim the outer array, you also have to redim the inner arrays for all the new elements of the outer array
I stumbled across this question while hitting this road block myself. I ended up writing a piece of code real quick to handle this ReDim Preserve on a new sized array (first or last dimension). Maybe it will help others who face the same issue.
So for the usage, lets say you have your array originally set as MyArray(3,5), and you want to make the dimensions (first too!) larger, lets just say to MyArray(10,20). You would be used to doing something like this right?
ReDim Preserve MyArray(10,20) '<-- Returns Error
But unfortunately that returns an error because you tried to change the size of the first dimension. So with my function, you would just do something like this instead:
MyArray = ReDimPreserve(MyArray,10,20)
Now the array is larger, and the data is preserved. Your ReDim Preserve for a Multi-Dimension array is complete. :)
And last but not least, the miraculous function: ReDimPreserve()
'redim preserve both dimensions for a multidimension array *ONLY
Public Function ReDimPreserve(aArrayToPreserve,nNewFirstUBound,nNewLastUBound)
ReDimPreserve = False
'check if its in array first
If IsArray(aArrayToPreserve) Then
'create new array
ReDim aPreservedArray(nNewFirstUBound,nNewLastUBound)
'get old lBound/uBound
nOldFirstUBound = uBound(aArrayToPreserve,1)
nOldLastUBound = uBound(aArrayToPreserve,2)
'loop through first
For nFirst = lBound(aArrayToPreserve,1) to nNewFirstUBound
For nLast = lBound(aArrayToPreserve,2) to nNewLastUBound
'if its in range, then append to new array the same way
If nOldFirstUBound >= nFirst And nOldLastUBound >= nLast Then
aPreservedArray(nFirst,nLast) = aArrayToPreserve(nFirst,nLast)
End If
Next
Next
'return the array redimmed
If IsArray(aPreservedArray) Then ReDimPreserve = aPreservedArray
End If
End Function
I wrote this in like 20 minutes, so there's no guarantees. But if you would like to use or extend it, feel free. I would've thought that someone would've had some code like this up here already, well apparently not. So here ya go fellow gearheads.
This is more compact and respect the intial first position in array and just use the inital bound to add old value.
Public Sub ReDimPreserve(ByRef arr, ByVal size1 As Long, ByVal size2 As Long)
Dim arr2 As Variant
Dim x As Long, y As Long
'Check if it's an array first
If Not IsArray(arr) Then Exit Sub
'create new array with initial start
ReDim arr2(LBound(arr, 1) To size1, LBound(arr, 2) To size2)
'loop through first
For x = LBound(arr, 1) To UBound(arr, 1)
For y = LBound(arr, 2) To UBound(arr, 2)
'if its in range, then append to new array the same way
arr2(x, y) = arr(x, y)
Next
Next
'return byref
arr = arr2
End Sub
I call this sub with this line to resize the first dimension
ReDimPreserve arr2, UBound(arr2, 1) + 1, UBound(arr2, 2)
You can add an other test to verify if the initial size is not upper than new array. In my case it's not necessary
Easiest way to do this in VBA is to create a function that takes in an array, your new amount of rows, and new amount of columns.
Run the below function to copy in all of the old data back to the array after it has been resized.
function dynamic_preserve(array1, num_rows, num_cols)
dim array2 as variant
array2 = array1
reDim array1(1 to num_rows, 1 to num_cols)
for i = lbound(array2, 1) to ubound(array2, 2)
for j = lbound(array2,2) to ubound(array2,2)
array1(i,j) = array2(i,j)
next j
next i
dynamic_preserve = array1
end function
Function Redim2d(ByRef Mtx As Variant, ByVal QtyColumnToAdd As Integer)
ReDim Preserve Mtx(LBound(Mtx, 1) To UBound(Mtx, 1), LBound(Mtx, 2) To UBound(Mtx, 2) + QtyColumnToAdd)
End Function
'Main Code
sub Main ()
Call Redim2d(MtxR8Strat, 1) 'Add one column
end sub
'OR
sub main2()
QtyColumnToAdd = 1 'Add one column
ReDim Preserve Mtx(LBound(Mtx, 1) To UBound(Mtx, 1), LBound(Mtx, 2) To UBound(Mtx, 2) + QtyColumnToAdd)
end sub
If you not want include other function like 'ReDimPreserve' could use temporal matrix for resizing. On based to your code:
Dim n As Integer, m As Integer, i as Long, j as Long
Dim arrTemporal() as Variant
n = 1
m = 0
Dim arrCity() As String
ReDim arrCity(n, m)
n = n + 1
m = m + 1
'VBA automatically adapts the size of the receiving matrix.
arrTemporal = arrCity
ReDim arrCity(n, m)
'Loop for assign values to arrCity
For i = 1 To UBound(arrTemporal , 1)
For j = 1 To UBound(arrTemporal , 2)
arrCity(i, j) = arrTemporal (i, j)
Next
Next
If you not declare of type VBA assume that is Variant.
Dim n as Integer, m As Integer

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