How to get index of Element in two dimensional array? - arrays

Question

I have an array with 2 dimensions.
I also have a For Each loops which loops with elements of these arrays.
How can i get a Index of vElement/vElement2 in the moment of my comment here in code?
I would be very, very thankful if You can help me.
For Each vElement In Table1
For Each vElement2 In Table2
If ws_1.Cells(1, c) = vElement Then
For Row = 3 To lastRow
amountValue = amountValue + ws_1.Cells(Row, c).value
ws_2.Cells(row2, colIlosc) = amountValue
'Here i would love to have index of vElement for example. In my head it would be something like... Index(vElement) or Index(Table1(vElement))
ws_2.Cells(row2, columncodeprod) = vElement2
row2 = row2 + 1
amountValue = 0
Next Row
End If
Next vElement2
Next vElement

Answer 1

Show Indices of an element in a 2-dim Array - the complicated way
If I understand correctly, you are looping through a datafield array via a ►For Each construction and want to get the current row/column index pair of that same array.
In order to answer your question
"How to get indices of an element in a two dimensional array",
I leave aside that you would get these automatically in a more evident and usual way if you changed the logic by looping through array rows first and inside this loop eventually through array columns - see Addendum *).
To allow a reconstruction of e.g. the 6th array element in the example call below as referring to the current index pair (element i=6 ~> table1(3,2) ~> row:=3/column:=2) it would be necessary
to add an element counter i by incrementing its value by +1 each time you get the next element and
to pass this counter as argument (additionally to a reference to the datafield) to a help function getIndex()
returning results as another array, i.e. an array consisting only of two values: (1) the current array row, (2) the current array column:
Example call
Note: For better readibility and in order to condense the answer to the mimimum needed (c.f. MCVE) the following example call executes only one For Each loop over the table1 datafield array; you will be in the position to change this to your needs or to ask another question.
Option Explicit ' declaration head of your code module
Sub ShowIndicesOf2DimArray()
Dim table1 ' declare variant 1-based 2-dim datafield
table1 = Sheet1.Range("A2:B4") ' << change to sheets Code(Name)
Dim vElem, i As Long
Dim curRow As Long, curCol As Long ' current row/column number
For Each vElem In table1
i = i + 1 ' increment element counter
curRow = getIndex(table1, i)(1) ' <~ get row index via help function
curCol = getIndex(table1, i)(2) ' <~ get col index via help function
'optional debug info in VB Editors immediate window (here: Direktbereich)
Debug.Print i & ". " & _
" Table1(" & curRow & "," & curCol & ") = " & vElem & vbTab;
Debug.Print ", where curRow|curCol are " & Join(getIndex(table1, i), "|")
Next vElem
End Sub
Help function getIndex() called by above procedure
Function getIndex(table1, ByVal no As Long) As Variant
'Purpose: get 1-based 1-dim array with current row+column indices
ReDim tmp(1 To 2)
tmp(1) = (no - 1) Mod UBound(table1) + 1
tmp(2) = Int((no - 1) / UBound(table1) + 1)
getIndex = tmp
End Function
*) Addendum - "the simple way"
Just the other way round using row and column variables r and c as mentioned above; allows to refer to an item simply via table1(r,c) :
Sub TheSimpleWay()
Dim table1 ' declare variant 1-based 2-dim datafield
table1 = Sheet1.Range("A2:B4") ' << change to sheets Code(Name)
Dim vElem, i As Long
Dim r As Long, c As Long ' row and column counter
For r = 1 To UBound(table1) ' start by row 1 (1-based!) up to upper boundary in 1st dimension
For c = 1 To UBound(table1, 2) ' start by col 1 (1-based!) up to upper boundary in 2nd dimension
i = i + 1
Debug.Print i & ". " & _
" Table1(" & r & "," & c & ") = " & table1(r, c) & vbTab;
Debug.Print ", where row|col are " & r & "|" & c
Next c
Next r
End Sub

Answer 2

There is NO index in the case you put in discussion...
vElement and vElement2 variables are of the Variant type. They are not objects, to have an Index property.
When you use a For Each vElement In Table1 loop, VBA starts from the array first element, goes down up to the last row and then do the same for the next column.
When you need to know what you name arrays 'indexes' you must use For i = 1 To Ubound(Table1, 1) followed by For j = 1 To Ubound(Table1, 2). In such a case you will know the matching array element row and columns. We can consider them your pseudo-indexes...
If you really want/insist to extract such indexes in an iteration of type For Each vElement In Table1, you must build them. I will try en elocvent code example:
Sub testElemIndex()
Dim sh As Worksheet, Table1 As Variant, vElement As Variant
Dim i As Long, indexRow As Long, indexCol
Set sh = ActiveSheet
sh.Range("C6").value = "TestIndex"
Table1 = sh.Range("A1:E10").value
For Each vElement In Table1
i = i + 1:
If vElement = "TestIndex" Then
If i <= UBound(Table1, 1) Then
indexRow = i: indexCol = 1
Else
indexCol = Int(i / UBound(Table1, 1)) + 1
indexRow = i - Int(i / UBound(Table1, 1)) * UBound(Table1, 1)
End If
Debug.Print Table1(indexRow, indexCol), indexRow, indexCol: Stop
End If
Next
End Sub
You can calculate the rows and columns of the array element. And the code proves that using them, the returned array value is exactly the found one...
Is it a little more light on the array 'indexes'...?

Answer 3

Dim Table1() As Variant
Dim Table2() As Variant
Table1 = Range(Cells(2, 3), Cells(lastRow, vMaxCol))
Table2 = Range(Cells(2, 1), Cells(lastRow, 1))
Table1 is Variant(1 to 33, 1 to 9)
Table2 is Variant(1 to 33, 1 to 1)
This 33 and 9 is dynamic.