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
Related
I have a strings in column "C", starting at C2 (for example: Cat, Dog, Bird, etc...) and I don't know how many. So I am using a LRow function to find the last row with data. Currently, the last row is C63 but this is expected to be different if I run the subroutine next week or next month (Hence why I said "I don't know how many"). I want to create an array for example RTArr = Array("Cat", "Dog", "Bird", etc...) So... I was thinking something like:
Dim RTArr As Variant
LRow = r.End(xlDown).Offset(x, y).Row
With ActiveSheet
For i = 2 To LRow
str = .Range("C" & i).Value
Next i
End With
Can I populate the array with something like:
Dim RTArr As Variant
LRow = r.End(xlDown).Offset(x, y).Row
With ActiveSheet
For i = 2 To LRow
ArrNum = (i - 1)
str = .Range("C" & i).Value
RTArr(ArrNum) = str
Next i
End With
Or does this not work because of the unknown size of the array? Or do I have to use "amend" in the loop? Would I be better off using a "collection" in this case? Or going about it some other way? Can I simply set a range of cells as an array without needing to loop?
If you declare a dynamic array at first (without the size), you need to ReDim it to the needed size before populating it, which in your case will be the number of rows e.g. ReDim RTArr(numberofitems). Or use a two dimensional array ReDim RTArr(numbercolumns, numberrows).
Remember that standard arrays begin at element 0, but you can define it however you like.
Remember that when inputting ranges into array Excel creates by default a two-dimensional array
More advanced techniques are possible of course, you can do some more research about VBA arrays regarding those:
1) you could ReDim the array after each element added inside of the loop, but this is mostly useful for one dimensional arrays.
2) you could define a much bigger size of array than needed before populating it, populate it, and then shrink the array to the actual size needed.
3) note that when using two (or more) dimensions ReDim Preserve works only on the last dimension.
Pseudo code for the basic populating:
Dim arr() as Variant
'we know we want to populate array with 10 elements
ReDim arr(1 to 10)
For i = 1 to 10
'This part will insert the count from the loop into the count position in array
' eg. first element of array will be a 1, second a 2 etc. until 10
arr(i) = i
Next i
If your version of Excel supports the TEXTJOIN function:
Sub Kolumn2Array()
Dim r As Range
Dim N As Long
Dim RTArray
Dim comma As String
comma = ","
N = Cells(Rows.Count, "C").End(xlUp).Row
Set r = Range("C2:C" & N)
With Application.WorksheetFunction
RTArray = Split(.TextJoin(comma, True, r), comma)
End With
End Sub
I need a function to extract 2 dimensions from a multidimesion array. which 2 dimensions to extract depending on the choise of the user. and the index in the discarded dimensions where those 2 dimensions are picked also depending on the user.
For example, i have a 3 dimension array v(1 to 100, 1 to 20, 1 to 10). i would like to extrat dimension 1 and dimension 3 from v. and the index in the discared dimension 2 is 11.
sub extract
dim i1 as integer 'for loop through dimension 1
dim i2 as integer 'for loop through dimension 3
dim d1 as integer 'index in dimension 2
d1=11
redim vn(1 to ubound(v,1),1 to ubound (v,3))
for i1 = 1 to ubound(v,1)
for i2= 1 to ubound(v,3)
vn(i1,i2)=v(i1,d1,i2)
next i2
next i1
end sub
I can extract dimensions from array, if i know which dimensions i need and the index (d1) in the discarded dimensions. however, i need to leave that to the users to decide. what i want is a function like that:
function extract(i1 as integer, i2 as intger, paramarray ov()) as variant
=extract(the_first_dimension_to_keep,the_second_dimension_to_keep,[index_in_the_first_discard_dimension,index_in_the_second_discard_dimension,...])
Keeping in mind that the origional array can have more than 3 dimensions, so list all the possibility in the code is not possible.
Any solution?
The quickest way would be to read the array with a pointer and increment the pointer value by an algorithmic value based on the number of dimensions and number of elements in each. This site has an excellent tutorial on managing pointers to arrays: http://bytecomb.com/vba-internals-getting-pointers. However, it'd be one mighty coding task - just dimensioning the rgabounds of your SAFEARRAY for the memory read would be a task - and if your array values were Strings, it'd be of an order of magnitude mightier.
An easier, though doubtless slower, option would be to exploit the For Each looping method, which can be applied to an array. Its looping sequence is like so:
arr(1,1)
arr(2,1)
arr(3,1)
arr(1,2)
arr(2,2)
arr(3,2)
etc.
So you'd only need a simple odometer-style index counter.
You could basically iterate every element in the array and if the combination of indexes matched what you wanted, you'd read the element into your extraction array. That would be a much easier task. The code below shows you how you could do this on a multi-dimensional array of unknown dimensions.
Option Explicit
Private Type ArrayBounds
Lower As Long
Upper As Long
Index As Long
WantedDimension As Boolean
DiscardIndex As Long
End Type
Public Sub RunMe()
Dim arr As Variant
Dim result As Variant
arr = CreateDummyArray
result = Extract(arr, 1, 3, 11)
End Sub
Private Function Extract(arr As Variant, i1 As Integer, i2 As Integer, ParamArray ov() As Variant) As Variant
Dim d As Long
Dim bounds() As ArrayBounds
Dim i As Long
Dim v As Variant
Dim ovIndex As Long
Dim doExtract As Boolean
Dim result() As Variant
'Dimension the output array
ReDim result(LBound(arr, i1) To UBound(arr, i1), LBound(arr, i2) To UBound(arr, i2))
'Get no. of dimensions in array
d = GetDimension(arr)
'Now we know the number of dimensions,
'we can check that the passed parameters are correct
If (i1 < 1 Or i1 > d) Or (i2 < 1 Or i2 > d) Then
MsgBox "i1/i2 - out of range"
Exit Function
End If
If UBound(ov) - LBound(ov) + 1 <> d - 2 Then
MsgBox "ov - wrong number of args"
Exit Function
End If
'Resise and populate the bounds type array
ReDim bounds(1 To d)
ovIndex = LBound(ov)
For i = 1 To d
With bounds(i)
.Lower = LBound(arr, i)
.Upper = UBound(arr, i)
.Index = .Lower
.WantedDimension = (i = i1) Or (i = i2)
If Not .WantedDimension Then
.DiscardIndex = ov(ovIndex)
ovIndex = ovIndex + 1
'Check index is in range
If .DiscardIndex < .Lower Or .DiscardIndex > .Upper Then
MsgBox "ov - out of range"
Exit Function
End If
End If
End With
Next
'Iterate each member of the multi-dimensional array with a For Each
For Each v In arr
'Check if this combination of indexes is wanted for extract
doExtract = True
For i = 1 To d
With bounds(i)
If Not .WantedDimension And .Index <> .DiscardIndex Then
doExtract = False
Exit For
End If
End With
Next
'Write value into output array
If doExtract Then
result(bounds(i1).Index, bounds(i2).Index) = v
End If
'Increment the dimension index
For i = 1 To d
With bounds(i)
.Index = .Index + 1
If .Index > .Upper Then .Index = .Lower Else Exit For
End With
Next
Next
Extract = result
End Function
Private Function GetDimension(arr As Variant) As Long
'Helper function to obtain number of dimensions
Dim i As Long
Dim test As Long
On Error GoTo GotIt
For i = 1 To 60000
test = LBound(arr, i)
Next
Exit Function
GotIt:
GetDimension = i - 1
End Function
I would like to write a function, where the input is a string (e.g. ACGTTGCATGTCGCATGATGCATGAGAGCT) and an integer (e.g. 4). The function should be able to identify the sub-string(s) with the length of the integer that are the most commonly repeated in the input string. I think I am supposed to use reDim, and although I have researched how it is supposed to work, I just cannot figure out the proper syntax.
Code Challenge: Solve the Frequent Words Problem. Input: A string
Text and an integer k. Output: All most frequent k-mers in Text.
Function BIOINFO2(txt As String, k As Integer)
Dim FrequentPatterns As String
Dim ptrn As String
Dim n() As Integer
Dim i As Integer
Dim s As Integer
Dim j As Integer
Dim maxCount As Integer
For s = 1 To Len(txt) - k + 1
ptrn = Mid(txt, s, k)
For i = 1 To Len(txt) - Len(ptrn) + 1
If Mid(txt, i, Len(ptrn)) = ptrn Then
ReDim n(i)
n(i) = 0
n(i) = n(i) + 1
End If
Next i
maxCount = Application.Max(n(i))
For j = 1 To Len(txt) - k + 1
If n(i) = maxCount Then
FrequentPatterns = FrequentPatterns + " " + Mid(txt, s, k)
End If
Next j
Next s
BIOINFO2 = FrequentPatterns
End Function
This is just to clarify the ReDim issue, not the logic of your code.
You use ReDim to resize an array (losing the values stored in it), ReDim Preserve to resize an array while keeping the values.
If you know the required size of your array beforehand, you should allocate it with the correct size at the start, as in
Dim arr(1 To m) As Long
If you don't know the required size, you can resize it later on like this
Dim arr() As Long
'Do stuff and find out you need the array to hold m elements
ReDim Preserve arr(1 To m)
Your case is somewhat in between because you know the size right at the start of your function but it is not a constant size. You can Dim an array with a specific size only if it's constant so in this case you need to declare it without bounds first and then ReDim with the correct size.
Dim arr() as Long
ReDim arr(1 To m) As Long
Redim arr(m) is the same as Redim arr(base To m) where base is either 0 or 1. The default value is 0 but you can set it using Option Base 1 at the beginning of your module.
You can find out the highest and lowest array indices using the UBound and LBound functions so a loop that loops over all the values could look like this
For i = LBound(arr) To UBound(arr)
'Do something
Next i
That way you avoid problems with different bases.
It is usually better to do as little resizing of arrays as possible. In the worst case scenario, the system has to copy the whole array to another location in the memory where there is enough free space.
Just to show that VBA is not needed for a task like this:
(A1 would be your string and A2 the length)
=MID(A1,MATCH(MIN(LEN(SUBSTITUTE(A1,MID(A1,ROW(A1:INDEX(A:A,LEN(A1)-A2+1)),A2),""))),LEN(SUBSTITUTE(A1,MID(A1,ROW(A1:INDEX(A:A,LEN(A1)-A2+1)),A2),"")),0),A2)
This is an array formula and must be confirmed with ctrl+shift+enter!
At least to get the first possible solution.
I'm trying to get a 2D array of size [x][3] filled. X is just the size of the sheet (number of rows) and there are 3 columns which I am interested in. The columns are not near each other, for instance arr[i][0] should be filled from column AA, arr[i][1] should come from column K, and arr[i][2] needs to be from columns L.
I tried assigning it the following way, but got an error in the array value assignment.
Any help on this would be greatly appreciated!
Code:
Sub SOC_work()
'Trying to sort each of the disciplines further, by Stage of Construction
Dim ar_SOC() As Variant
Dim int_NumRows As Long
Dim i_counter As Long
Dim j_Counter As Long
Dim lite As Range
Application.Calculation = xlCalculationManual
Application.ScreenUpdating = False
Sheets("AVEVA_PBOM_PARTS").Select
'Redimension the array size to the amount of parts in the PBOM
int_NumRows = ActiveSheet.UsedRange.Rows.count - 1
ReDim ar_SOC(int_NumRows, 3)
'now assignt he range into the array space
lite = Range("AA2", Range("AA2").End(xlDown))
ar_SOC[][1]=lite
End Sub
Is there any way to do this without looping through the entire column?
As described in the comments, you can fill three 2-D arrays. You can then populate a fourth array from the three arrays, like below.
Sub populateArray()
Dim arrColOne() As Variant, arrColTwo() As Variant, arrColThree() As Variant
Dim arrAllData() As Variant
Dim i As Long
arrColOne = Range("A2:A" & lrow(1)) 'amend column number
arrColTwo = Range("D2:D" & lrow(4))
arrColThree = Range("G2:G" & lrow(7))
ReDim arrAllData(1 To UBound(arrColOne, 1), 2) As Variant
For i = 1 To UBound(arrColOne, 1)
arrAllData(i, 0) = arrColOne(i, 1)
arrAllData(i, 1) = arrColTwo(i, 1)
arrAllData(i, 2) = arrColThree(i, 1)
Next i
End Sub
Public Function lrow(colNum As Integer) As Long
lrow = Cells(Rows.Count, colNum).End(xlUp).Row
End Function
The above will require all 3 columns to be the same length (otherwise populating the last array will not work); this is due to the fourth array being redimensioned to contain the number of elements contained in the first array.
Testing with 250,000 rows of data, the fourth array populated in 0.43 seconds.
How lenient are you with the array you get in return? I can get you a Array(col)(row)-style array, without having to loop to get it, if that works. Note that's not Array(col, row), by the way. It's a single-dimensional array of columns, with each element containing a single-dimensional array of row values. If you're okay with that, you can do this:
Dim a(1 To 3)
a(1) = WorksheetFunction.Index(WorksheetFunction.Transpose(Range("AA2:AA10")), 1, 0)
a(2) = WorksheetFunction.Index(WorksheetFunction.Transpose(Range("K2:K10" )), 1, 0)
a(3) = WorksheetFunction.Index(WorksheetFunction.Transpose(Range("L2:L10" )), 1, 0)
Then you could access your array items like so:
Debug.Print UBound(a) ' Number of columns (3)
Debug.Print UBound(a(1)) ' Number of rows in column 1
Debug.Print a(1)(3) ' Value of column 1 (AA), row 3
The Index() function can return a 1D array but only in the rows direction. So, you need to combine it with Transpose() to return a 1D column array. That's all the code above is doing.
What about an array of arrays?
Sub NoLoop()
Dim R1 As Range, R2 As Range, R3 As Range
Dim Arr1() As Variant, Arr2() As Variant, Arr3() As Variant
Dim LR As Long
LR1 = Cells(Rows.Count, "AA").End(xlUp).Row
LR2 = Cells(Rows.Count, "K").End(xlUp).Row
LR3 = Cells(Rows.Count, "L").End(xlUp).Row
Set R1 = Range(Cells(1, "AA"), Cells(LR1, "AA"))
Set R2 = Range(Cells(1, "K"), Cells(LR2, "K"))
Set R3 = Range(Cells(1, "L"), Cells(LR3, "L"))
Arr1 = R1.Value
Arr2 = R2.Value
Arr3 = R3.Value
ArrArr = Array(Arr1, Arr2, Arr3)
End Sub
With this you can call your values using:
MyVal = ArrArr(0)(1,1)
MyVal = ArrArr(0)(2,1)
MyVal = ArrArr(1)(1,1)
Where the first number is for the array (starts from 0 and ends with 2) and the second number is for row/cell of the range used to fill array.
The third number is always 1 (because adding a range to an array returns a bidimensional array)
With this code you can also have different dimensions for each column so to save memory.
I have been given the job of fixing a holiday spreadsheet at work, the problem with the document is that there are direct references, indirect references to other worksheets and some parts non referenced, so if you want to put people in a different order it completely messes the whole thing up. So, what I have been trying to do is to populate an array with the peoples names, sort the array, then cross reference that with the original and find a new order so that this new order can be implemented throughout the worksheets without messing things up.
The problem is that I can't seem to get the arrays to populate, I have looked at the other answers on here, but I'm sure I'm using redim properly and that tends to be the problem.
So Previous is the original listing, Current is the sorted list and Position is the number that relates the previous to the current.
Sub Sorting()
Dim Previous() As Variant
Dim Current() As Variant
Dim maxrow As Long
Dim i As Long
Dim j As Long
Dim k As Long
Dim Position() As Long
Dim rng As Range
Dim strTemp As String
k = 0
i = 3
maxrow = 3
Do While Worksheets(1).Cells(i, 1).Value <> "STAT.HOL'S (ST)"
maxrow = maxrow + 1
i = i + 1
Loop
maxrow = maxrow - 1
ReDim Previous(0 To maxrow)
ReDim Position(0 To maxrow)
ReDim Current(0 To maxrow)
Previous = Range("a4", Range("a" & maxrow))
Current = Previous
For i = 0 To maxrow
For j = 0 To maxrow
If Current(i) > Current(j) Then
strTemp = Current(i)
Current(i) = Current(j)
Current(j) = strTemp
End If
Next j
Next i
For i = 0 To maxrow
For j = 0 To maxrow
If Previous(i) = Current(j).Value Then
Position(k) = j
k = k + 1
End If
Next j
Next i
End Sub
Thanks for your help.
Amy
You do populate the arrays, but you are adressing them the wrong way.
When you assign a range to an array, the array is automatically redimensioned to a two-dimensional array, one for the rows and one for the columns.
Since your range is just one columm, the second dimension is always 1. So after you run the line Previous = Range("a4", Range("a" & maxrow)) you have an array that is dimensioned as:
Previous(1 to 10, 1 To 1)
Also, your For loops could be changed to something like this:
For i = LBound(Current, 1) To UBound(Current, 1)
Next i
Excel ranges are always treated as having 2 dimensions even when they are a single column.
Also you don't need to redim your variant arrays - just assign the range to a plain variant variable and it will create a variant containing a 2-dimensional array:
Dim Previous as variant
Previous = Range("a4:a" & maxrow)