Using scripting dictionary to find/highlight skips in groups of repeating numbers in Column A using Excel VBA - vba

Question

I'm attempting to use a Scripting Dictionary in a way as to be able to find and ultimately highlight same values or groups of same values where there are inconsistencies (ie blanks or different values in between the two same values or groups of same values). Normally these same values will repeat, but what I'm trying to catch is when they do not repeat together (See example image below taken from my previous post).
Some context that will hopefully help this make a little more sense:
This is a follow-up of sorts to one of my previous questions here. I have a conditional formatting formula:
=NOT(AND(IFERROR(COUNTIF(OFFSET(A1,0,0,-COUNTIF($A$1:$A1,A2)),A2),0)=IFERROR(COUNTIF($A$1:$A1,A2),0),IFERROR(COUNTIF(OFFSET(A3,0,0,COUNTIF($A3:$A$5422,A2)),A2),0)=IFERROR(COUNTIF($A3:$A$5422,A2),0),A2<>""))
Which works perfectly. However, in my tinkering after receiving this formula as the answer to that previous question I realized that using conditional formatting of any sort for the amount of data I typically deal with (15000+ rows with 140 consistent columns) is an extremely slow endeavor, both when applying the formula and when filtering/adjusting afterwards. I've also tried applying this formula via the "helper column" route, but to no surprise, that is just as slow.
So, where I'm at now:
Essentially, I'm trying to translate that formula into a piece of code that does the same thing, but more efficiently, so that's where I starting thinking to use a Scripting Dictionary as a way to speed up my code execution time. I have the steps outlined, so I know what I need to do. However, I feel as though I am executing it wrong, which is why I'm here to ask for assistance. The following is my attempt at using a Scripting Dictionary to accomplish highlighting inconsistencies in Column A (my target column) along with the steps I figured out that I need to do to accomplish the task:
'dump column A into Array
'(Using Scripting.Dictionary) While cycling through check if duplicate
'IF duplicate check to make sure there is the same value either/or/both in the contiguous slot before/after the one being checked
'If not, then save this value (so we can go back and highlight all instances of this value at the end)
'Cycle through all trouble values and highlight all of their instances.
Sub NewandImprovedXIDCheck()
Dim d As Long, str As String, columnA As Variant
Dim dXIDs As Object
Application.ScreenUpdating = False
Set dXIDs = CreateObject("Scripting.Dictionary")
dXIDs.comparemode = vbTextCompare
With ActiveSheet
With .Cells(1, 1).CurrentRegion
With .Resize(.Rows.Count - 1, .Columns.Count).Offset(1, 0)
'.Value2 is faster than using .Value
columnA = .Columns(1).Value2
For d = LBound(columnA, 1) To UBound(columnA, 1)
str = columnA(d, 1)
If dXIDs.exists(str) Then
'the key exists in the dictionary
'Check if beside its like counterparts
If Not UBound(columnA, 1) Then
If (str <> columnA(d - 1, 1) And str <> columnA(d + 1, 1)) Or str <> columnA(d - 1, 1) Or str <> columnA(d + 1, 1) Then
'append the current row
dXIDs.Item(str) = dXIDs.Item(str) & Chr(44) & "A" & d
End If
End If
Else
'the key does not exist in the dictionary; store the current row
dXIDs.Add Key:=str, Item:="A" & d
End If
Next d
'reuse a variant var to provide row highlighting
Erase columnA
For Each columnA In dXIDs.keys
'if there is more than a single cell address, highlight all
If CBool(InStr(1, dXIDs.Item(columnA), Chr(44))) Then _
.Range(dXIDs.Item(columnA)).Interior.Color = vbRed
Next columnA
End With
End With
End With
dXIDs.RemoveAll: Set dXIDs = Nothing
Application.ScreenUpdating = True
End Sub
I feel like my logic is going wrong somewhere in my code execution, but can't seem to pinpoint where or how to correct it. Any help would be greatly appreciated. If you can provide any sort of code snippet that would also be a great help.

Answer 1

Here's one approach:
Sub HiliteIfGaps()
Dim rng As Range, arr, r As Long, dict As Object, v
Dim num As Long, num2 As Long
Set dict = CreateObject("scripting.dictionary")
With ActiveSheet
Set rng = .Range(.Range("A2"), .Cells(.Rows.Count, 1).End(xlUp))
End With
arr = rng.Value
For r = 1 To UBound(arr, 1)
v = arr(r, 1)
If Len(v) > 0 Then
If Not dict.exists(v) Then
num = Application.CountIf(rng, v) 'how many in total?
'all where expected?
num2 = Application.CountIf(rng.Cells(r).Resize(num, 1), v)
dict.Add v, (num2 < num)
End If
If dict(v) Then rng.Cells(r).Interior.Color = vbRed
Else
'highlight blanks
rng.Cells(r).Interior.Color = vbRed
End If
Next r
End Sub
EDIT: every time a new value is found (i.e. not already in the dictionary) then take a count of how many of those values in total there are in the range being checked. If all of those values are contiguous then they should all be found in the range rng.Cells(r).Resize(num, 1): if we find fewer than expected (num2<num) then that means the values are not contiguous so we insert True into the dictionary entry for that value, and start highlighting that value in the column.

Answer 2

#Tim Williams's approach did the job perfectly! I only made one slight alteration (to suit my needs). I changed
.Cells(.Rows.Count, 1).End(xlUp) to .Range("A" & .UsedRange.Rows.count)
Just because there are instances where the bottom-most row(s) might have missing values (be blank) and in this instance I feel safe enough using the .UsedRange reference because this snippet of code is one of the very first ones ran in my overall macro, so it (.UsedRange) is more likely to be accurate. I also added a Boolean operator (xidError, set to False) to be changed to True whenever we have to highlight. After I'm done looping through the Array I check xidError and if True I prompt the user to fix the error, then end the entire macro since there's no use in continuing until this particular error is corrected.
If xidError Then
'Prompt User to fix xid problem
MsgBox ("XID Error. Please fix/remove problematic XIDs and rerun macro.")
'Stop the macro because we can't continue until the xid problem has been sorted out
End
End If
Again, much thanks to Tim for his very efficient approach!