Two new range functions: Union and Subtract
This post discusses two functions I developed because of a recent need. The first is an enhanced version of the Excel Union method. The other is a Subtract function that operates on ranges.
The Union function
Those who use the Excel Application’s Union method with any sense of regularity know it doesn’t deal well with any argument being ‘nothing.’ Consequently, it is almost second nature to code
Sub RoundAboutCode()
Dim Rng1 As Range, Rng2 As Range, Rslt As Range
‘…
If Rng1 Is Nothing Then
Set Rslt = Rng2
ElseIf Rng2 Is Nothing Then
Set Rslt = Rng1
Else
Set Rslt = Application.Union(Rng1, Rng2)
End If
End Sub
Recently, I found myself writing the above code for the 3rd time in a few days. Annoyed at not having modularized it years ago, I did just that. It’s below.
Function Union(Rng1 As Range, Rng2 As Range) As Range
If Rng1 Is Nothing Then
Set Union = Rng2
ElseIf Rng2 Is Nothing Then
Set Union = Rng1
Else
Set Union = Application.Union(Rng1, Rng2)
End If
End Function
With the above, one can code the below without worrying about whether Rslt or NextCell is Nothing.
Set Rslt = Union(Rslt, NextCell)
True, unlike the Union method, the function accepts only 2 arguments. I did write a more generic function declared with a ParamArray argument only to discover that the Union method won’t accept a single variant (’argument not optional’ error) or an array (’type mismatch’ error). Maybe someone else can make the more generic case work.
The Range Subtract function
I also had reason to write a Subtract function. Given two ranges, Rng1, and Rng2, where Rng2 is a subset of Rng1, the result is Rng1 - Rng2, i.e., all those cells in Rng1 that are not part of Rng2.
I remembered a post by Tom Ogilvy from a long time ago that used the following method: In a temporary worksheet, in the range corresponding to the address of Rng1 enter some constant (say the value 1). Next, clear the cells corresponding to the address of Rng2, and finally, pick up the result with the SpecialCells method.
Function SubtractUsingWS(Rng As Range, RngToSubtract As Range)
If Application.Intersect(Rng, RngToSubtract).Address _
<> RngToSubtract.Address Then Exit Function
Dim OldEventsValue
OldEventsValue = Application.EnableEvents
Application.EnableEvents = False
On Error GoTo Finally1
With ThisWorkbook.Worksheets(”TempWS”)
.Cells.ClearContents
.Range(Rng.Address).Value = 1
.Range(RngToSubtract.Address).ClearContents
Set SubtractUsingWS = _
Rng.Parent.Range(.Cells.SpecialCells(xlCellTypeConstants).Address)
End With
ThisWorkbook.Saved = True
Finally1:
Application.EnableEvents = OldEventsValue
End Function
Sub testSubtract2()
MsgBox SubtractUsingWS(Range(”a1:C3″), Range(”b2″)).Address
End Sub
The above is probably as efficient as one can get but concerned about issues such as write privileges, multiple people accessing an add-in on a network drive, etc., I wrote up a solution from first principles, as it were.
Function SubtractFirstPrinciples(Rng1 As Range, Rng2 As Range) As Range
On Error Resume Next
If Application.Intersect(Rng1, Rng2).Address <> Rng2.Address Then _
Exit Function
On Error GoTo 0
Dim aCell As Range
For Each aCell In Rng1
Dim Rslt As Range
If Application.Intersect(aCell, Rng2) Is Nothing Then
Set Rslt = Union(Rslt, aCell)
End If
Next aCell
Set SubtractFirstPrinciples = Rslt
End Function
Sub testSubtractFirstPrinciples()
Debug.Print SubtractFirstPrinciples( _
Sheets(1).Range(”A1:f10″), _
Sheets(1).Range(”A1,b2,c3,d4:e5,f6″)).Address
End Sub
The advantage of working from first principles is that it works correctly irrespective of the shape of the two arguments Rng1 and Rng2. We don’t have to worry about whether they consist of multiple areas or not. The disadvantage, of course, is that it checks each cell in Rng1 and consequently might be slow under certain circumstances.
Before proceeding further, one should remember that the need for any optimization is unproven. I don’t know what, if any, problems the first solution will run into nor do I know how slow the solution based on first principles will be. So, the benefits of the optimizations below are somewhat uncertain. By contrast, it is certain that there will be some cost to developing the code, testing it, and maintaining it.
The first step in optimization would be to start small: a single area from which we want to subtract a single area. Clearly, in this case the result will be at the most four ranges as shown below. The first image shows the area that we want to subtract in yellow. The second image shows the four areas that will remain after the subtraction operation is completed.


The code below is a function that accepts two range arguments Rng1 and Rng2 and returns a range that corresponds to Rng1 - Rng2. It validates that each range consists of a single area. I don’t know what it means to subtract Rng2 from Rng1 if there is absolutely no overlap between the two ranges. So, I made the assumption that the result should be Rng1 itself. Before proceeding with the analysis, the code computes what part of Rng2 is actually within Rng1. Each of the four If statements enclose a block of code that calculates one of the four possible ranges in the result (see the above image). Finally, the code returns the result of the subtraction.
Function subtractOneArea(Rng1 As Range, inRng2 As Range) As Range
If Rng1.Areas.Count > 1 Then Exit Function
If inRng2.Areas.Count > 1 Then Exit Function
If Application.Intersect(Rng1, inRng2) Is Nothing Then
Set subtractOneArea = Rng1
Exit Function
End If
Dim Rng2 As Range
Set Rng2 = Application.Intersect(Rng1, inRng2)
Dim aRng As Range, OKRng As Range, Rslt As Range, WS As Worksheet
Set WS = Rng1.Parent
If Rng2.Row > Rng1.Row Then
Set Rslt = WS.Range(Rng1.Rows(1), Rng1.Rows(Rng2.Row - Rng1.Row))
End If
If Rng2.Row + Rng2.Rows.Count < Rng1.Row + Rng1.Rows.Count Then
Set Rslt = Union(Rslt, _
WS.Range(Rng1.Rows(Rng2.Row - Rng1.Row + Rng2.Rows.Count + 1), _
Rng1.Rows(Rng1.Rows.Count)))
End If
If Rng2.Column > Rng1.Column Then
Set Rslt = Union(Rslt, WS.Range(WS.Cells(Rng2.Row, Rng1.Column), _
WS.Cells(Rng2.Row + Rng2.Rows.Count - 1, Rng2.Column - 1)))
End If
If Rng2.Column + Rng2.Columns.Count < Rng1.Column + Rng1.Columns.Count Then
Set Rslt = Union(Rslt, _
WS.Range(WS.Cells(Rng2.Row, Rng2.Column + Rng2.Columns.Count), _
WS.Cells(Rng2.Row + Rng2.Rows.Count - 1, _
Rng1.Column + Rng1.Columns.Count - 1)))
End If
Set subtractOneArea = Rslt
End Function
With the building block in place, writing the Subtract function to calculate Rng1 - Rng2 is a lot easier. All we need to do is accumulate the result as we loop through each area of Rng1 and subtract from it each area of Rng2.
Function Subtract(Rng1 As Range, Rng2 As Range) As Range
On Error Resume Next
If Application.Intersect(Rng1, Rng2).Address <> Rng2.Address Then _
Exit Function
On Error GoTo 0
Dim Rslt As Range, Rng1Rslt As Range, J As Integer, I As Integer
For J = 1 To Rng1.Areas.Count
Set Rslt = subtractOneArea(Rng1.Areas(J), Rng2.Areas(1))
For I = 2 To Rng2.Areas.Count
Set Rslt = Application.Intersect( _
Rslt, subtractOneArea(Rng1.Areas(J), Rng2.Areas(I)))
Next I
Set Rng1Rslt = Union(Rng1Rslt, Rslt)
Next J
Set Subtract = Rng1Rslt
End Function
The code is used as in the following example:
Sub testSubtract()
Debug.Print Subtract( _
Sheets(1).Range(”A1:f10″), _
Sheets(1).Range(”A1,b2,c3,d4:e5,f6″)).Address
Debug.Print Subtract( _
Sheets(1).Range(”A1:f9,a10:f10″), _
Sheets(1).Range(”A1,b2,c3,d4:e5,f6″)).Address
Debug.Print Subtract( _
Sheets(1).Range(”$I$1:$K$4,$L$4:$N$8,$K$7:$K$13″), _
Sheets(1).Range(”$K$4:$L$4,$K$8:$L$8″)).Address
End Sub
At some point a variant of the above will show up in the “publications and training” section of my web site.
And, that concludes all I have to share on this subject…at least for the time being.
Andy Pope:
Hi Tushar,
How about this for a multi argument union?
Dim rngA As Range
Dim rngB As Range
Dim rngC As Range
Dim rngD As Range
Set rngA = Range("A1:B5")
Set rngB = Range("A3:E4")
Set rngD = Range("D2:D10")
Union(rngA, rngB, rngC, rngD).Select
Debug.Print Selection.Address
End Sub
Function Union(ParamArray Rngs()) As Range
Dim rngUnion As Range
Dim vntX As Variant
For Each vntX In Rngs
If Not vntX Is Nothing Then
If rngUnion Is Nothing Then
Set rngUnion = vntX
Else
Set rngUnion = Application.Union(rngUnion, vntX)
End If
End If
Next
If Not rngUnion Is Nothing Then Set Union = rngUnion
End Function
Jon Peltier:
Minor technicality, I might have named the second function Remove or RangeRemove rather than Subtract, which is too strongly linked with the mathematical operation. The approach itself is nicely done.
17 August 2007, 6:56 amTushar Mehta:
Andy: Duh! Focused on making a single call to the Union method I overlooked the possibility of multiple calls. Thanks.
Jon: Subtraction is a valid set operation (http://en.wikipedia.org/wiki/Set) and thanks.
17 August 2007, 8:39 amMatt Vidas:
Tushar,
17 August 2007, 9:27 amUnrelated to the functions above, just wanted to send a quick hi; never realized before that we're neighbors (though I live downtown, I used to spend quite a bit of time in RV). Keep up the good work!
Jon Peltier:
Tushar -
Good point. Maybe I'm thinking in terms of collections, where you add and remove items.
17 August 2007, 1:27 pmkeepITcool:
Andy: You'll need more testing on the ParamArray, as users will no doubt pass the unexpected.
17 August 2007, 4:36 pmFor Each vRng In Rngs
If IsObject(vRng) Then
If Not vRng Is Nothing Then
If TypeOf vRng Is Range Then
Andy Pope:
keepITcool: Your right I did not include code to check arguments where indeed ranges.
18 August 2007, 5:22 amBut the post was just to point Tushar in the right direction
fzz:
FWIW, my version of Union with argument checking.
strict As Boolean, _
ParamArray a() As Variant _
) As Range
'-----------------------------
Dim v As Variant, r As Range
For Each v In a
If Not IsObject(v) Then
GoTo Fail
ElseIf v Is Nothing Then
If strict Then GoTo Fail
ElseIf Not TypeOf v Is Range Then
GoTo Fail
ElseIf fcnUnion Is Nothing Then
Set fcnUnion = v
Else
Set r = v
Set fcnUnion = Union(fcnUnion, r)
End If
Next v
Exit Function
Fail:
Set fcnUnion = Nothing
End Function
Arguable whether it should return an error value when it finds non-Range arguments rather than returning Nothing.
As for the pseudo nonsymmetric set difference of a range with a single area range, i.e., removing the single area range from another, more general range, A \ B = Intersect(A, Complement(B)), and the complement of a single area range B is the union of rows above B, rows below B, columns to the left of B and columns to the right of B. Also, if A and B are disjoint, A \ B = A.
'comparisons in order to please the fine blog software
Function fcnRangeRemoveSA(a As Range, b As Range) As Range
Dim i(1 To 4) As Long, w(1 To 2) As Long
Dim ct As Range, cb As Range, cl As Range, cr As Range
If Intersect(a, b) Is Nothing Then
Set fcnRangeRemoveSA = a
Exit Function
End If
w(1) = b.Parent.Rows.Count
w(2) = b.Parent.Columns.Count
i(1) = b.Row - 1
i(2) = b.Row + b.Rows.Count
i(3) = b.Column - 1
i(4) = b.Column + b.Columns.Count
With b.Parent
If Sgn(i(1)) = 1 Then _
Set ct = .Range(Cells(1, 1), Cells(i(1), w(2)))
If Sgn(w(1) - i(2)) = 1 Then _
Set cb = .Range(Cells(i(2), 1), Cells(w(1), w(2)))
If Sgn(i(3)) = 1 Then _
Set cl = .Range(Cells(1, 1), Cells(w(1), i(3)))
If Sgn(w(2) - i(4)) = 1 Then _
Set cb = .Range(Cells(1, i(4)), Cells(w(1), w(2)))
End With
Set fcnRangeRemoveSA = Intersect(a, _
fcnUnion(False, ct, cb, cl, cr))
End Function
The complement of a multiple area range is the intersection of the complements of each area.
19 August 2007, 6:02 pmfzz:
Just noticed the fine blog software doesn't display the VBA keyword With in blue text. Does it not consider With a VBA keyword?
19 August 2007, 6:08 pmfzz:
And a bug in the 4th If statement inside the With block in the second function. It should be
Set cr = .Range(Cells(1, i(4)), Cells(w(1), w(2)))
Dick Kusleika:
"Does it not consider With a VBA keyword?"
Fixed.
20 August 2007, 7:28 am