# Column Numbers To Letters

Chip Pearson posted this solution to the newsgroups for finding the column letters when you know the column numbers:

```Function ColLetter(ColNumber As Integer) As String ColLetter = Left(Cells(1, ColNumber).Address(False, False), _ 1 - (ColNumber > 26)) End Function ```

Very smart function, in my opinion. Sometime before the year 3,000, Microsoft will hopefully increase the number of columns in Excel (Hey, I can dream can’t I). The challenge before you is to write a function that converts a column number to its letter equivalent assuming columns go to ZZZZ. That’s about 450,000 columns – maybe more than I need.

I was just going to whip up a short function to do this and post it, but after 10 minutes I still couldn’t do it. Rather than admit defeat, I turned this post into a challenge for you. It’s Friday, you weren’t going to work anyway, were you?

## 58 thoughts on “Column Numbers To Letters”

1. Juan Pablo says:

Am I getting annoying ? :)

Function ColLetter(ColNumber As Long) As String
ColLetter = Application.Substitute(Cells(1, ColNumber).Address(False, False), “1?, “”)
End Function

2. Dick says:

Very nice! Needs some error checking if the number passed is too big, easy enough. Also, what if the activesheet is a chart sheet? What if all the sheets in the activeworkbook are chart sheets?

This is exactly the easy solution that I couldn’t find. Now if we can avoid the non-worksheet situation, it will be perfect.

3. Brian says:

Howsabout a kludgey untested formula for letter to column? (Yes, I did misread the original challenge. Ho hum.)
=IF(LEN(A1)<4,0,26^3*(1+CODE(MID(UPPER(A1),LEN(A1)-3,1))-CODE(“A”)))+IF(LEN(A1)<3,0,26^2*(1+CODE(MID(UPPER(A1),LEN(A1)-2,1))-CODE(“A”)))+IF(LEN(A1)<2,0,26*(1+CODE(MID(UPPER(A1),LEN(A1)-1,1))-CODE(“A”)))+IF(LEN(A1)

4. JWalk says:

This seems to work for column letters up to ZZ (702 columns).

Function ColLetter(ColNumber As Long) As String
Dim x2 As Double, x3 As Integer
Dim x4 As String, x5 As Integer
Dim x6 As String, x7 As String
If ColNumber > 702 Or ColNumber < 1 Then Exit Function
x2 = (ColNumber – 1) / 26
x3 = Int(x2)
x4 = Chr(x3 + 64)
x5 = ColNumber Mod 26
If x4 = “@” Then x6 = “” Else x6 = x4
If x5 = 0 Then x7 = “Z” Else x7 = Chr(x5 + 64)
ColLetter = x6 & x7
End Function

I figured this out using worksheet formulas, and then converted the formulas to VBA.

In Row 1, put the numbers 1-256
A2: =(A1-1)/26
A3: =TRUNC(A2)
A4: =CHAR(A3+64)
A5: =MOD(A1,26)
A6: =IF(A4=”@”,””,A4)
A7: =IF(A5=0,”Z”,CHAR(A5+64))
A8: =A6&A7

Then copy A2:A8 across.

It doesn’t solve the problem, but at least it gave me something to do for the past 15 minutes.

5. Juan Pablo says:

This works if the activesheet is a chart sheet or even if no workbook are visible:

Function ColLetter(ColNumber As Long) As String
On Error Resume Next
ColLetter = Application.Substitute(Application.ConvertFormula(“R1C” & ColNumber, _
xlR1C1, xlA1, 4), “1?, “”)
End Function

6. Tommy Bak says:

This is only testet to 702 columns :-)
Columnnumber is written in A1

7. Gary Waters says:

Sorry Gary, I’m changing your comment. I can’t post comments to this post anymore because the submit button is off the screen. To read Gary’s comment, go here

http://www.dicks-blog.com/excel/garycolletter.htm

8. Harald Staff says:

I believe this will work:

Function Num2Let(L As Long) As String
Dim s0 As String, s1 As String, S2 As String, s3 As String
If L > 18278 Then s0 = Chr((Int((L – 18279) / 17576) Mod 26) + 65)
If L > 702 Then s1 = Chr((Int((L – 703) / 676) Mod 26) + 65)
If L > 26 Then S2 = Chr(Num2Let & (Int((L – 27) / 26)) Mod 26 + 65)
s3 = Chr(((L – 1) Mod 26) + 65)
Num2Let = s0 & s1 & S2 & s3
End Function

9. J-Walk says:

Wow, I think Harald wins.

10. Harald Staff says:

Thanks John. I always wanted a Corvette :-)
I have no idea how the
“Num2Let & “
in line 5 survived the final tests. Please pretend it’s not there.

11. kevboy says:

I have a secret message for Harald, Dick, and anyone else who might be bored right now:

1000, 9, 217287, 535, 119, 253, 319778, 357181, 43661.

12. Dick says:

247, 131174, 17311, 217287, 254

13. Gary Waters says:

Very Good Harald. Here is my new improved formula that covers all positive long integers, based on Harald’s insightful fuction:

Function ConvertNumberToColumnLetter2(ByVal colNum As Long) As String
Dim i As Long, x As Long
For i = 6 To 0 Step -1
x = (1 – 26 ^ (i + 1)) / (-25) – 1 ‘ Geometric Series formula
If colNum > x Then
ConvertNumberToColumnLetter2 = ConvertNumberToColumnLetter2 & Chr(((colNum – x – 1) 26 ^ i) Mod 26 + 65)
End If
Next i
End Function

14. Gary Waters says:

Sorry Dick, I promise this will be my last say on this subject. I slightly improved on my last posting, using the Geometric Series Sum Formula, to limit the number of iterations:

Function ColumnLetter(ByVal colNum As Long) As String
Dim i As Long, x As Long
For i = Int(Log(CDbl(25 * (CDbl(colNum) + 1))) / Log(26)) – 1 To 0 Step -1
x = (26 ^ (i + 1) – 1) / 25 – 1
If colNum > x Then
ColumnLetter = ColumnLetter & Chr(((colNum – x – 1) 26 ^ i) Mod 26 + 65)
End If
Next i
End Function

15. Dick says:

Gary: Keep them coming, this is good stuff.

16. Harald Staf says:

Very nice, Gary. That’s a piece of art.
What we’d need now is the reverse function. All work and no play this week…

Function ColNum(Byval ColumnLetter as string) as Long
‘anyone ?

17. Harald Staff says:

Very nice, Gary. That’s a piece of art.
What we’d need now is the reverse function. All work and no play this week…

Function ColNum(Byval ColumnLetter as string) as Long
‘anyone ?

18. Roger Crawley says:

I used Harald’s Function to devise a function that could handle more columns. It is a more simplistic function than Gary’s and will go no further than column ZZZZZZ:

Function ColNum2Let(ColumnNumber As Long) As String
Dim CL As String, CN As Long, S As Long, N As Integer, Ns As Long, c As Integer, e As Integer
Let CN = ColumnNumber
Let N = 0
Let Ns = 0
Do
Let N = N + 1
Let Ns = Ns + 26 ^ N
Loop Until CN < = Ns
Let CL = “”
For c = 1 To N
Let S = 0
For e = 0 To c – 1
Let S = S + 26 ^ e
Next e
Let CL = Chr((Int((CN – S) / (26 ^ (c – 1))) Mod 26) + 65) & CL
Next c
ColNum2Let = CL
End Function

Then using this as a base I have written a reverse function (which can handle all positive long integers, i.e. up to column FXSHRXW):

Function ColLet2Num(ColumnLetter As String) As Long
Dim CL As String, CN As Long, N As Integer, S As Long, Se As Integer, c As Integer, t As String, ti As Integer, A As Long
Let CL = ColumnLetter
Let N = Len(CL)
Let S = 0
For Se = 0 To N – 1
Let S = S + 26 ^ Se
Next Se
Let t = “”
Let A = 0
For c = 1 To N
t = UCase(Mid(CL, c, 1))
Let ti = 65
Do Until Chr(ti) = t
Let ti = ti + 1
If ti > 90 Then
Let CN = “Error”
GoTo 1
End If
Loop
Let A = A + ((ti – 65) * (26 ^ (N – c)))
Next
Let CN = S + A
1 ColLet2Num = CN
End Function

19. Tushar Mehta says:

I am curious as to why people prefer Harald’s solution to Juan Pablo’s Substitute(ConvertFormula()) one-liner. It would appear the latter would cover *any* number of columns not just restrict itself to those with four letter designations.

20. The Contrarian says:

To really use a large number of columns — more than 10? :) — one really needs to change the addressing paradigm. I find that I often switch to R1C1 mode when I have to scroll horizontally to see the entire worksheet. Where the f*** is AM? Or GD?

And, that would take care of the need for functions that provide column letters. Wouldn’t it? ;-)

In fact, the A1 and R1C1 conventions are like the QWERTY and Dvorak keyboard layouts or VHS and Betamax. In each case, the latter is/was far superior but the former dominates/won the marketplace.

21. anonymous says:

Why don’t you just use the built-in address function to generate column names from row/column numbers?

22. Edd says:

Col_Letter = Left(Cells(1, Col_Num).Address(False, False), 1 + Fix(Col_Num ^ (1 / 26)))

I think that this works however many columns there are

23. Edd says:

ignore that its wrong!

24. Die_Another_Day says:

Function getColLet(ColumnNumber As Integer) As String
Dim colLet() As String
On Error Resume Next
If colLet(1) = “” Then
getColLet = “error”
MsgBox “This version of Excel does not support ” & ColumnNumber & ” Columns”
Else
getColLet = colLet(1)
End If
On Error GoTo 0
End Function

…except for the non-worksheet problem. Hmmm.

Die_Another_Day

25. Die_Another_Day says:

And now for people who want to do this when the Active Sheet is a chart…

Function getColLet(ColumnNumber As Integer) As String
Dim colLet() As String
Dim aWorkbook As Workbook ‘Active Workbook
Dim nWorkbook As Workbook ‘New Workbook

Application.ScreenUpdating = False ‘Hide screen updates while program is running

Set aWorkbook = ActiveWorkbook ‘Get the active workbook so it can be restored before exiting function
‘Create new workbook so we can be sure we’re looking at a worksheet
nWorkbook.Sheets(1).Activate ‘Activate Sheet1 on the new workbook
‘Don’t pause on error in event that we passed a column that isn’t supported in Excel
On Error Resume Next
colLet = Split(Cells(1, ColumnNumber).Address, “\$”) ‘Split Address into array using the “\$” as a delimiter
If colLet(1) = “” Then ‘error handler
getColLet = “error”
MsgBox “This version of Excel does not support ” & ColumnNumber & ” Columns”
Else
getColLet = colLet(1)
End If
On Error GoTo 0 ‘Reset error handling to default
nWorkbook.Close (False) ‘Close new workbook without saving
aWorkbook.Activate ‘Activate the previously active workbook

Application.ScreenUpdating = True ‘Turn Screen updating back on

End Function

26. Jon Peltier says:

How does the overhead of opening and closing a workbook improve on the earlier proposed solutions? It also limits one to 256 columns, or letters only up to IV.

27. Die_Another_Day says:

Actually John, this is not specifically limited to 256 columns. Technically it is limited to the max number of columns for the current version of Excel. If a future version of Excel is capable of > 256 columns it won’t return an error this way I can make sure the column really does exist. However I’m not sure if future versions of Excel will be backwards compatible with the code.

Die_Another_Day

P.S. Thanks for taking the time to look at my code

28. Jon Peltier says:

Yeah, okay, it’s not limited to 256, it’s limited to the current column count, which is 256. Many of the other approaches don’t have this limit.

I was really more concerned with opening and closing a workbook just to generate a text string. Worksheets.Add would be less of a performance hit, I’m sure. But I’d investigate one of the other approaches. I might even convert to R1C1 notation.

29. Die_Another_Day says:

Jon, here’s a more “unique” way of doing this… I view it as a number base problem and found it to be quite simple to go both ways.

Function ColNumToLet(ByVal ColumnNumber As Integer) As String
Dim Remain As Double
Dim Quot As Integer
If ColumnNumber 26
ColNumToLet = Chr(Quot Mod 26 + 64) & ColNumToLet
Quot = WorksheetFunction.RoundDown(Quot / 26, 0)
Loop
ColNumToLet = Chr(Quot + 64) & ColNumToLet
End Function

Function ColLetToNum(ByVal otherBaseNumber As String) As Double
Dim index As Double
Dim digits As String
Dim digitValue As Double

digits = “ABCDEFGHIJKLMNOPQRSTUVWXYZ”

For index = 1 To Len(otherBaseNumber)
digitValue = InStr(1, digits, Mid\$(otherBaseNumber, index, 1), vbTextCompare)
ColLetToNum = ColLetToNum * 26 + digitValue
Next

End Function

this has no limits on number of columns which could be a problem…

Let me know what you think. feel free to email me if you wish. chick65stang@yahoo.com

30. Die_Another_Day says:

Crap forgot to update the Argument names for the ColLetToNum Function…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
Dim index As Double
Dim digits As String
Dim digitValue As Double

digits = “ABCDEFGHIJKLMNOPQRSTUVWXYZ”

For index = 1 To Len(ColumnLetter)
digitValue = InStr(1, digits, Mid\$(ColumnLetter, index, 1), vbTextCompare)
ColLetToNum = ColLetToNum * 26 + digitValue
Next

End Function

31. Jon Peltier says:

Yeah, number base is the way to go.

32. Rob van Gelder says:

I’ve sort of been following this thread and remembered I had something on my website.
Number to Column Letter:

33. chandu says:

how can I cahnge the column numbers back to letters?

34. Die_Another_Day says:

Jon, my previous code had some flaws in it. I believe this code should work, that is unless you have columns in excess of 3.267 Quadrillion :) I think it must be the limit of the double precision number. Note that I had to make my own “mod” function, this is due to the fact that the built in VBA mod function is a long data type and would only support columns up to 32,767.

Charles

Function ColNumToLet(ByVal ColumnNumber As Double) As String
If ColumnNumber 3.267215265265262E+15 Then
ColNumToLet = “Error”
Else
ColNumToLet = “”
If ColumnNumber 26
If myMod(ColumnNumber, 26) = 0 Then
ColNumToLet = “Z” & ColNumToLet
ColumnNumber = ColumnNumber – 1
Else
ColNumToLet = Chr(myMod(ColumnNumber, 26) + 64) & ColNumToLet
End If
ColumnNumber = WorksheetFunction.RoundDown(ColumnNumber / 26, 0)
Loop
If Not ColumnNumber = 0 Then ColNumToLet = Chr(ColumnNumber + 64) & ColNumToLet
End If
End If
End Function

Function myMod(Numerator As Double, Denominator As Double) As Double
myMod = Numerator – (WorksheetFunction.RoundDown(Numerator / Denominator, 0) * Denominator)
End Function

35. Die_Another_Day says:

Sorry, something got messed up when I posted previously.

Function ColNumToLet(ByVal ColumnNumber As Double) As String
If ColumnNumber 3.26721526526526E+15 Then
ColNumToLet = “Error”
Else
ColNumToLet = “”
If ColumnNumber 26
If myMod(ColumnNumber, 26) = 0 Then
ColNumToLet = “Z” & ColNumToLet
ColumnNumber = ColumnNumber – 1
Else
ColNumToLet = Chr(myMod(ColumnNumber, 26) + 64) & ColNumToLet
End If
ColumnNumber = WorksheetFunction.RoundDown(ColumnNumber / 26, 0)
Loop
If Not ColumnNumber = 0 Then ColNumToLet = Chr(ColumnNumber + 64) & ColNumToLet
End If
End If
End Function

36. Die_Another_Day says:

Ok, I think some weird HTML thing is killing my greater than less than stuff.

If ColumnNumber “Less Than” 1 or ColumnNumber “Greater Than” 3.26721526526526E+15 Then

37. Eric says:

Based on Rob van Gelder’s above, the following function gets the column letter of any type of range (including cells):

‘ Description: Returns the column letter for a cell or column input Range; returns empty string (“”) for a row input Range
Private Function rangeLet(ByVal inputRange As Range) As String

‘to understand how this works, here are examples of the inputRange.Address(RowAbsolute:=True, ColumnAbsolute:=False) for:
‘ – a cell (C1) “C\$1?
‘ – a column (C) “C:C”
‘ – a row (1) “\$1:\$1?
‘it gets all characters in the string before the first “\$” from everything before the (first) “:”
rangeLet = Split(Split(inputRange.Address(RowAbsolute:=True, ColumnAbsolute:=False), “:”)(0), “\$”)(0)

End Function

38. Zach says:

I’m glad this got resurrected. I just wrote the same function today for probably the 4th time. I thought my latest implementation was slick, but I don’t compare to the one liners. I think Juan Pablo is the winner here. So I feel a little bit better that I can correct even his sweet one line by changing the application.subsitute to a simple replace.

It’s funny how there are so many ways to tackle the same issue. I’ll bet every single one of these approaches differs from the other in execution time by no more than .001 seconds.

39. keepITcool says:
Function ColumnLetter(ByVal c As Long) As String
Dim p As Long
While c
p = 1 + (c – 1) Mod 26
c = (c – p) 26
ColumnLetter = Chr\$(64 + p) &amp; ColumnLetter
Wend
End Function

Elegantly pure math. None faster.

40. phil says:

need help to figure out formula to make 2 letters equal quantities and then take that value of which ever letter you put in that cell and use it in another formula. (i.e. column A reads 2000 and column B could read “S” which should equal “1? or “P” which should equal “0? then column C takes 2000 times “s”(which equals 1) and calculates 2000 in column C or 2000 times “p” (which equals 0) then it would calculate 0 in column C

41. Elias says:

Try this,

C1 =A1*(B1=”S”)

Regards

42. doglips says:

A slight improvement to keepITcool’s solution (eliminates an addition inside a loop):

Function ColumnLetter(ByVal c As Long) As String
Dim p As Long
While c
p = (c – 1) Mod 26
c = (c – p) 26
ColumnLetter = Chr\$(65 + p) &amp;#038 ColumnLetter   ‘Ampersand for concatenation
Wend
End Function
43. John Davies says:

Without a UDF, this will work for any value in A1 that represents the number of a single or two character column that is available on your version of Excel, i.e. in 2007 (untested), it should go up to ZZ:

Regards,

44. incognito says:

how to convert more than 2147483647, e.g, 21474836470
all above solutions do not work

45. Sean says:

@ incognito. You would have to define a user defined variable type that will be bigger than double. This will allow for enough accuracy to use this solution from above

Function ColNumToLet(ByVal ColumnNumber As Double) As String
If ColumnNumber 3.267215265265262E+15 Then
ColNumToLet = “Error”
Else
ColNumToLet = “”
If ColumnNumber 26
If myMod(ColumnNumber, 26) = 0 Then
ColNumToLet = “Z” & ColNumToLet
ColumnNumber = ColumnNumber – 1
Else
ColNumToLet = Chr(myMod(ColumnNumber, 26) + 64) & ColNumToLet
End If
ColumnNumber = WorksheetFunction.RoundDown(ColumnNumber / 26, 0)
Loop
If Not ColumnNumber = 0 Then ColNumToLet = Chr(ColumnNumber + 64) & ColNumToLet
End If
End If
End Function

Function myMod(Numerator As Double, Denominator As Double) As Double
myMod = Numerator – (WorksheetFunction.RoundDown(Numerator / Denominator, 0) * Denominator)
End Function

46. Rick Rothstein (MVP - Excel) says:

I know this is an old thread; but, unless I missed it, it doesn’t look like anyone ever posted this extremely simple function…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
ColLetToNum = Range(ColumnLetter & “1?).Column
End Function

47. Rick Rothstein (MVP - Excel) says:

I guess even simpler would be this…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
ColLetToNum = Columns(ColumnLetter).Column
End Function

48. Rick Rothstein (MVP - Excel) says:

Never mind… I just re-read the challenge and my code doesn’t really address it. Sorry.

49. Rick Rothstein (MVP - Excel) says:

Okay, for the original BLOG challenge to convert column letters from A up to ZZZZ to their numerical column number, this function should do that…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
L = Split(StrConv(Right(“@@@@” & UCase(ColumnLetter), 4), vbUnicode), Chr(0))
ColLetToNum = 17576# * Asc(L(0)) + 626# * Asc(L(1)) + 26# * Asc(L(2)) + Asc(L(3)) – 1166656#
End Function
50. Rick Rothstein (MVP - Excel) says:

Damn! Damn! Damn! A couple of bad constants. Here is the correct letters (A to ZZZZ) to column number function…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
L = Split(StrConv(Right(“@@@@” & UCase(ColumnLetter), 4), vbUnicode), Chr(0))
ColLetToNum = 17576# * Asc(L(0)) + 676# * Asc(L(1)) + 26# * Asc(L(2)) + Asc(L(3)) – 1169856#
End Function

51. Rick Rothstein (MVP - Excel) says:

General Case Solution

The following macro with accurately return the column number for a column letter input of up to 20 letters! So put the following into the indicated cells…

A1: ZZZZZZZZZZZZZZZZZZZZ

A2 =ColLetToNum(A1)

and cell A2 will display the column number of 20725274851017785518433805270. Of course, this value had to be returned to the cell as a text string. Anyway, here is the function…

Function ColLetToNum(ByVal ColumnLetter As String) As Variant
Power = 1
NoChars = String(Len(ColumnLetter), “@”)
L = Split(StrConv(Right(NoChars & UCase(ColumnLetter), Len(NoChars)), vbUnicode), Chr(0))
For X = 0 To Len(NoChars) – 1
ColLetToNum = ColLetToNum + Power * (Asc(L(Len(NoChars) – 1 – X)) – 64)
Power = Power * CDec(26)
Next
ColLetToNum = CStr(ColLetToNum)
End Function
52. Jorgen says:

Wouldn’t this work?

//Jorgen

53. Rick Rothstein (MVP - Excel) says:

@Jorgen,

This is from the original blog article… “The challenge before you is to write a function that converts a column number to its letter equivalent assuming columns go to ZZZZ. That’s about 450,000 columns – maybe more than I need.” So, no, for that question, your code line would not work.

54. Bob Lorenz says:

Function ColumnNumberToLetter(ByVal intRowNum As Integer) As String
Dim strTemp As String

Do While intRowNum > 0
strTemp = Chr\$((((intRowNum – 1) Mod 26) + 1) + 64) & strTemp
intRowNum = Int((intRowNum – 1) / 26)
Loop
ColumnNumberToLetter = strTemp
End Function

55. Amedee Van Gasse says:

Bob,

Your function will work up to 32.767 (AVLG), however according to MSDN documentation the column number isn’t an Integer but a Long, so one small change will make it work up to 2.147.483.647 (FXSHRXW).

56. frank says:

450,000 columns? Isn’t 16,384 more than enough?

=SUBSTITUTE(ADDRESS(1, 16384, 4), “1”, “”) returns XFD, that’s all I’l ever need.

Posting code? Use <pre> tags for VBA and <code> tags for inline.