Column Numbers To Letters

By in Uncategorized on .

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

    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

    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

    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

    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

    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

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

    =LEFT(ADRESS(1;A1;4);FIND(1;ADDRESS(1;A1;4))-1)

  7. Harald Staff

    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

  8. Harald Staff

    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.

  9. kevboy

    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.

  10. Gary Waters

    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

  11. Gary Waters

    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

  12. Harald Staf

    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 ?

  13. Harald Staff

    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 ?

  14. Roger Crawley

    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

  15. Tushar Mehta

    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.

  16. The Contrarian

    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.

  17. anonymous

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

  18. Edd

    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

  19. Die_Another_Day

    Function getColLet(ColumnNumber As Integer) As String
    Dim colLet() As String
    On Error Resume Next
    colLet = Split(Cells(1, ColumnNumber).Address, “$”)
    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

  20. Die_Another_Day

    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
    Set nWorkbook = Application.Workbooks.Add
    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

  21. Jon Peltier

    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.

  22. Die_Another_Day

    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

  23. Jon Peltier

    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.

  24. Die_Another_Day

    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

  25. Die_Another_Day

    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

  26. Rob van Gelder

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

    strColumn = Split(Columns(lngColumn).Address(, False), “:”)(1)

  27. Die_Another_Day

    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

  28. Die_Another_Day

    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

  29. Die_Another_Day

    Ok, I think some weird HTML thing is killing my greater than less than stuff.
    line 2 should read:

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

  30. Eric

    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

  31. Zach

    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.

  32. keepITcool
    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.

  33. phil

    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

  34. doglips

    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
  35. John Davies

    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:
    =SUBSTITUTE(LEFT(ADDRESS(1,A1),3),”$”,””)

    Regards,

  36. Sean

    @ 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

  37. Rick Rothstein (MVP - Excel)

    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

  38. Rick Rothstein (MVP - Excel)

    I guess even simpler would be this…

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

  39. Rick Rothstein (MVP - Excel)

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

  40. Rick Rothstein (MVP - Excel)

    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
  41. Rick Rothstein (MVP - Excel)

    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

  42. Rick Rothstein (MVP - Excel)

    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
  43. Rick Rothstein (MVP - Excel)

    @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.

  44. Bob Lorenz

    How about this one? I think it will work for any positive integer.

    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

  45. Amedee Van Gasse

    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).

  46. frank

    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.

  47. Pingback: How do I find the Excel column name that corresponds to a given integer? - Excel Solutions - Developers Q & A

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