## Column Numbers To Letters

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

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?

Am I getting annoying ?

Function ColLetter(ColNumber As Long) As String

ColLetter = Application.Substitute(Cells(1, ColNumber).Address(False, False), “1?, “”)

End Function

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.

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)

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.

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

This is only testet to 702 columns

Columnnumber is written in A1

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

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

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

Wow, I think Harald wins.

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.

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.

247, 131174, 17311, 217287, 254

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

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

Gary: Keep them coming, this is good stuff.

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 ?

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 ?

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

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.

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.

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

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

ignore that its wrong!

[...] So I’m grateful for the existence of Dick’s Blog and in particular this post. [...]

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

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

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.

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

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.

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

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

Yeah, number base is the way to go.

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)

how can I cahnge the column numbers back to letters?

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

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

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

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

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.

Dim p As Long

While c

p = 1 + (c – 1) Mod 26

c = (c – p) 26

ColumnLetter = Chr$(64 + p) & ColumnLetter

Wend

End Function

Elegantly pure math. None faster.

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

Try this,

C1 =A1*(B1=”S”)

Regards

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

Dim p As Long

While c

p = (c – 1) Mod 26

c = (c – p) 26

ColumnLetter = Chr$(65 + p) &#038 ColumnLetter ‘Ampersand for concatenation

Wend

End Function

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,

how to convert more than 2147483647, e.g, 21474836470

all above solutions do not work

@ 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

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

I guess even simpler would be this…

Function ColLetToNum(ByVal ColumnLetter As String) As Double

ColLetToNum = Columns(ColumnLetter).Column

End Function

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

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…

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

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

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…

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

Wouldn’t this work?

//Jorgen

@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.

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

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

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).

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.

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