Convert Spreadsheet Column Index into Character Sequence

Convert column index into corresponding column letter

I wrote these a while back for various purposes (will return the double-letter column names for column numbers > 26):

function columnToLetter(column)
{
  var temp, letter = '';
  while (column > 0)
  {
    temp = (column - 1) % 26;
    letter = String.fromCharCode(temp + 65) + letter;
    column = (column - temp - 1) / 26;
  }
  return letter;
}

function letterToColumn(letter)
{
  var column = 0, length = letter.length;
  for (var i = 0; i < length; i++)
  {
    column += (letter.charCodeAt(i) - 64) * Math.pow(26, length - i - 1);
  }
  return column;
}

Convert spreadsheet number to column letter

start_index = 1   #  it can start either at 0 or at 1
letter = ''
while column_int > 25 + start_index:   
    letter += chr(65 + int((column_int-start_index)/26) - 1)
    column_int = column_int - (int((column_int-start_index)/26))*26
letter += chr(65 - start_index + (int(column_int)))

Function to convert column number to letter?

This function returns the column letter for a given column number.

Function Col_Letter(lngCol As Long) As String
    Dim vArr
    vArr = Split(Cells(1, lngCol).Address(True, False), "$")
    Col_Letter = vArr(0)
End Function

testing code for column 100

Sub Test()
    MsgBox Col_Letter(100)
End Sub

Convert an excel or spreadsheet column letter to its number in Pythonic fashion

There is a way to make it more pythonic (works with three or more letters and uses less magic numbers):

def col2num(col):
    num = 0
    for c in col:
        if c in string.ascii_letters:
            num = num * 26 + (ord(c.upper()) - ord('A')) + 1
    return num

And as a one-liner using reduce (does not check input and is less readable so I don't recommend it):

col2num = lambda col: reduce(lambda x, y: x*26 + y, [ord(c.upper()) - ord('A') + 1 for c in col])

How to convert a column number (e.g. 127) into an Excel column (e.g. AA)

Here's how I do it:

private string GetExcelColumnName(int columnNumber)
{
    string columnName = "";

    while (columnNumber > 0)
    {
        int modulo = (columnNumber - 1) % 26;
        columnName = Convert.ToChar('A' + modulo) + columnName;
        columnNumber = (columnNumber - modulo) / 26;
    } 

    return columnName;
}

Translate a column index into an Excel Column Name

The answer I came up with is to get a little recursive. This code is in VB.Net:

Function ColumnName(ByVal index As Integer) As String
        Static chars() As Char = {"A"c, "B"c, "C"c, "D"c, "E"c, "F"c, "G"c, "H"c, "I"c, "J"c, "K"c, "L"c, "M"c, "N"c, "O"c, "P"c, "Q"c, "R"c, "S"c, "T"c, "U"c, "V"c, "W"c, "X"c, "Y"c, "Z"c}

        index -= 1 ' adjust so it matches 0-indexed array rather than 1-indexed column

        Dim quotient As Integer = index \ 26 ' normal / operator rounds. \ does integer division, which truncates
        If quotient > 0 Then
               ColumnName = ColumnName(quotient) & chars(index Mod 26)
        Else
               ColumnName = chars(index Mod 26)
        End If
End Function

And in C#:

string ColumnName(int index)
{
    index -= 1; //adjust so it matches 0-indexed array rather than 1-indexed column

    int quotient = index / 26;
    if (quotient > 0)
        return ColumnName(quotient) + chars[index % 26].ToString();
    else
        return chars[index % 26].ToString();
}
private char[] chars = new char[] {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};

The only downside it that it uses 1-indexed columns rather than 0-indexed.

Convert Column To Field Order

You are really close. You are going to want to use the ASC() function which returns the ASCII value of a character. When you subtract 64, it will get you the correct column index.

Sub Test()
Dim colletter As String
    colletter = "C"
    Debug.Print Asc(colletter) - 64
End Sub

*EDIT: I've added some controls for multiple letters and to make sure that the letters are upper case. This does, however, limit it to only having two letters, meaning column "ZZ" is your last column, but hopefully your user doesn't have more than 702 columns. :)

Sub Test()
Dim colLetter As String
Dim colNumber As Integer
Dim multiplier As Integer

colLetter = "AB"
multiplier = 0

'If there is more than one letter, that means it's gone through the whole alphabet
If Len(colLetter) > 1 Then
    multiplier = Asc(Left(UCase(colLetter), 1)) - 64
End If

colNumber = (multiplier * 26) + Asc(Right(UCase(colLetter), 1)) - 64
Debug.Print colNumber

End Sub

What is the algorithm to convert an Excel Column Letter into its Number?

public static int ExcelColumnNameToNumber(string columnName)
{
    if (string.IsNullOrEmpty(columnName)) throw new ArgumentNullException("columnName");

    columnName = columnName.ToUpperInvariant();

    int sum = 0;

    for (int i = 0; i < columnName.Length; i++)
    {
        sum *= 26;
        sum += (columnName[i] - 'A' + 1);
    }

    return sum;
}

How do I convert a spreadsheet letternamed column coordinate to an integer?

I have found particularly neat way to do this conversion:

def index_from_column_name(colname)
  s=colname.size
  (colname.to_i(36)-(36**s-1).div(3.5)).to_s(36).to_i(26)+(26**s-1)/25-1
end

Explanation why it works

(warning spoiler ;) ahead). Basically we are doing this

(colname.to_i(36)-('A'*colname.size).to_i(36)).to_s(36).to_i(26)+('1'*colname.size).to_i(26)-1

which in plain English means, that we are interpreting colname as 26-base number. Before we can do it we need to interpret all A's as 1, B's as 2 etc. If only this is needed than it would be even simpler, namely

(colname.to_i(36) - '9'*colname.size).to_i(36)).to_s(36).to_i(26)-1

unfortunately there are Z characters present which would need to be interpreted as 10(base 26) so we need a little trick. We shift every digit 1 more then needed and than add it at the end (to every digit in original colname)
`

---

Related Topics