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

Question

Is there a more pythonic way of converting excel-style columns to numbers (starting with 1)?

Working code up to two letters:

Answer 1

This should do, in VBA, what you're looking for:

Function columnNumber(colLetter As String) As Integer

    Dim colNumber As Integer
    Dim i As Integer

    colLetter = UCase(colLetter)
    colNumber = 0
    For i = 1 To Len(colLetter)
        colNumber = colNumber + (Asc(Mid(colLetter, Len(colLetter) - i + 1, 1)) - 64) * 26 ^ (i - 1)
    Next

    columnNumber = colNumber

End Function

You can use it as you would an Excel formula--enter column, in letters, as a string (eg, "AA") and should work regardless of column length.

Your code breaks when dealing with three letters because of the way you're doing the counting--you need to use base 26.

Answer 2

I'm not sure I understand properly, do you want to "translate" the referenced C# code to python? If so, you were on the right track; just modify it so:

def column_to_number(c):
  """Return number corresponding to excel-style column."""
  sum = 0
  for l in c:
    if not l in string.ascii_letters:
      return False
    sum*=26
    sum+=ord(l.upper())-64
  return sum

Answer 3

You could also do it by a series of multiplies and adds as follows. Here "A" will equal to 1. Running time is O(n) where n is the length of the column, col.

import functools
def spreadsheet_column_encoding(col):
    return functools.reduce(
        lambda result, char: result * 26 + ord(char) - ord("A") + 1, col, 0
    )

E.g ZZ = 702:

0 * 26 + 90 - 65 + 1 = 26
26 * 26 + 90 - 65 + 1 = 702

P.S: ord('Z') = 90

To convert number to column letter, kindly see my answer here. You get to do the opposite using division and modulus calculations.

Answer 4

After reading this, I decided to find a way to do it directly in Excel cells. It even accounts for columns after Z.

Just paste this formula into a cell of any row of any column and it will give you the corresponding number.

=IF(LEN(SUBSTITUTE(ADDRESS(ROW(),COLUMN(),4),ROW(),""))=2,
 CODE(LEFT(SUBSTITUTE(ADDRESS(ROW(),COLUMN(),4),ROW(),""),1))-64*26)+
 CODE(RIGHT(SUBSTITUTE(ADDRESS(ROW(),COLUMN(),4),ROW(),""),1)-64),
 CODE(SUBSTITUTE(ADDRESS(ROW(),COLUMN(),4),ROW(),""))-64)

The theme here was to grab the letter of the column, get the Code() of it and subtract 64, based on the fact that the ASCII character code for letter A is 64.

Answer 5

just do :

print ws.Range("E2").Column

call example :

from win32com import client
xl = client.Dispatch("Excel.Application")
wb = xl.Workbooks.Open("c:/somePath/file.xls")
xl.Visible = 1
ws = wb.Sheets("sheet 1")
print ws.Range("E2").Column

result :

>>5

Answer 6

For index that starts from zero (e.g. A = 0, B = 1, and so on):

def col_to_index(col):
    A = ord('A')
    return sum(i * 26 + (ord(c) - A) for i, c in enumerate(col[::-1].upper()))