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

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])

Answer 2

You could use this oneliner using comprehension and string that is fairly easy to use:

sum([string.ascii_lowercase.index(c) + 26 ** i for i,c in enumerate(col_letters)])

Answer 3

You could just add the following to the console after installing the openpyxl module:

import openpyxl
from openpyxl.utils import get_column_letter, column_index_from_string
workbook = openpyxl.load_workbook('your_workbook.xlsx')
sheet = wb.get_sheet_by_name('your_sheet_from_workbook')
print(get_column_letter(1))
print(column_index_from_string('A'))

Just change the letters and number to suit your needs or create a for-loop to do the job for a whole project. I hope this helps. Cheers.

Answer 4

Here's what I use (wrote before I found this page):

def col_to_index(col):
    return sum((ord(c) - 64) * 26**i for i, c in enumerate(reversed(col))) - 1

And some runs:

>>> col_to_index('A')
1
>>> col_to_index('AB')
28
>>> col_to_index('ABCD')
19010

Answer 5

Using openpyxl

import openpyxl
(column_string, row) = openpyxl.cell.coordinate_from_string(address)
column = openpyxl.cell.column_index_from_string(column_string) 

Answer 6

Coincise and elegant Ruby version:

def col_num(col_name)
    col_name.split(//).inject(0) { |n, c| n * 26 + c.upcase.ord - "A".ord + 1 }
end