getting the row and column numbers from coordinate value in openpyxl

There is a method in the openpyxl.utils.cell module that meets the desired functionality. The method, openpyxl.utils.cell.coordinate_to_tuple(), takes as input the alphanumeric excel coordinates as a string and returns these coordinates as a tuple of integers.

openpyxl.utils.cell.coordinate_to_tuple('B1')
>> (1, 2)

This provides a cleaner, one-line solution using the specified lib.

openpyxl has a function called get_column_letter that converts a number to a column letter.

from openpyxl.utils import get_column_letter
print(get_column_letter(1))

1 --> A

50 --> AX

1234-- AUL

I have been using it like:

from openpyxl import Workbook
from openpyxl.utils import get_column_letter

#create excel type item
wb = Workbook()
# select the active worksheet
ws = wb.active

counter = 0
for column in range(1,6):
    column_letter = get_column_letter(column)
    for row in range(1,11):
        counter = counter +1
        ws[column_letter + str(row)] = counter

wb.save("sample.xlsx")

What you want is openpyxl.utils.coordinate_from_string() and openpyxl.utils.column_index_from_string()

from openpyxl.utils.cell import coordinate_from_string, column_index_from_string
xy = coordinate_from_string('A4') # returns ('A',4)
col = column_index_from_string(xy[0]) # returns 1
row = xy[1]

This is building off of Nathan's answer. Basically, his answer does not work properly when the row and/or column is more than one character wide. Sorry - I went a little over board. Here is the full script:

def main():
    from sys import argv, stderr

    cells = None

    if len(argv) == 1:
        cells = ['Ab102', 'C10', 'AFHE3920']
    else:
        cells = argv[1:]

    from re import match as rematch

    for cell in cells:
        cell = cell.lower()

        # generate matched object via regex (groups grouped by parentheses)
        m = rematch('([a-z]+)([0-9]+)', cell)

        if m is None:
            from sys import stderr
            print('Invalid cell: {}'.format(cell), file=stderr)
        else:
            row = 0
            for ch in m.group(1):
                # ord('a') == 97, so ord(ch) - 96 == 1
                row += ord(ch) - 96
            col = int(m.group(2))

            print('Cell: [{},{}] '.format(row, col))

if __name__ == '__main__':
    main()

Tl;dr with a bunch of comments...

# make cells with multiple characters in length for row/column
# feel free to change these values
cells = ['Ab102', 'C10', 'AFHE3920']

# import regex
from re import match as rematch

# run through all the cells we made
for cell in cells:
    # make sure the cells are lower-case ... just easier
    cell = cell.lower()

    # generate matched object via regex (groups grouped by parentheses)
    ############################################################################
    # [a-z] matches a character that is a lower-case letter
    # [0-9] matches a character that is a number
    # The + means there must be at least one and repeats for the character it matches
    # the parentheses group the objects (useful with .group())
    m = rematch('([a-z]+)([0-9]+)', cell)

    # if m is None, then there was no match
    if m is None:
        # let's tell the user that there was no match because it was an invalid cell
        from sys import stderr
        print('Invalid cell: {}'.format(cell), file=stderr)
    else:
        # we have a valid cell!
        # let's grab the row and column from it

        row = 0

        # run through all of the characters in m.group(1) (the letter part)
        for ch in m.group(1):
            # ord('a') == 97, so ord(ch) - 96 == 1
            row += ord(ch) - 96
        col = int(m.group(2))

        # phew! that was a lot of work for one cell ;)
        print('Cell: [{},{}] '.format(row, col))

print('I hope that helps :) ... of course, you could have just used Adam\'s answer,\
but that isn\'t as fun, now is it? ;)')