excel-style label conversion

excel_col_names.py

"""MIT license"""

import xlrd
import re

def col2index(name):
    name = name.upper()
    col = -1
    for i, letter in enumerate(name[::-1]):
        col += (ord(letter) - ord('A') + 1) * 26**i
    return col

def row2index(name):
    return int(name) - 1

def cell2index(name):
    colname, rowname, _ = re.split(r'(\d+)', name)
    return row2index(rowname), col2index(colname)

for i in range(1000):
    assert i == col2index(xlrd.colname(i))

print cell2index('AB1243')