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

I made this one-liner:

colNameToNum = lambda cn: sum([((ord(cn[-1-pos]) - 64) * 26 ** pos) for pos in range(len(cn))])

It works by iterating through the letters in reverse order and multiplying by 1, 26, 26 * 26 etc, then summing the list. This method would be compatible with longer strings of letters, too.

I call it with:

print(colNameToNum("AA")) # 27

or

print(colNameToNum("XFD")) # the highest column allowed, I believe. Result = 16384

Answer 2

Use:

LETTERS = list(string.ascii_uppercase)
def column_number(column_id):
    return sum([(LETTERS.index(j)+1)*(26**i) for i,j in enumerate(column_id[::-1])])

There are several parts to this one-liner, so here's the explanation:

column_id[::-1]: reverses the string, e.g. converts 'AZ' to 'ZA', there's a good reason to do so, which we will see in a bit.

enumerate(): produces a iterable, e.g. (0, 'Z'), (1, 'A')

With some observation:

 A -> 1  = (26**0)*1              # ** is the exponential operator
 B -> 2  = (26**0)*2 
 Z -> 26 = (26**0)*26
AA -> 27 = (26**0)*1  + (26**1)*1
AB -> 28 = (26**0)*2  + (26**1)*1
AZ -> 52 = (26**0)*26 + (26**1)*1  # recall that we have (0, 'Z'), (1, 'A')

Reversing the column_id and enumerate() allows us to use the index as the exponent for 26. The rest is now trivial.

LETTERS.index(j): gives us the index of the letter in LETTERS

sum(): takes a list of numbers and returns the total.

Answer 3

One-liners tested in Python 2.7.1 and 3.5.2

excel_col_num = lambda a: 0 if a == '' else 1 + ord(a[-1]) - ord('A') + 26 * excel_col_num(a[:-1])

excel_col_name = lambda n: '' if n <= 0 else excel_col_name((n - 1) // 26) + chr((n - 1) % 26 + ord('A'))

Multi-liners likewise

def excel_column_name(n):
    """Number to Excel-style column name, e.g., 1 = A, 26 = Z, 27 = AA, 703 = AAA."""
    name = ''
    while n > 0:
        n, r = divmod (n - 1, 26)
        name = chr(r + ord('A')) + name
    return name

def excel_column_number(name):
    """Excel-style column name to number, e.g., A = 1, Z = 26, AA = 27, AAA = 703."""
    n = 0
    for c in name:
        n = n * 26 + 1 + ord(c) - ord('A')
    return n

def test (name, number):
    for n in [0, 1, 2, 3, 24, 25, 26, 27, 702, 703, 704, 2708874, 1110829947]:
        a = name(n)
        n2 = number(a)
        a2 = name(n2)
        print ("%10d  %-9s  %s" % (n, a, "ok" if a == a2 and n == n2 else "error %d %s" % (n2, a2)))

test (excel_column_name, excel_column_number)
test (excel_col_name, excel_col_num)

All tests print

         0             ok
         1  A          ok
         2  B          ok
         3  C          ok
        24  X          ok
        25  Y          ok
        26  Z          ok
        27  AA         ok
       702  ZZ         ok
       703  AAA        ok
       704  AAB        ok
   2708874  EXCEL      ok
1110829947  COLUMNS    ok

Answer 4

Here is one way to do it. It is a variation on code in the XlsxWriter module:

def col_to_num(col_str):
    """ Convert base26 column string to number. """
    expn = 0
    col_num = 0
    for char in reversed(col_str):
        col_num += (ord(char) - ord('A') + 1) * (26 ** expn)
        expn += 1

    return col_num


>>> col_to_num('A')
1
>>> col_to_num('AB')
28
>>> col_to_num('ABA')
729
>>> col_to_num('AAB')
704

Answer 5

Here is a recursive solution:

def column_string_to_num(s):
    n = ord(s[-1]) - 64
    if s[:-1]:
        return 26 * (column_string_to_num(s[:-1])) + n
    else:
        return n
    
column_string_to_num("AB")
#output: 28

The inverse can also be defined recursively, in a similar way:

def column_num_to_string(n):
    n, rem = divmod(n - 1, 26)
    next_char = chr(65 + rem)
    if n:
        return column_string(n) + next_char
    else:
        return next_char

column_num_to_string(28)
#output: 'AB'