In Python, how can I naturally sort a list of alphanumeric strings such that alpha characters sort ahead of numeric characters?

Question

This is a fun little challenge that confronted me recently. I\'ll provide my answer below, but I\'m curious to see whether there are more elegant or efficient solutions.

Answer 1

re_natural = re.compile('[0-9]+|[^0-9]+')

def natural_key(s):
    return [(1, int(c)) if c.isdigit() else (0, c.lower()) for c in re_natural.findall(s)] + [s]

for case in test_cases:
    print case[1]
    print sorted(case[0], key=natural_key)

['a', 'b', 'c']
['a', 'b', 'c']
['A', 'b', 'C']
['A', 'b', 'C']
['a', 'B', 'r', '0', '9']
['a', 'B', 'r', '0', '9']
['a1', 'a2', 'a100', '1a', '10a']
['a1', 'a2', 'a100', '1a', '10a']
['alp1', 'alp2', 'alp10', 'ALP11', 'alp100', 'GAM', '1', '2', '100']
['alp1', 'alp2', 'alp10', 'ALP11', 'alp100', 'GAM', '1', '2', '100']
['A', 'a', 'b', 'r', '0', '9']
['A', 'a', 'b', 'r', '0', '9']
['ABc', 'Abc', 'abc']
['ABc', 'Abc', 'abc']

Edit: I decided to revisit this question and see if it would be possible to handle the bonus case. It requires being more sophisticated in the tie-breaker portion of the key. To match the desired results, the alpha parts of the key must be considered before the numeric parts. I also added a marker between the natural section of the key and the tie-breaker so that short keys always come before long ones.

def natural_key2(s):
    parts = re_natural.findall(s)
    natural = [(1, int(c)) if c.isdigit() else (0, c.lower()) for c in parts]
    ties_alpha = [c for c in parts if not c.isdigit()]
    ties_numeric = [c for c in parts if c.isdigit()]
    return natural + [(-1,)] + ties_alpha + ties_numeric

This generates identical results for the test cases above, plus the desired output for the bonus case:

['A', 'a', 'A0', 'a0', '0', '00', '0A', '00A', '0a', '00a']

Answer 2

This function makes no claims as to performance at this time:

def alpha_before_numeric_natural_sensitive(unsorted_list):
    """presorting the list should work because python stable sorts; see:
    http://wiki.python.org/moin/HowTo/Sorting/#Sort_Stability_and_Complex_Sorts"""
    presorted_list = sorted(unsorted_list)
    return alpha_before_numeric_natural(presorted_list)

def alpha_before_numeric_natural(unsorted_list):
    """splice each string into tuple like so:
    'abc100def' -> ('a', 'b', 'c', 100, 'd', 'e', 'f') ->
    (ord('a'), ord('b'), ord('c'), ord('z') + 1 + 100, ...) then compare
    each tuple"""
    re_p = "([0-9]+|[A-za-z])"
    ordify = lambda s: ord('z') + 1 + int(s) if s.isdigit() else ord(s.lower())
    str_to_ord_tuple = lambda key: [ordify(c) for c in re.split(re_p, key) if c]
    return sorted(unsorted_list, key=str_to_ord_tuple)

It's based off the insight provided by this natural sort solution and this function that I wrote:

def alpha_before_numeric(unsorted_list):
    ord_shift = lambda c: c.isdigit() and chr(ord('z') + int(c.lower()) + 1) or c.lower()
    adjust_word = lambda word: "".join([ord_shift(c) for c in list(word)])

    def cmp_(a, b):
        return cmp(adjust_word(a), adjust_word(b))

    return sorted(unsorted_list, cmp_)

For full test script here that compares different functions, see http://klenwell.com/is/Pastebin20120829

Answer 3

Here's one that works for the bonus test too:

def mykey(s):
    lst = re.findall(r'(\d+)|(\D+)', s)
    return [(0,a.lower()) if a else (1,int(n)) for n, a in lst]\
        + [a for n, a in lst if a]\
        + [len(n) for n, a in lst if n]

def mysort(lst):
    return sorted(lst, key=mykey)

With this type of pattern, re.findall breaks the string to a list of tuples, e.g.

>>> re.findall(r'(\d+)|(\D+)', 'ab12cd')
[('', 'ab'), ('12', ''), ('', 'cd')]