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.
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']