sort order for strings of digits

[djc]

I learn lots of useful things from the list, some not always welcome. No 
sooner had I found a solution to a minor inconvenience in my code, than 
a recent thread here drew my attention to the fact that it will not work 
for python 3. So suggestions please:

     TODO 2012-10-22: sort order numbers first then alphanumeric
 >>> n
('1', '10', '101', '3', '40', '31', '13', '2', '2000')
 >>> s
('a', 'ab', 'acd', 'bcd', '1a', 'a1', '222 bb', 'b a 4')

 >>> sorted(n)
['1', '10', '101', '13', '2', '2000', '3', '31', '40']
 >>> sorted(s)
['1a', '222 bb', 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
 >>> sorted(n+s)
['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40', 
'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']



Possibly there is a better way but for Python 2.7 this gives the 
required result

Python 2.7.3 (default, Sep 26 2012, 21:51:14)

 >>> sorted(int(x) if x.isdigit() else x for x in n+s)
[1, 2, 3, 10, 13, 31, 40, 101, 2000, '1a', '222 bb', 'a', 'a1', 'ab', 
'acd', 'b a 4', 'bcd']


[str(x) for x in sorted(int(x) if x.isdigit() else x for x in n+s)]
['1', '2', '3', '10', '13', '31', '40', '101', '2000', '1a', '222 bb', 
'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']


But not for Python 3
Python 3.2.3 (default, Oct 19 2012, 19:53:16)

 >>> sorted(n+s)
['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40', 
'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']

 >>> sorted(int(x) if x.isdigit() else x for x in n+s)
Traceback (most recent call last):
   File "<stdin>", line 1, in <module>
TypeError: unorderable types: str() < int()
 >>>

The best I can think of is to split the input sequence into two lists, 
sort each and then join them.


-- 
djc