Stupid idea: make a first pass to divide all the different items in groups that can be compared between each other, sort the individual groups and finally concatenate them. I assume that an item is comparable to all members of a group, if it is comparable with the first member of a group. Something like this (Python3):
import itertools
def python2sort(x):
it = iter(x)
groups = [[next(it)]]
for item in it:
for group in groups:
try:
item < group[0] # exception if not comparable
group.append(item)
break
except TypeError:
continue
else: # did not break, make new group
groups.append([item])
print(groups) # for debugging
return itertools.chain.from_iterable(sorted(group) for group in groups)
This will have quadratic running time in the pathetic case that none of the items are comparable, but I guess the only way to know that for sure is to check all possible combinations. See the quadratic behavior as a deserved punishment for anyone trying to sort a long list of unsortable items, like complex numbers. In a more common case of a mix of some strings and some integers, the speed should be similar to the speed of a normal sort. Quick test:
In [19]: x = [0, 'one', 2.3, 'four', -5, 1j, 2j, -5.5, 13 , 15.3, 'aa', 'zz']
In [20]: list(python2sort(x))
[[0, 2.3, -5, -5.5, 13, 15.3], ['one', 'four', 'aa', 'zz'], [1j], [2j]]
Out[20]: [-5.5, -5, 0, 2.3, 13, 15.3, 'aa', 'four', 'one', 'zz', 1j, 2j]
It seems to be a ‘stable sort’ as well, since the groups are formed in the order the incomparable items are encountered.