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You won't like it but...sounds like you are using the wrong language for the task in hand. If you want to get some huge leaps in speed then a port to a compiled language would be the next step to go. Compiled languages run much, much faster than a scripting language ever will so you'll see your processing time drop off.

Additionally you might be able to dump the data into the DB using a build in command. Postgres had one (Dump? Load? something like that) which would read in a tab delimited text file who's columns matched up with the columns in the table. That would allow you to just focus on getting a text file in the right format and then spitting it into DB with one command and let it handle the optimisation of that rather than yourself.

You've done the right thing with knocking the ORM on the head, splitting the file should not be needed though as your text file reader should just use a buffer internally so it "should" not matter but I'm not a *nix guy so could be wrong on that front.

We've done something similar with a .net app that chomps through 20Gb of files every morning doing RegExp on every line, keeps a in memory hash for unique records and then pokes new ones into a DB. From that we then spit out 9000+ JS files using a Ruby Script for ease (this is the slowest part). We used to have the importer written in Ruby too and the whole thing took 3+ hours, re-write in .net runs the whole process in about 30-40 mins and 20 of that is the slow Ruby script (not worth optimising that anymore though it does the job well enough).

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Use multiple inheritance, so your class with common tests doesn't itself inherit from TestCase.

import unittest""class CommonTests(object):"    def testCommon(self):"        print 'Calling BaseTest:testCommon'"        value = 5"        self.assertEquals(value, 5)""class SubTest1(unittest.TestCase, CommonTests):""    def testSub1(self):"        print 'Calling SubTest1:testSub1'"        sub = 3"        self.assertEquals(sub, 3)"""class SubTest2(unittest.TestCase, CommonTests):""    def testSub2(self):"        print 'Calling SubTest2:testSub2'"        sub = 4"        self.assertEquals(sub, 4)""if __name__ == '__main__':"    unittest.main()

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What are you looking for -- all (the countable infinity of) possible solutions? The "shortest" (by some measure) non-empty solution, or the set of equal-shortest ones, or...?

Because, if any solution will do, setting U and V both to [] meets all the stated conditions, and is O(1) to boot;-).

Edit: ok, so, joking apart, here's a nicely symmetrical solution to print the first ten of the countably-infinite non-empty solutions:

import itertools as it"import collections""L = ['a', 'ab', 'bba']"M = ['baa', 'aa', 'bb']""def cmbs(L=L, M=M):"  Ucans = collections.defaultdict(list)"  Vcans = collections.defaultdict(list)"  sides = (L, Vcans, Ucans), (M, Ucans, Vcans)"  for i in it.count(1):"    for k, (G, Ocans, Tcans) in enumerate(sides):"      for u in it.product(G, repeat=i):"        j = ''.join(u)"        if j in Ocans:"          for samp in Ocans[j]:"            result = samp, u"            yield result[1-k], result[k]"        Tcans[j].append(u)""if __name__ == '__main__':"  for x, y in it.islice(cmbs(), 10):"    print x, y, ''.join(x), ''.join(y)

which emits

('a', 'a') ('aa',) aa aa"('bba', 'a') ('bb', 'aa') bbaa bbaa"('a', 'a', 'a', 'a') ('aa', 'aa') aaaa aaaa"('a', 'a', 'bba', 'a') ('aa', 'bb', 'aa') aabbaa aabbaa"('a', 'ab', 'a', 'a') ('aa', 'baa') aabaa aabaa"('a', 'ab', 'bba', 'a') ('aa', 'bb', 'baa') aabbbaa aabbbaa"('bba', 'a', 'a', 'a') ('bb', 'aa', 'aa') bbaaaa bbaaaa"('bba', 'ab', 'a', 'a') ('bb', 'aa', 'baa') bbaabaa bbaabaa"('bba', 'ab', 'bba', 'a') ('bb', 'aa', 'bb', 'baa') bbaabbbaa bbaabbbaa"('bba', 'a', 'bba', 'a') ('bb', 'aa', 'bb', 'aa') bbaabbaa bbaabbaa

I'm not sure what's meant by O(N) in the context of a problem with countably infinite solutions -- what's N supposed to be here?!-)

Edit 2: changed to use (default)dicts of lists to ensure it finds all solutions even when the same joined-string can be made in > 1 ways from one of the input collections (a condition which did not occur in the sample input, so the sample output is unaffected); for example, if L is ['a', 'aa'] clearly any joined string with > 1 a can be made in multiple ways -- the current solution will emit all of those multiple ways when such a joined string matches one made for M, while the previous one just emitted one of them.

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