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copied!<p>You can use <a href="http://docs.python.org/library/itertools.html#itertools.product" rel="nofollow noreferrer">itertools.product</a> for this. It returns all possible combinations.</p> <p>For example</p> <pre><code>for a1, a2, b in itertools.product(optionlist1,optionlist1,optionlist2): do_something(a1,a2,b) </code></pre> <p>This will produce "doubles" as [a1,a1,b2] and [a2,a3,b2],[a3,a2,b2]. You can fix this with a filter. The following prevents any doubles*: </p> <pre><code>for a1,a2,b in itertools.ifilter(lambda x: x[0]<x[1], itertools.product(optionlist1,optionlist1,optionlist2)): do_something(a1,a2,b) </code></pre> <p>(*) This assumes that the options have some natural ordering which will be the case with all primitive values.</p> <p><a href="https://stackoverflow.com/users/572606/shang">shang</a>'s <a href="https://stackoverflow.com/questions/4906041/testing-all-combinations-in-python/4906054#4906054">answer</a> is also very good. I wrote some code to compare them: </p> <pre><code>from itertools import ifilter, product import random from timeit import repeat def generator_way(list1, list2): def combinations(list1, list2): return ([opt1, opt2, opt3] for i,opt1 in enumerate(list1) for opt2 in list1[i+1:] for opt3 in list2) count = 0 for a1,a2,b in combinations(list1,list2): count += 1 return count def itertools_way(list1,list2): count = 0 for a1,a2,b in ifilter(lambda x: x[0] < x[1], product(list1,list1,list2)): count += 1 return count list1 = range(0,100) random.shuffle(list1) list2 = range(0,100) random.shuffle(list2) print sum(repeat(lambda: generator_way(list1,list2),repeat = 10, number=1))/10 print sum(repeat(lambda: itertools_way(list1,list2),repeat = 10, number=1))/10 </code></pre> <p>And the result is: </p> <pre><code>0.189330005646 0.428138256073 </code></pre> <p>So the generator method is faster. However, speed is not everything. Personally I find my code 'cleaner', but the choice is yours!</p> <p>(Btw, they give both identical counts, so both are equally correct.)</p>

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