Optimizing Loop. Faster Resultlist.append( [ C, D, C[1]/d[1]] )? Array? Map?

The following works well but I'd like to make it faster. The actual application could process Tuple1 and Tuple2 each with 30,000 elements and 17 nested sequences per element. I s

Solution 1:

Simply switching from an explicit for loop to a list comprehension and using a tuple instead of a list can significantly speed up your operation:

from itertools import product

ResultList =[(c, d,c[1]/ d[1])for a, b in product(t1, t2)forc, d in product(a, b)]#             ^Use a tuple here instead of a list

As can be shown by the timeit trial below:

>>>from timeit import Timer>>>original_test = Timer('original_version(Tuple1, Tuple2)', '''\...Tuple1 = (...    ((1, 2.2, 3), (2, 3.3, 4)), ...    ((5, 6.6, 7), (6, 7.7, 8))...)...Tuple2 = (...    ((10, 11, 12), (11, 12, 13), (12, 13, 14)), ...    ((20, 21, 22), (21, 22, 23), (22, 23, 24))...)...def original_version(t1, t2):...    ResultList = []...    for a in t1:...        for b in t2:...            for c in a:...                for d in b:...                    ResultList.append([c, d, c[1] / d[1]])''')>>>improved_test = Timer('improved_version(Tuple1, Tuple2)', '''\...from itertools import product...Tuple1 = (...    ((1, 2.2, 3), (2, 3.3, 4)), ...    ((5, 6.6, 7), (6, 7.7, 8))...)...Tuple2 = (...    ((10, 11, 12), (11, 12, 13), (12, 13, 14)), ...    ((20, 21, 22), (21, 22, 23), (22, 23, 24))...)...def improved_version(t1, t2):...    return [(c, d, c[1] / d[1]) for a, b in product(t1, t2) for c, d in product(a, b)]''')>>>original_time = original_test.timeit()>>>improved_time = improved_test.timeit()>>>print'Improved version is %{} faster'.format(...    (original_time - improved_time) / original_time * 100...)
Improved version is %30.0181954314 faster
>>>