Marco Buttu added the comment: By reading the Ronald's comment, I realized it is better to keep it simple, so I agree with him.
The "extremely inefficient" reason seems to be less important (Python 3.3): $ python -m timeit -s "a=['a']*10000; b=['b']*10000; a+b" 100000000 loops, best of 3: 0.00831 usec per loop $ python -m timeit -s "a=['a']*10000; b=['b']*10000; sum([a, b], [])" 100000000 loops, best of 3: 0.0087 usec per loop ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue18424> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
