Message192818
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 |
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Date |
User |
Action |
Args |
2013-07-10 15:21:31 | marco.buttu | set | recipients:
+ marco.buttu, ronaldoussoren, ezio.melotti, r.david.murray, docs@python |
2013-07-10 15:21:31 | marco.buttu | set | messageid: <1373469691.51.0.592655347063.issue18424@psf.upfronthosting.co.za> |
2013-07-10 15:21:31 | marco.buttu | link | issue18424 messages |
2013-07-10 15:21:31 | marco.buttu | create | |
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