On 02/13/2014 11:43 AM, Peter Otten wrote:
forman.si...@gmail.com wrote:
I ran across this and I thought there must be a better way of doing it,
but then after further consideration I wasn't so sure.
if key[:1] + key[-1:] == '<>': ...
Some possibilities that occurred to me:
if key.startswith('<') and key.endswith('>'): ...
and:
if (key[:1], key[-1:]) == ('<', '>'): ...
I haven't run these through a profiler yet, but it seems like the original
might be the fastest after all?
$ python -m timeit -s 's = "<alpha>"' 's[:1]+s[-1:] == "<>"'
1000000 loops, best of 3: 0.37 usec per loop
$ python -m timeit -s 's = "<alpha>"' 's[:1] == "<" and s[-1:] == ">"'
1000000 loops, best of 3: 0.329 usec per loop
$ python -m timeit -s 's = "<alpha>"' 's.startswith("<") and
s.endswith(">")'
1000000 loops, best of 3: 0.713 usec per loop
The first is too clever for my taste.
The second is fast and easy to understand. It might attract "improvements"
replacing the slice with an index, but I trust you will catch that with your
unit tests ;)
Personally, I'm willing to spend the few extra milliseconds and use the
foolproof third.
For completeness:
# the slowest method from Peter
$ python -m timeit -s 's = "<alpha>"' 's.startswith("<") and s.endswith(">")'
1000000 loops, best of 3: 0.309 usec per loop
# the re method from Roy
$ python -m timeit -s "import re;pattern=re.compile(r'^<.*>$');s = '<alpha>'"
"pattern.match(s)"
1000000 loops, best of 3: 0.466 usec per loop
--
~Ethan~
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