Ron Adam wrote:
> Ron Adam wrote:
>> Alex Martelli wrote:
>>> Ron Adam <[EMAIL PROTECTED]> wrote:
>>> ...
>>>> Considering the number time I sort keys after getting them, It's the
>>>> behavior I would prefer. Maybe a more dependable dict.sortedkeys()
>>>> method would be nice. ;-)
>>>
>>> sorted(d) is guaranteed to do exactly the same thing as sorted(d.keys())
>>> AND to be faster (would be pretty weird if it weren't faster...!).
>>>
>>> E.g., ...:
>>>
>>> helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
>>> 'sorted(d.keys())'
>>> 100000 loops, best of 3: 6.82 usec per loop
>>>
>>> helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
>>> 'sorted(d)'
>>> 100000 loops, best of 3: 5.98 usec per loop
>>>
>>>
>>> Alex
>>
>>
>> Yes, it did decrease it. And simplified it as well. ;)
>>
>> def psort11(s1, s2):
>> d = dict(zip(s2, s1))
>> assert len(d) == len(s1)
>> sorted(d)
>> s1[:] = d.values()
Dictionaries are not ordered, the "sorted" line does nothing except produce
a sorted list of the dictionary's keys which is ignored.
> This probably should be:
>
> def psort11(s1, s2):
> d = dict(zip(s2,s1))
> assert len(d) == len(s1)
> s1[:] = list(d[v] for v in sorted(d))
You could do this to avoid all of those lookups:
def psort_xx(s1, s2):
pairs = sorted(dict(zip(s2, s1)).iteritems())
assert len(pairs) == len(s1)
s1[:] = [s1_value for s2_value, s1_value in pairs]
--
-Scott David Daniels
[EMAIL PROTECTED]
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