On 29 Mar 2005 23:41:15 -0800, [EMAIL PROTECTED] (OPQ) wrote:
[...]
>> > for k in hash.keys()[:]: # Note : Their may be a lot of keys here
>> > if len(hash[k])<2:
>> > del hash[k]
>>
>> - Try if it isn't faster to iterate using items instead of iterating
>> over keys
>
>items are huge lists of numbers. keys are simple small strings. And
>even if it is faster, how can I find the key back, in order to delete
>it ?
items() returns a list of (k,v) tuples, so you can unpack them
into the for assignment target, e.g.,
for k,v in hashh.items():
if len(v)<2:
del hassh[k]
>for v in hashh.items():
> if len(v)<2:
> del ???????
>
but to avoid a big temporary items() list, perhaps (untested)
for k in [k for k,v in hashh.iteritems() if len(v)<2]:
del hassh[k]
UIAM that should build a temporary safe list
only of the keys selected for deletion.
Regards,
Bengt Richter
--
http://mail.python.org/mailman/listinfo/python-list
