On 12/10/14, Oleg Broytman <[email protected]> wrote:
> On Wed, Dec 10, 2014 at 01:09:50AM +0100, "Fetchinson ."
> <[email protected]> wrote:
>> On 12/10/14, Oleg Broytman <[email protected]> wrote:
>> > you'd better implement your own
>> > caching with proper locking.
>>
>> Well, that was exactly the question, how would I do that? :)
>> Precisely for the reason you mention above I was wary of using a more
>> or less standard caching decorator. Can't I attach the computed value
>> to the instance itself somehow by setting a new property?
>
> The classical approach is:
>
> def __init__(self):
> self.__cache = None
> self.__cache_lock = Lock()
>
> def _get_value(self):
> if self.__cache is not None:
> return self.__cache
> self.__cache_lock.acquire()
> try:
> if self.__cache is not None: # Calculated in another thread
> return self.__cache
> self.__cache = ...do expensive calculation once...
> return self.__cache
> finally: # finally works both for exceptions and returns
> self.__cache_lock.release()
Great, thanks a lot, this will get me started.
But I was thinking, maybe I'm doing things not in an optimal way: the
reason I thought about caching is that my expensive property really
takes a long time. I have 'collections' which can contain 'articles'
and 'articles' can be 'tagged'. So there is a many-to-many
relationship between 'articles' and 'tags' done by RelatedJoin in both
directions. What I'd like to have is for each 'collection' have a list
of [ ( 'tag1', 'number-of-articles' ), ( 'tag2', 'number-of-articles'
), ..... ] so for each 'collection' a list of all tags and the number
of articles associated with each tag. And for a collection of even
moderate size, let's say about 1000 articles and about 100 tags, this
take on the order of 2-3 seconds. This is my current approach:
class collection( SQLObject ):
articles = MultipleJoin( 'article' )
class article( SQLObject ):
tags = RelatedJoin( 'tag' )
class tag( SQLObject ):
name = StringCol( )
articles = RelatedJoin( 'article' )
def _get_articlecount( self ):
"""Count number of articles with given tag."""
# maybe this could be done better?
return tag.select( tag.q.name == self.name ).throughTo.items.count( )
coll = collection.get( 1 )
# get list of all articles, then list of all distinct tags of all articles
# I guess this could be done also better?
tags = { }
for art in coll.articles:
for tag in art.tags:
tags[tag] = 1
tags = tags.keys( )
# the above is quite slow, but the following is slower still
# how could I speed this up?
result = [ ( tag, tag.articlecount ) for tag in tags ]
Cheers,
Daniel
> Oleg.
> --
> Oleg Broytman http://phdru.name/ [email protected]
> Programmers don't die, they just GOSUB without RETURN.
>
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