It runs noticeably faster even on localhost! Can't wait to get it on
production :)
It has not broken anything I am currently using so far. It's excellent!
On Wednesday, August 22, 2012 6:54:49 AM UTC+8, Massimo Di Pierro wrote:
>
> Looks like we finally nailed it. Thanks to Jonathan and Omar from
> lang.python mailing list.
>
> 1) class Storage has been rewritten and accessing request.something is
> now 5x-10x faster.
> Storage is used everywhere it may have a big effect on some web2py
> apps.
>
> 2) class Row, class Table and class DAL have been written using the same
> principles. (*)
>
> As result of the changes in 2, the time to define a table (20 fields)
> dropped from ~2ms to ~1ms (2x)
> This is without using lazy tables which are also in trunk.
> With lazy tables and not special attributes, the time dropped to ~3e-2ms
> (~80x)
>
> The time to retrieve a row value dropped from ~6e-3ms to ~3e-4ms (20x)
> This is important because this is what you do when you loop over rows and
> print the values in a view.
>
> It would be nice to have some benchmarks for real life applications.
>
> (*) what's the principle? The magic below:
>
> class Storage(dict):
> def __init__(self, ...):
> self.__dict__ = self
>
> The allows storage of dict-like elements in the same place as attributes
> without need to overload getattr and getattr and it is faster. It does not
> seem to break any existing APIs.
>
> PLEASE CHECK IT. If nothing breaks this is going into 2.0 next week and it
> will be hard to undo it.
>
> Massimo
>
>
> My benchmark code:
> import time
> n = 1000
>
> db=DAL(lazy_tables=False)
> fields = [Field('name')]+[Field('f%s'%k) for k in range(20)]
> db.define_table('t',*fields)
> t0 = time.time()
> for k in range(n):
> db.define_table('t%s'%k,*fields,**dict(migrate=False))
> print (time.time()-t0)/n
>
> db.t.insert(name='max')
>
> t0 = time.time()
> for k in range(n):
> row = db(db.t).select().first()
> print (time.time()-t0)/n
>
> t0 = time.time()
> for k in range(n):
> row.name=row.name # time to retrieve data
> print (time.time()-t0)/n
>
>
--
