Thanks very much Niphlod. I was mostly wondering whether this was common knowledge or not. But I've also never taken the time (pun intended) to learn how to time functions, so your sample code is really helpful. It has gone in my snippets files.
Your test confirms what I thought - that for the kind of apps I'm writing right now the speed difference in using Storage objects and the dal is too small to affect user experience. And as you say the design and maintenance gains are much more significant. If I'm going to speed up my app I'm much better served by looking at the usual webdev suspects -- number of file and image downloads, resolution of image files, server optimization, etc. Good to know. Ian On Thursday, October 4, 2012 3:52:24 PM UTC-4, Niphlod wrote: > > Not directed only to monotasker: aren't all you tired of reading > benchmarks when the sourcecode is there ? > Just set-up your own test and see how it performs. Let's test it. Below > the code to do a simple test. > > PS: my results are > set storage takes 1.86146702766 > set dict takes 0.960257053375 > get storage takes 6.44219303131 > get dict takes 1.02610206604 > > This is for 1 million repetitions. I can allow this time difference if my > code is more readable. > To make a simple analogy, let's say I lost 7 seconds from the use of > storage instead of plain dict, for 1 million repetitions. Let's say I can > skip 100 repetitions using dict instead of Storage for serving one page (a > function in my controller). I'm going to gain 7 seconds every 10 thousands > of pages served. > Let's assume that in the same function I have 3 simple queries. Let's be > super-optimistic, I have a super-fast db (emphatize on super-fast). Every > query is returned in 20ms (please note that on average db response times > are higher). > I can easily gain 20ms if I cut a simple query from it (let's say I do a > join instead of 2 separate fetches). For every page I'm cutting 20ms. It > means that for 1000 pages I gained 20 seconds. For 10000 the gain is 200 > seconds. Lost 7 seconds in the process choosing storage over dict ? Let be > it! > > > from gluon.storage import Storage > > def set_storage(): > st = Storage() > st.test1 = 'test1' > st.test2 = 'test2' > st.test3 = 'test3' > st.test4 = 'test4' > st.test5 = 'test5' > st.test6 = 'test6' > st.test7 = 'test7' > st.test8 = 'test9' > st.test9 = 'test9' > st.test10 = 'test10' > return st > > def set_dict(): > st = dict() > st['test1'] = 'test1' > st['test2'] = 'test2' > st['test3'] = 'test3' > st['test4'] = 'test4' > st['test5'] = 'test5' > st['test6'] = 'test6' > st['test7'] = 'test7' > st['test8'] = 'test8' > st['test9'] = 'test9' > st['test10'] = 'test10' > return st > > def get_storage(st): > return st.test1 + st.test2 + st.test3 + st.test4 + st.test5 +\ > st.test6 + st.test7 + st.test8 + st.test9 + st.test10 == > 'test1test2test3test4test5test6test7test9test9test10' > > def get_dict(st): > return st['test1'] + st['test2'] + st['test3'] + st['test4'] + st[ > 'test5'] +\ > st['test6'] + st['test7'] + st['test8'] + st['test9'] + st['test10'] > == 'test1test2test3test4test5test6test7test8test9test10' > > if __name__ == '__main__': > from timeit import Timer > t0 = Timer(setup='from __main__ import set_storage', > stmt='set_storage()') > t1 = Timer(setup='from __main__ import set_dict', > stmt='set_dict()') > > t2 = Timer(setup="from __main__ import set_storage, get_storage; st = > set_storage()", stmt="get_storage(st)") > t3 = Timer(setup="from __main__ import set_dict, get_dict; st = > set_dict()", stmt="get_dict(st)") > > print 'set storage takes ', t0.timeit(number=1000000) > print 'set dict takes', t1.timeit(number=1000000) > print 'get storage takes', t2.timeit(number=1000000) > print 'get dict takes', t3.timeit(number=1000000) > > > > > On Thursday, October 4, 2012 7:53:13 PM UTC+2, monotasker wrote: >> >> Has anyone looked at the speed differences between operations performed >> with a Storage object and the equivalent object with a dictionary? I wonder >> how these would compare? >> >> bob = MyStorageObject.name >> >> bob = MyDictionary['name'] >> >> I suspect that the difference with one lookup would be trivial, but I'm >> wondering whether it is enough that it could make a noticeable difference >> if we're working with a long list of nested Storage objects or nested >> dicts. E.g.: >> >> allrows = db(db.mytable.id > 0).select() >> allrows.find(lambda row: [n for n in row.tags[0].names if n in >> list_of_names]) >> >> allrows_list = allrows.as_list() >> allrows_list = [d for d in allrows_list if [n for n in >> d['tags'][0]['names'] in list_of_names]] >> >> Does anyone have an idea whether there will be much speed difference? >> >> Ian >> > --
