Ok, wasn't know if it was direct Python operator or web2py implementation, the documentation is not clear about that : http://web2py.com/books/default/chapter/29/06/the-database-abstraction-layer?search=union
It seems, it is a python operation Richard On Tue, Oct 22, 2013 at 3:07 PM, Anthony <[email protected]> wrote: > Looks like a bug in the "|" operator method -- I think I have a fix. > > Anthony > > > On Tuesday, October 22, 2013 1:20:15 PM UTC-4, Richard wrote: > >> Ok, actually, it the "|" that lead to the "problem". >> >> len(rows1) >> >>> 1 >> len(rows2) >> >>> 3 >> >> rows3 = rows1&rows2 >> >> len(row3) >> >>> 4 >> >> len(rows1) >> >>> 1 >> len(rows2) >> >>> 3 >> >> But >> >> rows4 = rows1|rows2 >> >> len(rows4) >> >>> 4 (in my use case) >> >> len(rows1) >> >>> 4 >> >> len(rows2) >> >>> 3 >> >> Richard >> >> >> >> >> On Tue, Oct 22, 2013 at 1:01 PM, Richard Vézina <[email protected]>wrote: >> >>> Hello Anthony, >>> >>> Not sure if it is an issue or web2py (or python) is intent to work like >>> so... But let's say I have 2 rows sets : >>> >>> rows1 = query1 >>> rows2 = query2 >>> >>> rows_combined = rows1 & rows2 >>> >>> rows_combined and rows1 seems to be equal, I mean rows1 seems to be >>> containing rows2, so I can't do rows1 & rows3 and not having rows2 since >>> they are already in rows1 case the preceding rows1 & rows2... >>> >>> Thanks >>> >>> Richard >>> >>> >>> On Tue, Oct 22, 2013 at 12:21 PM, Richard Vézina >>> <[email protected]>wrote: >>> >>>> Yes sir!! >>>> >>>> I didn't think about it, I don't need to matter with duplicate, since >>>> each component of the union are mutually exclusive... >>>> >>>> At least it was working with or, but very long... >>>> >>>> Thanks Anthony!! >>>> >>>> Richard >>>> >>>> >>>> On Tue, Oct 22, 2013 at 11:45 AM, Anthony <[email protected]> wrote: >>>> >>>>> On Tuesday, October 22, 2013 11:28:07 AM UTC-4, Richard wrote: >>>>> >>>>>> Hello, >>>>>> >>>>>> I try to do union as explain the book, but it keeps running and >>>>>> running... on row1|row2, I hava complexe query though with a lot of join >>>>>> and subrequest... >>>>>> >>>>> >>>>> It loops through row2 and checks whether each record already exists in >>>>> row1, so if there are a lot of records in either and the records are >>>>> complex, I suppose that could take a while. If you just do row1&row2 >>>>> instead, that will just combine the two sets of records without removing >>>>> duplicates -- so if you don't need to worry about duplicates, that should >>>>> be a lot faster. >>>>> >>>>> Anthony >>>>> >>>>> -- >>>>> Resources: >>>>> - http://web2py.com >>>>> - http://web2py.com/book (Documentation) >>>>> - http://github.com/web2py/**web2py >>>>> <http://github.com/web2py/web2py>(Source code) >>>>> - >>>>> https://code.google.com/p/**web2py/issues/list<https://code.google.com/p/web2py/issues/list>(Report >>>>> Issues) >>>>> >>>>> --- >>>>> You received this message because you are subscribed to the Google >>>>> Groups "web2py-users" group. >>>>> To unsubscribe from this group and stop receiving emails from it, send >>>>> an email to web2py+un...@**googlegroups.com. >>>>> >>>>> For more options, visit >>>>> https://groups.google.com/**groups/opt_out<https://groups.google.com/groups/opt_out> >>>>> . >>>>> >>>> >>>> >>> >> -- > Resources: > - http://web2py.com > - http://web2py.com/book (Documentation) > - http://github.com/web2py/web2py (Source code) > - https://code.google.com/p/web2py/issues/list (Report Issues) > --- > You received this message because you are subscribed to the Google Groups > "web2py-users" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > For more options, visit https://groups.google.com/groups/opt_out. > -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/groups/opt_out.
