[sqlalchemy] Re: Efficient dictificationof result sets
-Original Message- From: sqlalchemy@googlegroups.com [mailto:sqlalch...@googlegroups.com] On Behalf Of Andreas Jung Sent: 19 December 2008 06:30 To: sqlalchemy@googlegroups.com Subject: [sqlalchemy] Re: Efficient dictificationof result sets On 19.12.2008 2:57 Uhr, Michael Bayer wrote: On Dec 18, 2008, at 3:04 PM, Andreas Jung wrote: Does SA contain some official API to introspect the list of defined synonyms for a particular class? The goal is to take values defined as a synonym also into account for the dictification (for backward compatiblity reasons for an existing codebase). the mapper's get_property() method includes a resolve_synonyms keyword arg that indicates a given key which points to a synonym should return the actual referenced property, so a recipe that builds upon this would look like: set([mapper.get_property(p.key, resolve_synonyms=True) for p in mapper.iterate_properties]) However this does not apply when using the declarative layer. Any options within such a context? Andreas I haven't been following this discussion closely, so I'm probably wrong, but that statement doesn't sound right to me. As far as I'm aware, the declarative layer is just a convenience for setting up Tables and mapped classes at the same time. The end result is exactly the same as if you created the tables and classes in the traditional way. I would be very surprised if the above expression didn't work. You can get the mapper for a mapped class or object using the class_mapper and object_mapper functions. Hope that helps, Simon --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups sqlalchemy group. To post to this group, send email to sqlalchemy@googlegroups.com To unsubscribe from this group, send email to sqlalchemy+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
[sqlalchemy] Re: Efficient dictificationof result sets
exactly, the declarative layer is syntax sugar, i.e. shorter way to say same thing, same as elixir and dbcook - they just differ in how many things each one automates/hides. after that, i.e. after mappers compiled, it's all plain SA - sessions, queries etc... i dont know about elixir, although dbcook does have some extra conveniences around query and overall model'metadata, they do not change the overall idea - it's another shorter way to say same thing. ciao svil On Friday 19 December 2008 12:02:07 King Simon-NFHD78 wrote: -Original Message- From: sqlalchemy@googlegroups.com [mailto:sqlalch...@googlegroups.com] On Behalf Of Andreas Jung Sent: 19 December 2008 06:30 To: sqlalchemy@googlegroups.com Subject: [sqlalchemy] Re: Efficient dictificationof result sets On 19.12.2008 2:57 Uhr, Michael Bayer wrote: On Dec 18, 2008, at 3:04 PM, Andreas Jung wrote: Does SA contain some official API to introspect the list of defined synonyms for a particular class? The goal is to take values defined as a synonym also into account for the dictification (for backward compatiblity reasons for an existing codebase). the mapper's get_property() method includes a resolve_synonyms keyword arg that indicates a given key which points to a synonym should return the actual referenced property, so a recipe that builds upon this would look like: set([mapper.get_property(p.key, resolve_synonyms=True) for p in mapper.iterate_properties]) However this does not apply when using the declarative layer. Any options within such a context? Andreas I haven't been following this discussion closely, so I'm probably wrong, but that statement doesn't sound right to me. As far as I'm aware, the declarative layer is just a convenience for setting up Tables and mapped classes at the same time. The end result is exactly the same as if you created the tables and classes in the traditional way. I would be very surprised if the above expression didn't work. You can get the mapper for a mapped class or object using the class_mapper and object_mapper functions. Hope that helps, Simon --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups sqlalchemy group. To post to this group, send email to sqlalchemy@googlegroups.com To unsubscribe from this group, send email to sqlalchemy+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
[sqlalchemy] Re: Efficient dictificationof result sets
On 10.12.2008 20:36 Uhr, Michael Bayer wrote:
On Dec 10, 2008, at 2:27 PM, Andreas Jung wrote:
Hi there,
is there some more efficient way for dictifying a resultset other than
lst = list()
for row in session.query(...).all():
d = self.__dict__.copy()
for k in d.keys():
if k.startswith('_sa'):
del d[k]
lst.append(d)
Especially the loop of the keys takes pretty long for large result
sets
and tables with lots of columns.
the most efficient way would be to use a ResultProxy instead of the
ORM and to call dict() on each row.Otherwise a more succinct way
of what you're doing there, not necessarily much faster, might be
list = [
dict((k, v) for k, v in obj.__dict__.iteritems() if not
k.startswith('_sa'))
for obj in session.query.all()
Does SA contain some official API to introspect the list of defined
synonyms for a particular class? The goal is to take values defined as a
synonym also into account for the dictification (for backward
compatiblity reasons for an existing codebase).
Andreas
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begin:vcard
fn:Andreas Jung
n:Jung;Andreas
org:ZOPYX Ltd. Co. KG
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[sqlalchemy] Re: Efficient dictificationof result sets
On Dec 18, 2008, at 3:04 PM, Andreas Jung wrote:
On 10.12.2008 20:36 Uhr, Michael Bayer wrote:
On Dec 10, 2008, at 2:27 PM, Andreas Jung wrote:
Hi there,
is there some more efficient way for dictifying a resultset other
than
lst = list()
for row in session.query(...).all():
d = self.__dict__.copy()
for k in d.keys():
if k.startswith('_sa'):
del d[k]
lst.append(d)
Especially the loop of the keys takes pretty long for large result
sets
and tables with lots of columns.
the most efficient way would be to use a ResultProxy instead of the
ORM and to call dict() on each row.Otherwise a more succinct way
of what you're doing there, not necessarily much faster, might be
list = [
dict((k, v) for k, v in obj.__dict__.iteritems() if not
k.startswith('_sa'))
for obj in session.query.all()
Does SA contain some official API to introspect the list of defined
synonyms for a particular class? The goal is to take values defined
as a
synonym also into account for the dictification (for backward
compatiblity reasons for an existing codebase).
the mapper's get_property() method includes a resolve_synonyms
keyword arg that indicates a given key which points to a synonym
should return the actual referenced property, so a recipe that builds
upon this would look like:
set([mapper.get_property(p.key, resolve_synonyms=True) for p in
mapper.iterate_properties])
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[sqlalchemy] Re: Efficient dictificationof result sets
On 19.12.2008 2:57 Uhr, Michael Bayer wrote:
On Dec 18, 2008, at 3:04 PM, Andreas Jung wrote:
On 10.12.2008 20:36 Uhr, Michael Bayer wrote:
On Dec 10, 2008, at 2:27 PM, Andreas Jung wrote:
Hi there,
is there some more efficient way for dictifying a resultset other
than
lst = list()
for row in session.query(...).all():
d = self.__dict__.copy()
for k in d.keys():
if k.startswith('_sa'):
del d[k]
lst.append(d)
Especially the loop of the keys takes pretty long for large result
sets
and tables with lots of columns.
the most efficient way would be to use a ResultProxy instead of the
ORM and to call dict() on each row.Otherwise a more succinct way
of what you're doing there, not necessarily much faster, might be
list = [
dict((k, v) for k, v in obj.__dict__.iteritems() if not
k.startswith('_sa'))
for obj in session.query.all()
Does SA contain some official API to introspect the list of defined
synonyms for a particular class? The goal is to take values defined
as a
synonym also into account for the dictification (for backward
compatiblity reasons for an existing codebase).
the mapper's get_property() method includes a resolve_synonyms
keyword arg that indicates a given key which points to a synonym
should return the actual referenced property, so a recipe that builds
upon this would look like:
set([mapper.get_property(p.key, resolve_synonyms=True) for p in
mapper.iterate_properties])
However this does not apply when using the declarative layer. Any
options within such a context?
Andreas
--~--~-~--~~~---~--~~
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fn:Andreas Jung
n:Jung;Andreas
org:ZOPYX Ltd. Co. KG
adr;quoted-printable:;;Charlottenstr. 37/1;T=C3=BCbingen;;72070;Germany
email;internet:i...@zopyx.com
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url:www.zopyx.com
version:2.1
end:vcard
[sqlalchemy] Re: Efficient dictificationof result sets
On Dec 10, 2008, at 2:27 PM, Andreas Jung wrote:
Hi there,
is there some more efficient way for dictifying a resultset other than
lst = list()
for row in session.query(...).all():
d = self.__dict__.copy()
for k in d.keys():
if k.startswith('_sa'):
del d[k]
lst.append(d)
Especially the loop of the keys takes pretty long for large result
sets
and tables with lots of columns.
the most efficient way would be to use a ResultProxy instead of the
ORM and to call dict() on each row.Otherwise a more succinct way
of what you're doing there, not necessarily much faster, might be
list = [
dict((k, v) for k, v in obj.__dict__.iteritems() if not
k.startswith('_sa'))
for obj in session.query.all()
]
--~--~-~--~~~---~--~~
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