Re: [sqlalchemy] Select the hybrid method result on the query

[Leonardo L. P. da Mata]

Thanks mike, that was a helpful explanation.


2017-03-06 18:59 GMT-03:00 mike bayer :

>
>
> On 03/06/2017 04:47 PM, Leonardo  L. P. da Mata wrote:
>
>> Hello, thanks for the help.
>>
>> This makes sense but it looks like that the value is calculated twice,
>> one in the query and the other accessing the property.
>>
>> My idea is to have the result on the query return itself.
>>
>>
> "lat" and "lng" here are variables that are in your program and are not
> related to the model you have.   So you're asking that a particular object
> be returned with additional values stuck onto it.
>
> The Session uses an identity map that ensures that an object of a
> particular primary key is only present once; if you make two queries which
> both return the same primary key, you get the same object back.
>
> If you did two queries with different values of "lat" / "lng" that both
> ultimately return the same PartnerAddress object, would the old values be
> erased?  It's not clear.
>
> Instead, this "distance" you're getting is really a value associated not
> just with the object but this specific query.   You can get it back by just
> adding it as a column:
>
>  query = session.query(PartnerV3, PartnerAddress.distance(lat,
> lng)).join(PartnerAddress).order_by(PartnerAddress.distance(lat, lng))
>
>
>  for partnerv3, distance in query:
>  print distance
>
>
>
>
>>
>> 2017-03-06 18:31 GMT-03:00 mike bayer > >:
>>
>>
>>
>> On 03/06/2017 04:16 PM, Leonardo Mata wrote:
>>
>> Hello, My applications does some ordering using the distance from
>> latitude and longitude haversine distance, i was able to
>> calculate this
>> using @hybrid.method and @.*expression, but i can't output the
>> calculated distance:
>>
>> /class PartnerAddress(db.Model, WithTimestampsModel,
>> SerializeMixin):/
>> //
>> /# Columns/
>> /id = db.Column(db.Integer(), primary_key=True,
>> nullable=False)/
>> /
>> /
>> /partner_id = db.Column(/
>> /db.Integer(),/
>> /db.ForeignKey('partner.id '),/
>> /nullable=False/
>> /)/
>> /
>> /
>> /latitude = db.Column(db.Numeric(precision=9, scale=7),
>> nullable=False)/
>> /longitude = db.Column(db.Numeric(precision=10, scale=7),
>> nullable=False)/
>> /   /
>> /
>> /
>> /@hybrid_method/
>> /def distance(self, lat, lng):/
>> //
>> /return math.acos(math.cos(math.radians(self.latitude)) *
>> math.cos(math.radians(lat)) */
>> /   math.cos(math.radians(self.longitude) -
>> math.radians(lng)) +/
>> /   math.sin(math.radians(self.la
>> titude)) *
>>
>> math.sin(math.radians(lat))) * 6371/
>> /
>> /
>> /@distance.expression/
>> /def distance(cls, lat, lng):/
>> /return func.acos(func.cos(func.radians(cls.latitude)) *
>> func.cos(func.radians(lat)) */
>> / func.cos(func.radians(lng) -
>> func.radians(cls.longitude)) +/
>> / func.sin(func.radians(cls.latitude)) *
>> func.sin(func.radians(lat))) * 6371/
>> /
>> /
>> /
>> /
>> /class PartnerV3(db.Model, WithTimestampsModel,/
>> /SoftDeletableModel, SerializeMixin):/
>> //
>> /
>> /
>> /name = db.Column(db.String(128), nullable=False)/
>> //
>> /
>> /
>> /address = db.relationship(/
>> /'mustafar.partner.v3.models.ad
>> dress.PartnerAddress',/
>>
>> /backref='partner',/
>> /uselist=False/
>> /)/
>>
>>
>> when querying like this:
>>
>> query =
>> PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress
>> .distance(lat,
>> lng)).paginate(1,10, False)
>>
>> I can paginate the result, but it returns only the PartnerV3
>> object, i
>> can't access the distance.
>>
>> How do I access this distance property?
>>
>>
>> you have a PartnerV3 there so the PartnerAddress is on the .address
>> property:
>>
>> query =
>> PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress
>> .distance(lat,
>> lng))
>>
>>
>> some_partner = query.first()
>>
>> partner_address = some_partner.address
>> distance = partner_address.distance(lat, lng)
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> --
>> SQLAlchemy -
>> The Python SQL Toolkit and Object Relational Mapper
>>
>> 

Re: [sqlalchemy] Select the hybrid method result on the query

[mike bayer]

On 03/06/2017 04:47 PM, Leonardo  L. P. da Mata wrote:

Hello, thanks for the help.

This makes sense but it looks like that the value is calculated twice,
one in the query and the other accessing the property.

My idea is to have the result on the query return itself.



"lat" and "lng" here are variables that are in your program and are not 
related to the model you have.   So you're asking that a particular 
object be returned with additional values stuck onto it.


The Session uses an identity map that ensures that an object of a 
particular primary key is only present once; if you make two queries 
which both return the same primary key, you get the same object back.


If you did two queries with different values of "lat" / "lng" that both 
ultimately return the same PartnerAddress object, would the old values 
be erased?  It's not clear.


Instead, this "distance" you're getting is really a value associated not 
just with the object but this specific query.   You can get it back by 
just adding it as a column:


 query = session.query(PartnerV3, PartnerAddress.distance(lat, 
lng)).join(PartnerAddress).order_by(PartnerAddress.distance(lat, lng))



 for partnerv3, distance in query:
 print distance






2017-03-06 18:31 GMT-03:00 mike bayer >:



On 03/06/2017 04:16 PM, Leonardo Mata wrote:

Hello, My applications does some ordering using the distance from
latitude and longitude haversine distance, i was able to
calculate this
using @hybrid.method and @.*expression, but i can't output the
calculated distance:

/class PartnerAddress(db.Model, WithTimestampsModel,
SerializeMixin):/
//
/# Columns/
/id = db.Column(db.Integer(), primary_key=True, nullable=False)/
/
/
/partner_id = db.Column(/
/db.Integer(),/
/db.ForeignKey('partner.id '),/
/nullable=False/
/)/
/
/
/latitude = db.Column(db.Numeric(precision=9, scale=7),
nullable=False)/
/longitude = db.Column(db.Numeric(precision=10, scale=7),
nullable=False)/
/   /
/
/
/@hybrid_method/
/def distance(self, lat, lng):/
//
/return math.acos(math.cos(math.radians(self.latitude)) *
math.cos(math.radians(lat)) */
/   math.cos(math.radians(self.longitude) -
math.radians(lng)) +/
/   math.sin(math.radians(self.la
titude)) *
math.sin(math.radians(lat))) * 6371/
/
/
/@distance.expression/
/def distance(cls, lat, lng):/
/return func.acos(func.cos(func.radians(cls.latitude)) *
func.cos(func.radians(lat)) */
/ func.cos(func.radians(lng) -
func.radians(cls.longitude)) +/
/ func.sin(func.radians(cls.latitude)) *
func.sin(func.radians(lat))) * 6371/
/
/
/
/
/class PartnerV3(db.Model, WithTimestampsModel,/
/SoftDeletableModel, SerializeMixin):/
//
/
/
/name = db.Column(db.String(128), nullable=False)/
//
/
/
/address = db.relationship(/
/'mustafar.partner.v3.models.ad
dress.PartnerAddress',/
/backref='partner',/
/uselist=False/
/)/


when querying like this:

query =

PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress.distance(lat,
lng)).paginate(1,10, False)

I can paginate the result, but it returns only the PartnerV3
object, i
can't access the distance.

How do I access this distance property?


you have a PartnerV3 there so the PartnerAddress is on the .address
property:

query =
PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress.distance(lat,
lng))


some_partner = query.first()

partner_address = some_partner.address
distance = partner_address.distance(lat, lng)










--
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The Python SQL Toolkit and Object Relational Mapper

http://www.sqlalchemy.org/

To post example code, please provide an MCVE: Minimal, Complete, and
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 for a full
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Re: [sqlalchemy] Select the hybrid method result on the query

[Leonardo L. P. da Mata]

Hello, thanks for the help.

This makes sense but it looks like that the value is calculated twice, one
in the query and the other accessing the property.

My idea is to have the result on the query return itself.



2017-03-06 18:31 GMT-03:00 mike bayer :

>
>
> On 03/06/2017 04:16 PM, Leonardo Mata wrote:
>
>> Hello, My applications does some ordering using the distance from
>> latitude and longitude haversine distance, i was able to calculate this
>> using @hybrid.method and @.*expression, but i can't output the
>> calculated distance:
>>
>> /class PartnerAddress(db.Model, WithTimestampsModel, SerializeMixin):/
>> //
>> /# Columns/
>> /id = db.Column(db.Integer(), primary_key=True, nullable=False)/
>> /
>> /
>> /partner_id = db.Column(/
>> /db.Integer(),/
>> /db.ForeignKey('partner.id'),/
>> /nullable=False/
>> /)/
>> /
>> /
>> /latitude = db.Column(db.Numeric(precision=9, scale=7),
>> nullable=False)/
>> /longitude = db.Column(db.Numeric(precision=10, scale=7),
>> nullable=False)/
>> /   /
>> /
>> /
>> /@hybrid_method/
>> /def distance(self, lat, lng):/
>> //
>> /return math.acos(math.cos(math.radians(self.latitude)) *
>> math.cos(math.radians(lat)) */
>> /   math.cos(math.radians(self.longitude) -
>> math.radians(lng)) +/
>> /   math.sin(math.radians(self.latitude)) *
>> math.sin(math.radians(lat))) * 6371/
>> /
>> /
>> /@distance.expression/
>> /def distance(cls, lat, lng):/
>> /return func.acos(func.cos(func.radians(cls.latitude)) *
>> func.cos(func.radians(lat)) */
>> / func.cos(func.radians(lng) -
>> func.radians(cls.longitude)) +/
>> / func.sin(func.radians(cls.latitude)) *
>> func.sin(func.radians(lat))) * 6371/
>> /
>> /
>> /
>> /
>> /class PartnerV3(db.Model, WithTimestampsModel,/
>> /SoftDeletableModel, SerializeMixin):/
>> //
>> /
>> /
>> /name = db.Column(db.String(128), nullable=False)/
>> //
>> /
>> /
>> /address = db.relationship(/
>> /'mustafar.partner.v3.models.address.PartnerAddress',/
>> /backref='partner',/
>> /uselist=False/
>> /)/
>>
>>
>> when querying like this:
>>
>> query =
>> PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress
>> .distance(lat,
>> lng)).paginate(1,10, False)
>>
>> I can paginate the result, but it returns only the PartnerV3 object, i
>> can't access the distance.
>>
>> How do I access this distance property?
>>
>
> you have a PartnerV3 there so the PartnerAddress is on the .address
> property:
>
> query = 
> PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress.distance(lat,
> lng))
>
>
> some_partner = query.first()
>
> partner_address = some_partner.address
> distance = partner_address.distance(lat, lng)
>
>
>
>
>
>
>>
>>
>>
>>
>> --
>> SQLAlchemy -
>> The Python SQL Toolkit and Object Relational Mapper
>>
>> http://www.sqlalchemy.org/
>>
>> To post example code, please provide an MCVE: Minimal, Complete, and
>> Verifiable Example. See http://stackoverflow.com/help/mcve for a full
>> description.
>> ---
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>> .
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>> For more options, visit https://groups.google.com/d/optout.
>>
>
> --
> SQLAlchemy - The Python SQL Toolkit and Object Relational Mapper
>
> http://www.sqlalchemy.org/
>
> To post example code, please provide an MCVE: Minimal, Complete, and
> Verifiable Example.  See  http://stackoverflow.com/help/mcve for a full
> description.
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>



-- 
Leonardo Luiz Padovani da Mata
barr...@gmail.com

"May the force be with you, always"
"Nerd Pride... eu tenho. Voce tem?"

-- 
SQLAlchemy - 
The Python SQL Toolkit and Object Relational Mapper

http://www.sqlalchemy.org/

To post example code, please provide an MCVE: Minimal, Complete, and Verifiable 
Example.  See  http://stackoverflow.com/help/mcve for a full description.
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Re: [sqlalchemy] Select the hybrid method result on the query

[mike bayer]

On 03/06/2017 04:16 PM, Leonardo Mata wrote:

Hello, My applications does some ordering using the distance from
latitude and longitude haversine distance, i was able to calculate this
using @hybrid.method and @.*expression, but i can't output the
calculated distance:

/class PartnerAddress(db.Model, WithTimestampsModel, SerializeMixin):/
//
/# Columns/
/id = db.Column(db.Integer(), primary_key=True, nullable=False)/
/
/
/partner_id = db.Column(/
/db.Integer(),/
/db.ForeignKey('partner.id'),/
/nullable=False/
/)/
/
/
/latitude = db.Column(db.Numeric(precision=9, scale=7), nullable=False)/
/longitude = db.Column(db.Numeric(precision=10, scale=7),
nullable=False)/
/   /
/
/
/@hybrid_method/
/def distance(self, lat, lng):/
//
/return math.acos(math.cos(math.radians(self.latitude)) *
math.cos(math.radians(lat)) */
/   math.cos(math.radians(self.longitude) -
math.radians(lng)) +/
/   math.sin(math.radians(self.latitude)) *
math.sin(math.radians(lat))) * 6371/
/
/
/@distance.expression/
/def distance(cls, lat, lng):/
/return func.acos(func.cos(func.radians(cls.latitude)) *
func.cos(func.radians(lat)) */
/ func.cos(func.radians(lng) -
func.radians(cls.longitude)) +/
/ func.sin(func.radians(cls.latitude)) *
func.sin(func.radians(lat))) * 6371/
/
/
/
/
/class PartnerV3(db.Model, WithTimestampsModel,/
/SoftDeletableModel, SerializeMixin):/
//
/
/
/name = db.Column(db.String(128), nullable=False)/
//
/
/
/address = db.relationship(/
/'mustafar.partner.v3.models.address.PartnerAddress',/
/backref='partner',/
/uselist=False/
/)/


when querying like this:

query =
PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress.distance(lat,
lng)).paginate(1,10, False)

I can paginate the result, but it returns only the PartnerV3 object, i
can't access the distance.

How do I access this distance property?


you have a PartnerV3 there so the PartnerAddress is on the .address 
property:


query = 
PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress.distance(lat, 
lng))



some_partner = query.first()

partner_address = some_partner.address
distance = partner_address.distance(lat, lng)











--
SQLAlchemy -
The Python SQL Toolkit and Object Relational Mapper

http://www.sqlalchemy.org/

To post example code, please provide an MCVE: Minimal, Complete, and
Verifiable Example. See http://stackoverflow.com/help/mcve for a full
description.
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--
SQLAlchemy - 
The Python SQL Toolkit and Object Relational Mapper


http://www.sqlalchemy.org/

To post example code, please provide an MCVE: Minimal, Complete, and Verifiable 
Example.  See  http://stackoverflow.com/help/mcve for a full description.
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