Renaming Columns When Querying With Sqlalchemy Into Pandas Dataframe

Is there a way to retain the SqlAlchemy attribute names when you query the data into a pandas dataframe? Here's a simple mapping of my database. For the school table, I've renamed

Solution 1:

I am not a SQLAlchemy expert by any means, but I have come up with a more generalized solution (or at least a start).

Caveats

• Will not handle mapped columns with the same name across different Models. You should deal with this by adding suffix, or you could modify my answer below to create pandas columns as <tablename/model name>.<mapper column name>.

It involves four key steps:

1. Qualify your query statement with labels, which will result in column names in pandas of <table name>_<column name>:

df = pd.read_sql(query.statement, query.session.bind).with_labels()

2. Separate table name from (actual) column name

table_name, col = col_name.split('_', 1)

3. Get the Model based on tablename (from this question's answers)

forcin Base._decl_class_registry.values():if hasattr(c,'__tablename__') and c.__tablename__ == tname:returnc

4. Find the correct mapped name

for k, v in sa_class.__mapper__.columns.items():
        if v.name == col:
            return k

Bringing it all together, this is the solution I have come up with, with the main caveat being it will result in duplicate column names in your dataframe if you (likely) have duplicate mapped names across classes.

from sqlalchemy import Column, Integer, String
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

classSchool(Base):
    __tablename__ = 'DimSchool'id = Column('SchoolKey', Integer, primary_key=True)
    name = Column('SchoolName', String)
    district = Column('SchoolDistrict', String)


classStudentScore(Base):
    __tablename__ = 'FactStudentScore'

    SchoolKey = Column('SchoolKey', Integer, ForeignKey('DimSchool.SchoolKey'), primary_key = True)
    PointsPossible = Column('PointsPossible', Integer)
    PointsReceived = Column('PointsReceived', Integer)

    school = relationship("School", backref='studentscore')


defmapped_col_name(col_name):
    ''' Retrieves mapped Model based on
    actual table name (as given in pandas.read_sql)
    '''defsa_class(table_name):
        for c in Base._decl_class_registry.values():
            ifhasattr(c, '__tablename__') and c.__tablename__ == tname:
                return c

    table_name, col = col_name.split('_', 1)
    sa_class = sa_class(table_name)

    for k, v in sa_class.__mapper__.columns.items():
        if v.name == col:
            return k

query = session.query(StudentScore, School).join(School)
df = pd.read_sql(query.statement, query.session.bind).with_labels()
df.columns = map(mapped_col_name, df.columns)

Solution 2:

This is the kind of solution I would bitterly complain about if I had to maintain the code afterwards. But your question has so many constraints that I cannot find anything better.

First you construct a dictionary with the equivalences of schema and class columns using introspection like this (I'm using the first example you've posted):

In [132]:

def add_to_dict(c_map, t_map, table):
    name = table.__tablename__
    t_map[name] = table.__name__
    #print name
    c_map[name] = {}
    for column in dir(table):
        c_schema_name = table.__mapper__.columns.get(column)
        if isinstance(c_schema_name, Column):
            #print column, c_schema_name.name
            c_map[name][c_schema_name.name] = column

c_map = {}
t_map = {}
add_to_dict(c_map, t_map, School)
add_to_dict(c_map, t_map, StudentScore)
print c_map['DimSchool']['SchoolKey']
print c_map['FactStudentScore']['SchoolKey']
print t_map['DimSchool']
id
SchoolKey
School

[EDIT: clarifications on the way to build the dictionary with introspection

• c_map is the dictionary of correspondences of column names
• t_map is the dictionary of correspondences of table names
• needs to be called for each class of each table
• for table names the correspondence is easy, since it is just attributes of the class of the table
• for the column names of the class, 1st iterate the attributes of the class using dir
• for each of the attributes of the class (which will be the columns of the table, but also many other things) try to get the database column name using sqlalchemymapper
• the mapper will return a Column object only if the attribute is really a column
• thus for the Column objects, add them to the column names dictionary. The database name is obtained with .name and the other is just the attribute

Run this just once after creating all the objects in the database, calling it once per table class.]

Then you take your sql statement and build up a list of the translation of the columns you are going to get:

In [134]:

df_columns = []
for column in str(query.statement).split('FROM')[0].split('SELECT')[1].split(','):
    table = column.split('.')[0].replace('"', '').strip()
    c_schema = column.split('.')[1].replace('"', '').strip()
    df_columns += [t_map[table] + '.' + eq[table][c_schema]]
print df_columns
​
['StudentScore.SchoolKey', 'StudentScore.PointsPossible', 'StudentScore.PointsReceived', 'School.id', 'School.name', 'School.district']

Finally, you read the dataframe as in your question and change the names of the columns:

In [137]:

df.columns = df_columns
In [138]:

df
Out[138]:
StudentScore.SchoolKey  StudentScore.PointsPossible StudentScore.PointsReceived School.id   School.name School.district
011None1   School1 None

(The data is just a silly register I've created).

Hope it helps!