python sqlalchemy結合兩個表的查詢

Question

我有兩個不相關的表：

hurst = Table('Hurst', metadata,
    Column('symbol_id' , Integer, primary_key=True), 
    Column('Date'      , String(20), nullable=False),
    Column('symbol'    , String(40), nullable=False),
    Column('HurstExp'  , Float,      nullable=False),
)

fundamental = Table('Fundamental', metadata,
    Column('symbol_id' , Integer,    primary_key=True), 
    Column('Date'      , String(20), nullable=False),
    Column('symbol'    , String(40), nullable=False),
    Column('MarketCap' , Float,      nullable=False),
)

以下每個查詢都可以正常工作。 如何將它們結合起來，所以我可以得到只有那些價值超過500億以上公司的赫斯特？

# -*- coding: utf-8 -*-
"""
Created on Sun Dec 13 19:22:35 2015

@author: idf
"""

from sqlalchemy import *

def run(stmt):
    rs = stmt.execute()
    return rs

# Let's re-use the same database as before
dbh = create_engine('sqlite:///hurst.db')
dbf = create_engine('sqlite:///fundamental.db')

dbh.echo = True  # We want to see the SQL we're creating
dbf.echo = True  # We want to see the SQL we're creating

metadatah = MetaData(dbh)
metadataf = MetaData(dbf)

# The users table already exists, so no need to redefine it. Just
# load it from the database using the "autoload" feature.
hurst = Table('Hurst',       metadatah, autoload=True)
funda = Table('Fundamental', metadataf, autoload=True)

hurstQ = hurst.select(hurst.c.HurstExp < .5)
run(hurstQ)

fundaQ = funda.select(funda.c.MarketCap > 50000000000)
run(fundaQ)

如果我嘗試使用連接，我會收到錯誤：

j = join(hurst, funda, hurst.c.symbol == funda.c.symbol)
stmt = select([hurst]).select_from(j)
theJoin = run(stmt)


Traceback (most recent call last):
  File "/home/idf/anaconda3/lib/python3.5/site-packages/sqlalchemy/engine/base.py", line 1139, in _execute_context
    context)
  File "/home/idf/anaconda3/lib/python3.5/site-packages/sqlalchemy/engine/default.py", line 450, in do_execute
    cursor.execute(statement, parameters)

   cursor.execute(statement, parameters)
sqlite3.OperationalError: no such table: Fundamental

我甚至不能做簡單的版本

# This will return more results than you are probably expecting.
s = select([hurst, funda])
run(s)

Answer 1

目前沒有環境可以對此進行測試，但是應該采取以下措施：

j = join(husrt, funda, (hurst.c.symbol_id == funda.c.symbol_id) & (funda.c.MarketCap > 50000000000))
stmt = select([hurst]).select_from(j)
run(stmt)

在這個SQL Alchemy文檔頁面上查看sqlalchemy.sql.expression.join 。

Answer 2

我不相信你可以加入兩個數據庫 。

您的一個表位於一個數據庫中，另一個位於另一個數據庫中：

dbh = create_engine('sqlite:///hurst.db')
dbf = create_engine('sqlite:///fundamental.db')

您應該將所有表放在同一個database.db文件中，並使用一個db = create_engine('sqlite:///database.db') 。

更正：您可以這樣做： 在sqlalchemy中交叉數據庫加入

但是你真的想將每個表放在一個單獨的數據庫中嗎？