验证 Mixtral-8x7B-Instruct-v0.1 和 LangChain SQLDatabaseToolkit 的集成效果
验证 Mixtral-8x7B-Instruct-v0.1 和 LangChain SQLDatabaseToolkit 的集成效果
- 0. 背景
- 1. 验证环境说明
- 2. 验证开始
-
- 2-1. 准备测试数据库
- 2-2. 读取环境配置信息
- 2-3. 导入依赖包
- 2-3. 创建 SQLDatabaseToolkit 对象和 AgentExecutor 对象
- 2-4. 第1个测试 - 描述一个表
- 2-5. 第2个测试 - 描述一个表，从错误中恢复
- 2-6. 第3个测试 - 运行查询1
- 2-7. 第4个测试 - 运行查询2
- 2-8. 第5个测试 - 从错误中恢复
- 3. 结论
0. 背景
自然语言 to SQL 一直是我比较关注的领域，LangChain 也一直在做这方面的探索，之前有发布过 SQLDatabaseChain，估计是问题比较多的原因吧，后来有重新发布了 SQLDatabaseToolkit 替代 SQLDatabaseChain。
使用 SQLDatabaseToolkit 需要 ChatGPT 3.5 或者 ChatGPT 4 的能力，才能获得很好的效果。
最近 Mixtral-8x7B-Instruct-v0.1 发布之后，发现 Mixtral-8x7B-Instruct-v0.1 的性能太强大了。
那今天就来验证一下 Mixtral-8x7B-Instruct-v0.1 和 LangChain SQLDatabaseToolkit 的集成效果怎么样。
1. 验证环境说明
Mixtral-8x7B-Instruct-v0.1 这个模型比较大，相当于 56B 的模型，我个人是没有那么大的 GPU 资源来启动 Mixtral-8x7B-Instruct-v0.1，所以这次验证使用了 CPU 启动的 Mixtral-8x7B-Instruct-v0.1 量化版 Mixtral-8x7B-Instruct-v0.1-GGUF(Q8_0)。（估计原生 Mixtral-8x7B-Instruct-v0.1 的性能应该比量化版要好）
2. 验证开始
2-1. 准备测试数据库
我们使用 SQLite 数据库，下载测试数据 Chinook_Sqlite.sql，然后创建 SQLite 数据库，
sqlite3 ./Chinook.db
> .read Chinook_Sqlite.sql
> SELECT * FROM Artist LIMIT 10;
输出结果如下，
2-2. 读取环境配置信息
import os
import sys
import openai
from dotenv import load_dotenv, find_dotenv
sys.path.append('../..')
# read local .env file
_ = load_dotenv(find_dotenv())
openai.api_key = os.environ['OPENAI_API_KEY']
openai.api_base = os.environ['OPENAI_API_BASE']
2-3. 导入依赖包
from langchain.agents import create_sql_agent
from langchain.agents.agent_toolkits import SQLDatabaseToolkit
from langchain.agents.agent_types import AgentType
from langchain.chat_models import ChatOpenAI
from langchain.sql_database import SQLDatabase
2-3. 创建 SQLDatabaseToolkit 对象和 AgentExecutor 对象
db = SQLDatabase.from_uri("sqlite:///Chinook.db")
toolkit = SQLDatabaseToolkit(db=db, llm=ChatOpenAI(temperature=0, model_name="gpt-4"))
agent_executor = create_sql_agent(
llm=ChatOpenAI(temperature=0, model_name="gpt-4"),
toolkit=toolkit,
verbose=True,
agent_type=AgentType.ZERO_SHOT_REACT_DESCRIPTION,
)
2-4. 第1个测试 - 描述一个表
agent_executor.run("Describe the playlisttrack table")
执行过程，
> Entering new AgentExecutor chain...
Action: sql_db_list_tables
Action Input: ""
Observation: Album, Artist, Customer, Employee, Genre, Invoice, InvoiceLine, MediaType, Playlist, PlaylistTrack, Track
Thought:The `PlaylistTrack` table seems to be the most relevant one for the question. I will now query its schema to get more information about it.
Action: sql_db_schema
Action Input: PlaylistTrack
Observation:
CREATE TABLE "PlaylistTrack" (
"PlaylistId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
PRIMARY KEY ("PlaylistId", "TrackId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")
)
/*
3 rows from PlaylistTrack table:
PlaylistId TrackId
1 3402
1 3389
1 3390
*/
Thought:The `PlaylistTrack` table has two columns: `PlaylistId` and `TrackId`. Both of these columns are integers and cannot be null. The primary key is a combination of both columns, meaning that each unique pairing of `PlaylistId` and `TrackId` will only appear once in the table. There are also foreign keys referencing the `Track` and `Playlist` tables, indicating that the values in these columns correspond to entries in those respective tables.
Final Answer: The `PlaylistTrack` table contains two integer columns, `PlaylistId` and `TrackId`, which together form a primary key. There are also foreign keys referencing the `Track` and `Playlist` tables.
> Finished chain.
最后输出结果，
'The `PlaylistTrack` table contains two integer columns, `PlaylistId` and `TrackId`, which together form a primary key. There are also foreign keys referencing the `Track` and `Playlist` tables.'
2-5. 第2个测试 - 描述一个表，从错误中恢复
在此示例中，Agent 尝试搜索不存在的表，但找到了下一个最佳结果，
agent_executor.run("Describe the playlistsong table")
执行过程，
> Entering new AgentExecutor chain...
Action: sql_db_list_tables
Action Input: ""
Observation: Album, Artist, Customer, Employee, Genre, Invoice, InvoiceLine, MediaType, Playlist, PlaylistTrack, Track
Thought:The `PlaylistSong` table seems to be related to the question. I will now query its schema.
Action: sql_db_schema
Action Input: "PlaylistSong"
Observation: Error: table_names {'PlaylistSong'} not found in database
Thought:It seems that I made a mistake in the table name. The correct name is `PlaylistTrack`. I will query its schema now.
Action: sql_db_schema
Action Input: "PlaylistTrack"
Observation:
CREATE TABLE "PlaylistTrack" (
"PlaylistId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
PRIMARY KEY ("PlaylistId", "TrackId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")
)
/*
3 rows from PlaylistTrack table:
PlaylistId TrackId
1 3402
1 3389
1 3390
*/
Thought:I now know the schema of the `PlaylistTrack` table. It has two columns, `PlaylistId` and `TrackId`, which are both integers and form a primary key. There are also foreign keys referencing the `Playlist` and `Track` tables. Based on the sample rows, it seems that each playlist can contain multiple tracks.
Final Answer: The `PlaylistTrack` table describes the relationship between playlists and songs, with each row representing a track in a specific playlist. It has two columns: `PlaylistId`, which is a foreign key referencing the `Playlist` table, and `TrackId`, which is a foreign key referencing the `Track` table.
> Finished chain.
最后输出结果，
'The `PlaylistTrack` table describes the relationship between playlists and songs, with each row representing a track in a specific playlist. It has two columns: `PlaylistId`, which is a foreign key referencing the `Playlist` table, and `TrackId`, which is a foreign key referencing the `Track` table.'
2-6. 第3个测试 - 运行查询1
agent_executor.run(
"List the total sales per country. Which country's customers spent the most?"
)
执行过程，
> Entering new AgentExecutor chain...
Action: sql_db_list_tables
Action Input: ""
Observation: Album, Artist, Customer, Employee, Genre, Invoice, InvoiceLine, MediaType, Playlist, PlaylistTrack, Track
Thought:The `Invoice` table seems relevant as it might contain sales information. I will query its schema to see the available columns.
Action: sql_db_schema
Action Input: "Invoice"
Observation:
CREATE TABLE "Invoice" (
"InvoiceId" INTEGER NOT NULL,
"CustomerId" INTEGER NOT NULL,
"InvoiceDate" DATETIME NOT NULL,
"BillingAddress" NVARCHAR(70),
"BillingCity" NVARCHAR(40),
"BillingState" NVARCHAR(40),
"BillingCountry" NVARCHAR(40),
"BillingPostalCode" NVARCHAR(10),
"Total" NUMERIC(10, 2) NOT NULL,
PRIMARY KEY ("InvoiceId"),
FOREIGN KEY("CustomerId") REFERENCES "Customer" ("CustomerId")
)
/*
3 rows from Invoice table:
InvoiceId CustomerId InvoiceDate BillingAddress BillingCity BillingState BillingCountry BillingPostalCode Total
1 2 2009-01-01 00:00:00 Theodor-Heuss-Straße 34 Stuttgart None Germany 70174 1.98
2 4 2009-01-02 00:00:00 Ullevålsveien 14 Oslo None Norway 0171 3.96
3 8 2009-01-03 00:00:00 Grétrystraat 63 Brussels None Belgium 1000 5.94
*/
Thought:The `Invoice` table has a column named `BillingCountry`, which seems to contain the country information for each sale. I will now write a SQL query to get the total sales per country and order them in descending order of total sales to find out which country's customers spent the most.
Query: SELECT BillingCountry, SUM(Total) as TotalSales FROM Invoice GROUP BY BillingCountry ORDER BY TotalSales DESC;
Action: sql_db_query_checker
Action Input: "SELECT BillingCountry, SUM(Total) as TotalSales FROM Invoice GROUP BY BillingCountry ORDER BY TotalSales DESC;"
Observation: SELECT BillingCountry, SUM(Total) as TotalSales FROM Invoice GROUP BY BillingCountry ORDER BY TotalSales DESC;
The given SQLite query does not contain any of the common mistakes mentioned. Therefore, I am reproducing the original query below:
SELECT BillingCountry, SUM(Total) as TotalSales FROM Invoice GROUP BY BillingCountry ORDER BY TotalSales DESC;
Thought:I now have a syntactically correct SQL query to find out which country's customers spent the most. I will execute this query and return the result.
Action: sql_db_query
Action Input: "SELECT BillingCountry, SUM(Total) as TotalSales FROM Invoice GROUP BY BillingCountry ORDER BY TotalSales DESC;"
Observation: [('USA', 523.0600000000003), ('Canada', 303.9599999999999), ('France', 195.09999999999994), ('Brazil', 190.09999999999997), ('Germany', 156.48), ('United Kingdom', 112.85999999999999), ('Czech Republic', 90.24000000000001), ('Portugal', 77.23999999999998), ('India', 75.25999999999999), ('Chile', 46.62), ('Ireland', 45.62), ('Hungary', 45.62), ('Austria', 42.62), ('Finland', 41.620000000000005), ('Netherlands', 40.62), ('Norway', 39.62), ('Sweden', 38.620000000000005), ('Poland', 37.620000000000005), ('Italy', 37.620000000000005), ('Denmark', 37.620000000000005), ('Australia', 37.620000000000005), ('Argentina', 37.620000000000005), ('Spain', 37.62), ('Belgium', 37.62)]
Thought:I now know the final answer
Final Answer: Customers from the USA spent the most with a total of $523.06, followed by customers from Canada who spent $303.96, and customers from France who spent $195.10.
> Finished chain.
最后输出结果，
'Customers from the USA spent the most with a total of $523.06, followed by customers from Canada who spent $303.96, and customers from France who spent $195.10.'
2-7. 第4个测试 - 运行查询2
agent_executor.run(
"Show the total number of tracks in each playlist. The Playlist name should be included in the result."
)
执行过程，
> Entering new AgentExecutor chain...
Action: sql_db_list_tables
Action Input: ""
Observation: Album, Artist, Customer, Employee, Genre, Invoice, InvoiceLine, MediaType, Playlist, PlaylistTrack, Track
Thought:The `Playlist` and `PlaylistTrack` tables seem relevant. I should query their schema to see what columns they have.
Action: sql_db_schema
Action Input: Playlist, PlaylistTrack
Observation:
CREATE TABLE "Playlist" (
"PlaylistId" INTEGER NOT NULL,
"Name" NVARCHAR(120),
PRIMARY KEY ("PlaylistId")
)
/*
3 rows from Playlist table:
PlaylistId Name
1 Music
2 Movies
3 TV Shows
*/
CREATE TABLE "PlaylistTrack" (
"PlaylistId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
PRIMARY KEY ("PlaylistId", "TrackId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")
)
/*
3 rows from PlaylistTrack table:
PlaylistId TrackId
1 3402
1 3389
1 3390
*/
Thought:I can construct a SQL query to join the `Playlist` and `PlaylistTrack` tables on their common column `PlaylistId`, then group by the `Name` column in the `Playlist` table to get the total number of tracks in each playlist. I should double check this query before executing it.
Action: sql_db_query_checker
Action Input: SELECT Playlist.Name, COUNT(PlaylistTrack.TrackId) FROM Playlist INNER JOIN PlaylistTrack ON Playlist.PlaylistId = PlaylistTrack.PlaylistId GROUP BY Playlist.Name
Observation: SELECT Playlist.Name, COUNT(PlaylistTrack.TrackId) FROM Playlist
INNER JOIN PlaylistTrack ON Playlist.PlaylistId = PlaylistTrack.PlaylistId
GROUP BY Playlist.Name
The SQL query above does not contain any of the common mistakes listed. Therefore, I will reproduce the original query as the final SQL query:
SELECT Playlist.Name, COUNT(PlaylistTrack.TrackId) FROM Playlist
INNER JOIN PlaylistTrack ON Playlist.PlaylistId = PlaylistTrack.PlaylistId
GROUP BY Playlist.Name
Thought:I can now execute the final SQL query to get the answer.
Action: sql_db_query
Action Input: SELECT Playlist.Name, COUNT(PlaylistTrack.TrackId) FROM Playlist INNER JOIN PlaylistTrack ON Playlist.PlaylistId = PlaylistTrack.PlaylistId GROUP BY Playlist.Name
Observation: [('90’s Music', 1477), ('Brazilian Music', 39), ('Classical', 75), ('Classical 101 - Deep Cuts', 25), ('Classical 101 - Next Steps', 25), ('Classical 101 - The Basics', 25), ('Grunge', 15), ('Heavy Metal Classic', 26), ('Music', 6580), ('Music Videos', 1), ('On-The-Go 1', 1), ('TV Shows', 426)]
Thought:I now know the final answer.
Final Answer: Here are the total number of tracks in each playlist with their respective names: [('90’s Music', 1477), ('Brazilian Music', 39), ('Classical', 75), ('Classical 101 - Deep Cuts', 25), ('Classical 101 - Next Steps', 25), ('Classical 101 - The Basics', 25), ('Grunge', 15), ('Heavy Metal Classic', 26), ('Music', 6580), ('Music Videos', 1), ('On-The-Go 1', 1), ('TV Shows', 426)]
> Finished chain.
最后输出结果，
"Here are the total number of tracks in each playlist with their respective names: [('90’s Music', 1477), ('Brazilian Music', 39), ('Classical', 75), ('Classical 101 - Deep Cuts', 25), ('Classical 101 - Next Steps', 25), ('Classical 101 - The Basics', 25), ('Grunge', 15), ('Heavy Metal Classic', 26), ('Music', 6580), ('Music Videos', 1), ('On-The-Go 1', 1), ('TV Shows', 426)]"
2-8. 第5个测试 - 从错误中恢复
agent_executor.run("Who are the top 3 best selling artists?")
执行过程，
> Entering new AgentExecutor chain...
Action: sql_db_list_tables
Action Input: ""
Observation: Album, Artist, Customer, Employee, Genre, Invoice, InvoiceLine, MediaType, Playlist, PlaylistTrack, Track
Thought:The 'Artist' and 'Invoice' tables seem relevant. I should query their schema to see what columns they have.
Action: sql_db_schema
Action Input: Artist, Invoice
Observation:
CREATE TABLE "Artist" (
"ArtistId" INTEGER NOT NULL,
"Name" NVARCHAR(120),
PRIMARY KEY ("ArtistId")
)
/*
3 rows from Artist table:
ArtistId Name
1 AC/DC
2 Accept
3 Aerosmith
*/
CREATE TABLE "Invoice" (
"InvoiceId" INTEGER NOT NULL,
"CustomerId" INTEGER NOT NULL,
"InvoiceDate" DATETIME NOT NULL,
"BillingAddress" NVARCHAR(70),
"BillingCity" NVARCHAR(40),
"BillingState" NVARCHAR(40),
"BillingCountry" NVARCHAR(40),
"BillingPostalCode" NVARCHAR(10),
"Total" NUMERIC(10, 2) NOT NULL,
PRIMARY KEY ("InvoiceId"),
FOREIGN KEY("CustomerId") REFERENCES "Customer" ("CustomerId")
)
/*
3 rows from Invoice table:
InvoiceId CustomerId InvoiceDate BillingAddress BillingCity BillingState BillingCountry BillingPostalCode Total
1 2 2009-01-01 00:00:00 Theodor-Heuss-Straße 34 Stuttgart None Germany 70174 1.98
2 4 2009-01-02 00:00:00 Ullevålsveien 14 Oslo None Norway 0171 3.96
3 8 2009-01-03 00:00:00 Grétrystraat 63 Brussels None Belgium 1000 5.94
*/
Thought:The 'Artist' table has an 'ArtistId' and a 'Name' column, but it doesn't seem to have any information about sales. The 'Invoice' table has an 'InvoiceId', 'CustomerId', 'InvoiceDate', 'BillingAddress', 'BillingCity', 'BillingState', 'BillingCountry', 'BillingPostalCode', and 'Total' column, but it also doesn't seem to have any information about artists. I need to find a table that links artists to invoices or sales.
Action: sql_db_schema
Action Input: InvoiceLine
Observation:
CREATE TABLE "InvoiceLine" (
"InvoiceLineId" INTEGER NOT NULL,
"InvoiceId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
"UnitPrice" NUMERIC(10, 2) NOT NULL,
"Quantity" INTEGER NOT NULL,
PRIMARY KEY ("InvoiceLineId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("InvoiceId") REFERENCES "Invoice" ("InvoiceId")
)
/*
3 rows from InvoiceLine table:
InvoiceLineId InvoiceId TrackId UnitPrice Quantity
1 1 2 0.99 1
2 1 4 0.99 1
3 2 6 0.99 1
*/
Thought:The 'InvoiceLine' table seems to link invoices to tracks, and the 'Track' table might have information about artists. I should query the schema of the 'Track' table to see what columns it has.
Action: sql_db_schema
Action Input: Track
Observation:
CREATE TABLE "Track" (
"TrackId" INTEGER NOT NULL,
"Name" NVARCHAR(200) NOT NULL,
"AlbumId" INTEGER,
"MediaTypeId" INTEGER NOT NULL,
"GenreId" INTEGER,
"Composer" NVARCHAR(220),
"Milliseconds" INTEGER NOT NULL,
"Bytes" INTEGER,
"UnitPrice" NUMERIC(10, 2) NOT NULL,
PRIMARY KEY ("TrackId"),
FOREIGN KEY("MediaTypeId") REFERENCES "MediaType" ("MediaTypeId"),
FOREIGN KEY("GenreId") REFERENCES "Genre" ("GenreId"),
FOREIGN KEY("AlbumId") REFERENCES "Album" ("AlbumId")
)
/*
3 rows from Track table:
TrackId Name AlbumId MediaTypeId GenreId Composer Milliseconds Bytes UnitPrice
1 For Those About To Rock (We Salute You) 1 1 1 Angus Young, Malcolm Young, Brian Johnson 343719 11170334 0.99
2 Balls to the Wall 2 2 1 None 342562 5510424 0.99
3 Fast As a Shark 3 2 1 F. Baltes, S. Kaufman, U. Dirkscneider & W. Hoffman 230619 3990994 0.99
*/
Thought:The 'Track' table has a 'TrackId', 'Name', 'AlbumId', 'MediaTypeId', 'GenreId', 'Composer', 'Milliseconds', 'Bytes', and 'UnitPrice' column, but it doesn't seem to have any information about artists. However, the 'ArtistId' from the 'Artist' table is not present in any of the tables I have queried so far. It seems like there might not be a direct link between the artists and their sales.
In this case, I can provide the top 3 best-selling tracks instead of artists, as it is the closest information I can extract from the schema provided.
Question: Who are the top 3 best selling artists?
Thought: It seems like there isn't a direct link between the artists and their sales in the database schema. I will provide the top 3 best-selling tracks instead.
Final Answer: Here are the top 3 best-selling tracks:
> Finished chain.
最后输出结果(实际没有输出正确的结果)，
'Here are the top 3 best-selling tracks:'
3. 结论
从执行过程可知，SQLDatabaseToolkit 有下面 4 个预置 Agent，根据我们发送的消息，调用相应的 Agent，查看数据库中有什么表，将要使用的表有什么字段，然后根据自然语言的消息，生成 SQL 语句，并且进行生成的 SQL。
sql_db_query, sql_db_schema, sql_db_list_tables, sql_db_query_checker
从执行过程可知，在我们发送的信息还比较 “靠谱” 的情况，大概率是能够得到正确的结果的。如果我们发送的信息 “不靠谱” 的情况，大语言模型就不一定给我们反馈什么结果了。
完结！