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NaturalSQL-6.7B-v0.1

NaturalSQL is a series of models with state-of-the-art performance on Text to SQL instructions.

NaturalSQL is a LLM that can translate natural language queries to SQL based on your schema. It is finetuned on 8k text to PostgreSQL Natural Language <> SQL pairs. NaturalSQL matches the state of the art models in text to sql for it's size and produces the best in the field for complex questions.

Here is a write up, small test done here.

Table of Contents

  1. Benchmarks
  2. Future Improvements
  3. Usage
  4. SQL Generation Template
  5. Example SQL Output

Benchmarks

SQL-Eval on novel datasets not seen in training

Big thanks to the defog team for open sourcing sql-eval👏

SQL-Eval by SQL Category

NaturalSQL-6.7B-v0 matches or outperforms other industry leading models in multiple categories!

The date category will be a strong focus in the next iteration of v1.

Future Improvements

  • Much larger training set
  • More complex schemas, questions, and queries
  • Strong focus on Date Queries
  • Reward modeling via DPO

Usage

Make sure you have the correct version of the transformers library installed:

pip install transformers==4.35.2

Loading the Model

Use the following Python code to load the model:

import torch
from transformers import AutoModelForCausalLM, AutoTokenizer
tokenizer = AutoTokenizer.from_pretrained("cfahlgren1/NaturalSQL-6.7B-v0")
model = AutoModelForCausalLM.from_pretrained(
    "cfahlgren1/NaturalSQL-6.7B-v0",
    device_map="auto",
    torch_dtype=torch.float16,
)

Generating Text

To generate text, use the following Python code. Here is a full notebook with the SQL table prompt format to use.

messages=[
    { 'role': 'user', 'content': prompt}
]

inputs = tokenizer.apply_chat_template(messages, add_generation_prompt=True, return_tensors="pt").to(model.device)

# 32023 is the id of <|EOT|> token
outputs = model.generate(inputs, max_new_tokens=512, do_sample=False, top_k=50, top_p=0.95, num_return_sequences=1, eos_token_id=32023)

print(tokenizer.decode(outputs[0][len(inputs[0]):], skip_special_tokens=True))

SQL Generation Template

### Task 

Generate a SQL query to answer the following question: `{natural language question}` 

### Database Schema 

The query will run on a database with the following schema: 

'''
<SQL Table DDL Statements>
'''

### Answer 
Here is the SQL query that answers the question: `{natural language question}` 
'''sql

Example SQL Output

Example Schemas

CREATE TABLE users (
        user_id SERIAL PRIMARY KEY,
        username VARCHAR(50) NOT NULL,
        email VARCHAR(100) NOT NULL,
        password_hash TEXT NOT NULL,
        created_at TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP
    );

CREATE TABLE projects (
    project_id SERIAL PRIMARY KEY,
    project_name VARCHAR(100) NOT NULL,
    description TEXT,
    start_date DATE,
    end_date DATE,
    owner_id INTEGER REFERENCES users(user_id)
);

CREATE TABLE tasks (
    task_id SERIAL PRIMARY KEY,
    task_name VARCHAR(100) NOT NULL,
    description TEXT,
    due_date DATE,
    status VARCHAR(50),
    project_id INTEGER REFERENCES projects(project_id)
);

CREATE TABLE taskassignments (
    assignment_id SERIAL PRIMARY KEY,
    task_id INTEGER REFERENCES tasks(task_id),
    user_id INTEGER REFERENCES users(user_id),
    assigned_date DATE NOT NULL DEFAULT CURRENT_TIMESTAMP
);

CREATE TABLE comments (
    comment_id SERIAL PRIMARY KEY,
    content TEXT NOT NULL,
    created_at TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
    task_id INTEGER REFERENCES tasks(task_id),
    user_id INTEGER REFERENCES users(user_id)
);

Example SQL Outputs

Question: Show me the day with the most users joining

SELECT created_at::DATE AS day, COUNT(*) AS user_count
FROM users
GROUP BY day
ORDER BY user_count DESC
LIMIT 1;

Question: Show me the project that has a task with the most comments

SELECT p.project_name, t.task_name, COUNT(c.comment_id) AS comment_count
FROM projects p
JOIN tasks t ON p.project_id = t.project_id
JOIN comments c ON t.task_id = c.task_id
GROUP BY p.project_name, t.task_name
ORDER BY comment_count DESC
LIMIT 1;

Question: What is the ratio of users with gmail addresses vs without?

SELECT 
    SUM(CASE WHEN email ILIKE '%@gmail.com%' THEN 1 ELSE 0 END)::FLOAT / NULLIF(SUM(CASE WHEN email NOT ILIKE '%@gmail.com%' THEN 1 ELSE 0 END), 0) AS gmail_ratio
FROM 
    users;
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