README.md

---
base_model: deepseek-ai/deepseek-coder-6.7b-instruct
tags:
- instruct
- finetune
model-index:
- name: NaturalQuery-6.7B-v0.1
  results: []
license: other
license_name: deepseek
language:
- en
datasets:
- cfahlgren1/wiki-sql-codellama-expanded
- cfahlgren1/natural-sql
---

# **NaturalQuery-6.7B-v0.1**

**NaturalQuery** is a LLM that can translate natural language queries to SQL based on your schema.

NaturalQuery-v0.1 is finetuned on 8k text to PostgreSQL Natural Language <> SQL pairs.

**Future Improvements**:

- Much larger training set
- More complex schemas, questions, and queries
- Reward modeling via DPO
- Benchmarking

# **Usage**

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

```sh
pip install transformers==4.35.2
```

### **Loading the Model**

Use the following Python code to load the model:

```python
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](https://gist.github.com/cfahlgren1/ba17f01cf688c4229686dc3dfb4d4549) is a full notebook with the SQL table prompt format to use.

```python
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**

```sql
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)
);
```

**Question**: **Show me the day with the most users joining**
```sql
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**
```sql
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?**
```sql
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;
```