LangChain, PyScript, Auto-GPT, and more

Python news and tutorials from last month

Hi friends,

Welcome to another edition of my monthly newsletter. Let’s dive right in:

❓What can you expect?

This newsletter features tutorials, news, cool repos, and educational nuggets related to Python & AI.

📺📚Tutorials

Here’s a short selection of interesting tutorials from last month that are worth watching. The first three videos are from me:

• LangChain Crash Course - Build apps with language models. You can also find a written guide here.

• PyScript is officially here! Build web apps with Python & HTML

• Auto-GPT, BabyAGI, and 4 autonomous agents you must try

• 5 Steps to Build a Question Answering PDF Chatbot

• Complete Docker Course - From BEGINNER to PRO!

🐍 News from the Python Ecosystem

• The Python framework LangChain announced their $10M seed round

• PEP 684 has been accepted which introduces a per-interpreter GIL in Python 3.12

• Pandas 2.0 was released

• Anaconda Launches PyScript.com, a new platform to build web apps with Python and HTML

🤝Cool Repos

• Auto-GPT: Do I need to say something about this repo? 120K stars in 4 weeks🤯

• babyagi: An AI-powered task management system to solve tasks autonomously with the help of GPT-4. The code is intentionally kept minimal, so exploring the code is nice.

• llama-index: A project that provides a central interface to connect your LLM's with external data.

• weaviate: An open source vector database that stores both objects and vectors, allowing for combining vector search with structured filtering.

• lit-llama: An independent implementation of LLaMA based on nanoGPT that is fully open source under the Apache 2.0 license.

💡Python Tip

Do you know about the built-in any() function? any() returns True if any element of the iterable is true. Look at this example and how it can be used to refactor code:

numbers = [-1, -2, -4, 0, 3, -7]

# manually looping
has_positives = False
for n in numbers:
    if n > 0:
        has_positives = True
        break

# refactor using any()✅
has_positives = any(n > 0 for n in numbers)

I hope you enjoyed the read!

Patrick