jengland

uv has almost 2x the number of monthly downloads Poetry has.

- https://pypistats.org/packages/poetry - https://pypistats.org/packages/uv

In the 2024 Python developer survey, 18% of the ecosystem used Poetry. When I opened this manifold question[0], I'm pretty sure uv was about half of Poetry downloads.

Estimating from these numbers, probably about 30% of the ecosystem is using `uv` now. We'll get better numbers when the 2025 Python developer survey is published.

Also see this: https://biggo.com/news/202510140723_uv-overtakes-pip-in-ci-u...

[0]: https://manifold.markets/JeremiahEngland/will-uv-surpass-poe...

In my opinion Vitalik's post[0] about it was much more thoughtful and interesting.

[0]: https://vitalik.eth.limo/general/2023/07/24/biometric.html

I prompted GPT-4 with this and I think it solved it. The first time it assumed the words were generated with random letters and solved it like a math textbook problem. Then I prompted it with this:

Ok, I see you calculated the probably using randomly generated "words" from the letters of the English alphabet. I am interested in the actual probably of two real words in English that are 5 letters wrong share the first three characters.

I am a Python developer, so I will understand it if you give me a Python script.

I gave me this which looks right to me:

  import nltk
  from collections import defaultdict
  
  nltk.download('words')
  from nltk.corpus import words
  
  # Get the English words
  english_words = words.words()
  
  # Filter the words to get only five-letter words
  five_letter_words = [word for word in english_words if len(word) == 5]
  
  # Create a dictionary to store the count of words with the same first three letters
  words_dict = defaultdict(int)
  
  # Count the words with the same first three letters
  for word in five_letter_words:
      key = word[:3]
      words_dict[key] += 1
  
  # Calculate the number of pairs with the same first three letters
  same_first_three_letter_pairs = sum((count * (count - 1)) // 2 for count in words_dict.values())
  
  # Calculate the total number of pairs from the five-letter words list
  total_pairs = (len(five_letter_words) * (len(five_letter_words) - 1)) // 2
  
  # Calculate the probability
  probability = same_first_three_letter_pairs / total_pairs
  
  print(f"Probability: {probability:.4f} or {probability * 100:.2f}%")

Does anyone here know how it works and thinks it can be easily abused? The paper is here[0], but I would be satisfied with an explanation from anyone who just generally knows what "bridge-based ranking"[1] is. I'm pretty excited about the idea and I wonder if people mostly just don't know or if I am being too optimistic.

[0]: https://github.com/twitter/birdwatch/blob/main/birdwatch_pap...

[1]: https://www.belfercenter.org/publication/bridging-based-rank...

I heard that pol.is[0] is a neat technology for collective sense-making, and wanted to try it out! The problem is it's hard to find examples to try. So I thought I would create my own and submit it to Hacker News (and maybe other places depending on how it goes).

I have set up the poll to share the results and the visualization. Here is the poll description:

> If you have thoughts about the advantages and disadvantages about working fully remote, hybrid, or in-office, please submit them as comments or vote on existing comments that communicate your viewpoint.

> The results are open for anyone to view.

> To make a comment, you need to connect Twitter or Facebook. But to vote you do not need either.

I have put three "seed" questions there. But part of the point of pol.is is that the poll takers generate the best questions as they go. As people vote, the best statements bubble to the top (I think?). So please submit any thoughts you have as comments that are not already very well expressed in another comment/question.

[0]: https://pol.is

Just thankful for this resource as I consider names for my upcoming child and thought I would share.

FWIW, here's a graph[0] of the log of the data[1]. I find it's hard to eye-ball exponentials correctly without taking the log.

[0]: https://imgur.com/a/E7OBAX6

[1]: https://duckduckgo.com/traffic_data/direct.csv