StackOverflow Collective?

Stack Overflow has launched a new product today called Collectives, which seems like a way of aggregating multiple related tags for a specific product or technology together, and also add some more tools to moderate content/articles/links to project resources (and get some statistics): Empower and grow your tech community on Stack Overflow. It seems to me that it could be beneficial to create such a collective for Project Jupyter. I don’t think there is any need to point to it from existing community sites (it should not be a replacement for discourse), but rather to help organising the content and interactions which are already happening on Stack Overflow and pertain Jupyter; for one, we would have an ability to link to this discourse site. Any thoughts?

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I watched the intro video and it sounds like it’s not just about aggregating tags, it’s also about actively producing content to help developers and engage with the community. I agree with you that it shouldn’t be a replacement for discourse, but it’s probably worth investigating the pros and cons. If there was a Jupyter Stack Overflow Collective are there already people who could administer or moderate it?

I am actively trying to answer in JupyterLab tag, sometimes venturing to other tags as well. I see a few other Jupyter community members actively answering: 'jupyter' Top Users - Stack Overflow and there are many more in sub-project specific tags. A big downside for as for now appears to be a non-transparent costing policy: they say open-source projects will have much lower price but do not disclose it (https://meta.stackoverflow.com/a/408735/6646912).

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I assumed it was free- since it’s a paid service (albeit discounted) I think a big factor is whether you and other frequent StackOverflow contributors think it adds sufficient value.

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When first posting here I was assuming that to, I think they did not communicate it clearly. It is likely too early to say how useful this is and very hard to know cost/benefit ratio without a ballpark cost.