Overview of all the bundles
Hi All,
Do you know how to get an overview of all the bundles available?
What if i have 1000 bundles and I need an overview of all the bundles ?
Do you think it is possible to extract all the bundles info and have it in a table inside dataiku to have an overview?
Thank you.
Best Regards,
Orwin
Answers
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Hi,
You can programmatically list the exported bundles of a given project using the following code:
import dataiku client = dataiku.api_client() project = client.get_project("TEST") bundles = project.list_exported_bundles()
The output of this code (in the bundles variable) will be a list of dictionaries, each of those containing the detailed description of a distinct bundle. From there, you will be able to apply any filtering operation you need, or even write it into a Dataset.
Best,
Harizo
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Thank you HarizoR.
Do you have any idea why Dataiku did not include the overview of bundles in their rollouts ?
I find it very strange that the open source tools like DVC powered by GIT also has this overview of their versions but a closed source like Dataiku does not. Please share if you have any idea why Dataiku does not have this overview in their default features.
Thanks!
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Hi kim13,
As stated in my previous message, the overview of bundles is de facto available in Dataiku DSS via the API. Instead of embedding the access to this information in a UI, the ability to leverage the API allows you much more flexibility on how you want to process that information:
- you can combine it with the Datasets API in order to write that information in a Dataset directly accessible from the Flow using any connection you have available,
- you can create a custom UI thanks to webapps if you are looking for something more interactive,
- the fact that the API is RESTful even allows you to send the information to any 3rd-party tool that can send HTTP requests.
On a final note, we are actively maintaining the Python client that sits on top of the API, the client itself is open-source and available in this GitHub repository.
Best,
Harizo.