Update/delete score using python sdk

[msanand]

Hi team,

I noticed that when I use langfuse.score() to create a new score for a traceid, it appends a new score to the trace even if the score with the same name exists already. I'm looking to implement a behavior where only one score is maintained for a trace with a given score name, and that score gets updated on change instead of a new score getting added to the trace. Using the python SDK I could not find an option to update an existing score for a given trace with a given score name. I also couldn't delete all existing scores before creating a new one. Can you please guide me on how I can implement this with the python SDK? Thanks!

[simone-costamagna]

You can use the DELETE API found here: https://api.reference.langfuse.com/#delete-/api/public/scores/-scoreId-

When you create a new score, the function will return a scoreID. Make sure to save this ID. If you need to update the score later, you'll first need to delete the previous score using the DELETE API, and then create a new one.

[mjrusso]

I noticed in the docs that you can provide an id argument to score, which has the following comment:

    id="unique_id" # optional, can be used as an indempotency key to update the score subsequently

I'm assuming that if you use the same ID, it has the same effect without needing to manually delete?

[mjrusso]

I noticed in the docs that you can provide an id argument to score, which has the following comment:

    id="unique_id" # optional, can be used as an indempotency key to update the score subsequently

I'm assuming that if you use the same ID, it has the same effect without needing to manually delete?

Just tested this; looks like it works!

[FrankyCTY]

Yes, same for the API endpoint https://api.reference.langfuse.com/#post-/api/public/scores:~:text=the%20current%20limit-,Create%20a%20score,-POST%20/api/public

Internally, LangFuse is doing upsert operation based on if the score id is provided or not. If provided, it will try to perform update, otherwise it will create a new score.