You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
Description
I want to get the latest x runs of a job, for this is specify the limit parameter ("The number of runs to return"). Problem is that in the list_runs method a while loop sends not only one single request with the correct limit but instead it sends requests as long as there are runs found for a job. As a result instead of the expected x runs i get a ton of runs (probably all) which drastically slows down the code.
Reproduction
client.jobs.list_runs(job_id=<job_id>, limit=2). # for a job with more than 2 runs
--> returns not 2 runs but many more (probably all)
Expected behavior
list_runs returns as many runs as specified via limit parameter
Is it a regression?
This i don't know, tried 0.32.0 and 0.32.1
Other Information
OS: macOS
Version: 0.32.0 and 0.32.1
Additional context
The text was updated successfully, but these errors were encountered:
Description
I want to get the latest x runs of a job, for this is specify the
limit
parameter ("The number of runs to return"). Problem is that in thelist_runs
method a while loop sends not only one single request with the correct limit but instead it sends requests as long as there are runs found for a job. As a result instead of the expected x runs i get a ton of runs (probably all) which drastically slows down the code.Reproduction
client.jobs.list_runs(job_id=<job_id>, limit=2). # for a job with more than 2 runs
--> returns not 2 runs but many more (probably all)
Expected behavior
list_runs returns as many runs as specified via limit parameter
Is it a regression?
This i don't know, tried 0.32.0 and 0.32.1
Other Information
Additional context
The text was updated successfully, but these errors were encountered: