Hugo Kitano
07/28/2021, 6:15 PMSam Cook
07/28/2021, 6:19 PMKevin Kho
flow.set_reference_tasks([c])
where c is a task inside your Flow.Hugo Kitano
07/28/2021, 8:40 PMoutput = reference_task(input)
I have to set the reference task as
flow.set_reference_tasks(tasks = [output])
which to me, is a bit clunky. Is there a way to specify the task and not the output as an argument to the set_reference_task
method?Kevin Kho
Hugo Kitano
07/28/2021, 8:46 PMKevin Kho
Kevin Kho
with Flow("xxx") as flow:
a = abc(1)
b = abc(2, upstream_tasks=[a])
bcd(2)
flow.set_reference_tasks([flow.get_tasks(name="bcd")[0]])
Hugo Kitano
07/28/2021, 9:10 PM