I have a function sayHello
defined in one python file utils.py
. I also have a python file flow.py
that runs a simple metaflow. I want to import sayHello
from utils.py
and use sayHello
as a step function in flow.py
. Is it possible to do that? This might be hard because we need self
in the class. If it's possible, one question is that we how do pass output from the previous step into the function and pass the output to next step. The following is my attempt.
#utils.py
def sayHello():
print("hello world")
#flow.py
from metaflow import FlowSpec, step, Parameter
from metaflow import Metaflow
from utils import sayHello
def function(p):
return p
class BranchFlow(FlowSpec):
@step
def start(self):
self.next(self.a, self.b, self.sayHello)
@step
def a(self):
self.x = 1
self.next(self.join)
@step
def b(self):
self.x = 2
self.next(self.join)
@step
def join(self, inputs):
print('a is %s' % inputs.a.x)
print('b is %s' % inputs.b.x)
print('total is %d' % sum(input.x for input in inputs))
self.next(self.end)
@step
def end(self):
pass
if __name__ == '__main__':
BranchFlow()
According to the documentation for FlowSpec.next()
, the method you pass to self.next()
must be a member of the current class and decorated with @step
.
Probably the easiest way to use another function from another module is to wrap it in a member function:
class BranchFlow(FlowSpec):
# ...
@step
def sayHello(self):
sayHello()
Now you can do self.next(self.sayHello)
.