test_luigi.py

""" TEST Luigi processes """

import os

def python_function(filename1):
    # from filename1 create filename2
    if "." not in filename1:
      print "can't split"
      return None
    filename2 = filename1.split(".")[0]+"_2"+filename1.split(".")[1]
    print filename1, filename2
    r = os.system("cp "+filename1+" "+filename2)
    return filename2 

r_func = """
r_function <- function(filename2)
{ 
     #from filename2 create filename3
}
"""

import rpy2
from rpy2 import robjects
import luigi

class Task1(luigi.Task):
  filename = luigi.Parameter()
  def run():
    python_function(self.filename)
  def output():
    #how to specify filename2????
    
    
class Task2(luigi.Task)
  filename1 = luigi.Parameter()
  filename2 = None
  def requires():
    self.filename2 = Task1(self.filename1)
  def run():
    robjects.r[] #process R code to createe filename 3
  def output:
      #How to specify filename3???
  
if __name__ == '__main__':
    luigi.run()

Hey, so the output function should just return 'luigi.LocalTarget(filename)', where the filename is whatever you want the output to be called. Luigi will check if the file exists and only do he next task if it does. Not sure that answers your question?

yeah, that helps :)