How to output model pickle file to s3 in luigi?












0















I have a task which trains the model eg:



class ModelTrain(luigi.Task):
def output(self):
client = S3Client(os.getenv("CONFIG_AWS_ACCESS_KEY"),
os.getenv("CONFIG_AWS_SECRET_KEY"))
model_output = os.path.join(
"s3://", _BUCKET, exp.version + '_model.joblib')
return S3Target(model_output, client)

def run(self):
joblib.dump(model, '/tmp/model.joblib')
with open(self.output().path, 'wb') as out_file:
out_file.write(joblib.load('/tmp/model.joblib'))


FileNotFoundError: [Errno 2] No such file or directory: 's3://bucket/version_model.joblib'



Any pointers in this regard would be helpful










share|improve this question



























    0















    I have a task which trains the model eg:



    class ModelTrain(luigi.Task):
    def output(self):
    client = S3Client(os.getenv("CONFIG_AWS_ACCESS_KEY"),
    os.getenv("CONFIG_AWS_SECRET_KEY"))
    model_output = os.path.join(
    "s3://", _BUCKET, exp.version + '_model.joblib')
    return S3Target(model_output, client)

    def run(self):
    joblib.dump(model, '/tmp/model.joblib')
    with open(self.output().path, 'wb') as out_file:
    out_file.write(joblib.load('/tmp/model.joblib'))


    FileNotFoundError: [Errno 2] No such file or directory: 's3://bucket/version_model.joblib'



    Any pointers in this regard would be helpful










    share|improve this question

























      0












      0








      0








      I have a task which trains the model eg:



      class ModelTrain(luigi.Task):
      def output(self):
      client = S3Client(os.getenv("CONFIG_AWS_ACCESS_KEY"),
      os.getenv("CONFIG_AWS_SECRET_KEY"))
      model_output = os.path.join(
      "s3://", _BUCKET, exp.version + '_model.joblib')
      return S3Target(model_output, client)

      def run(self):
      joblib.dump(model, '/tmp/model.joblib')
      with open(self.output().path, 'wb') as out_file:
      out_file.write(joblib.load('/tmp/model.joblib'))


      FileNotFoundError: [Errno 2] No such file or directory: 's3://bucket/version_model.joblib'



      Any pointers in this regard would be helpful










      share|improve this question














      I have a task which trains the model eg:



      class ModelTrain(luigi.Task):
      def output(self):
      client = S3Client(os.getenv("CONFIG_AWS_ACCESS_KEY"),
      os.getenv("CONFIG_AWS_SECRET_KEY"))
      model_output = os.path.join(
      "s3://", _BUCKET, exp.version + '_model.joblib')
      return S3Target(model_output, client)

      def run(self):
      joblib.dump(model, '/tmp/model.joblib')
      with open(self.output().path, 'wb') as out_file:
      out_file.write(joblib.load('/tmp/model.joblib'))


      FileNotFoundError: [Errno 2] No such file or directory: 's3://bucket/version_model.joblib'



      Any pointers in this regard would be helpful







      python luigi






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 21 '18 at 14:51









      MussMuss

      276




      276
























          1 Answer
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          Could you try to remove .path in your open statement.



            def run(self):
          joblib.dump(model, '/tmp/model.joblib')
          with open(self.output(), 'wb') as out_file:
          out_file.write(joblib.load('/tmp/model.joblib'))





          share|improve this answer























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            1 Answer
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            active

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            active

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            active

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            0














            Could you try to remove .path in your open statement.



              def run(self):
            joblib.dump(model, '/tmp/model.joblib')
            with open(self.output(), 'wb') as out_file:
            out_file.write(joblib.load('/tmp/model.joblib'))





            share|improve this answer




























              0














              Could you try to remove .path in your open statement.



                def run(self):
              joblib.dump(model, '/tmp/model.joblib')
              with open(self.output(), 'wb') as out_file:
              out_file.write(joblib.load('/tmp/model.joblib'))





              share|improve this answer


























                0












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                0







                Could you try to remove .path in your open statement.



                  def run(self):
                joblib.dump(model, '/tmp/model.joblib')
                with open(self.output(), 'wb') as out_file:
                out_file.write(joblib.load('/tmp/model.joblib'))





                share|improve this answer













                Could you try to remove .path in your open statement.



                  def run(self):
                joblib.dump(model, '/tmp/model.joblib')
                with open(self.output(), 'wb') as out_file:
                out_file.write(joblib.load('/tmp/model.joblib'))






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 26 '18 at 12:36









                Hendrick Sabinorio AcostaHendrick Sabinorio Acosta

                62




                62
































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