How to output model pickle file to s3 in luigi?
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
add a comment |
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
add a comment |
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
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
python luigi
asked Nov 21 '18 at 14:51
MussMuss
276
276
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add a comment |
1 Answer
1
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oldest
votes
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'))
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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'))
add a comment |
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'))
add a comment |
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'))
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'))
answered Nov 26 '18 at 12:36
Hendrick Sabinorio AcostaHendrick Sabinorio Acosta
62
62
add a comment |
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