How to save sklearn model on s3 using joblib.dump?

Question

I have a sklearn model and I want to save the pickle file on my s3 bucket using joblib.dump

I used joblib.dump(model, \'model.pkl\') to save the model local

Answer 1

Here's a way that worked for me. Pretty straight forward and easy. I'm using joblib (it's better for storing large sklearn models) but you could use pickle too.
Also, I'm using temporary files for transferring to/from S3. But if you want, you could store the file in a more permanent location.

import tempfile
import boto3
import joblib

bucket_name = "my-bucket"
key = "model.pkl"

# WRITE
with tempfile.TemporaryFile() as fp:
    joblib.dump(model, fp)
    fp.seek(0)
    s3_resource.put_object(Body=fp.read(), Bucket=bucket_name, Key=key)

# READ
with tempfile.TemporaryFile() as fp:
    s3_resource.download_fileobj(Fileobj=fp, Bucket=bucket_name, Key=key)
    fp.seek(0)
    model = joblib.load(fp)

# DELETE
s3_resource.delete_object(Bucket=bucket_name, Key=key)

Answer 2

Use following code to dump your model to s3 location in .pkl or .sav format:

import tempfile
import boto3
s3 = boto3.resource('s3')

# you can dump it in .sav or .pkl format 
location = 's3://bucket_name/folder_name/'
model_filename = 'model.sav'  # use any extension you want (.pkl or .sav)
OutputFile = location + model_filename

# WRITE
with tempfile.TemporaryFile() as fp:
    joblib.dump(scikit_learn_model, fp)
    fp.seek(0)
    # use bucket_name and OutputFile - s3 location path in string format.
    s3.Bucket('bucket_name').put_object(Key= OutputFile, Body=fp.read())

Answer 3

Just correcting Sayali Sonawane's answer:

import tempfile
import boto3
s3 = boto3.resource('s3')

# you can dump it in .sav or .pkl format 
location = 'folder_name/' # THIS is the change to make the code work
model_filename = 'model.sav'  # use any extension you want (.pkl or .sav)
OutputFile = location + model_filename

# WRITE
with tempfile.TemporaryFile() as fp:
    joblib.dump(scikit_learn_model, fp)
    fp.seek(0)
    # use bucket_name and OutputFile - s3 location path in string format.
    s3.Bucket('bucket_name').put_object(Key= OutputFile, Body=fp.read())