Reputation: 121
I want to update my code of pyspark. In the pyspark, it must put the base model in a pipeline, the office demo of pipeline use the LogistictRegression as an base model. However, it seems not be able to use XGboost model in the pipeline api. How can I use the pyspark like this
from xgboost import XGBClassifier
...
model = XGBClassifier()
model.fit(X_train, y_train)
pipeline = Pipeline(stages=[..., model, ...])
...
It is convenient to use the pipeline api, so can anybody give some advices? Thanks.
Upvotes: 12
Views: 33801
Reputation: 191
There is an XBoost Implementation for Spark 2.4 and over here:
https://xgboost.readthedocs.io
Note that this is an external library but it should work easily with spark.
Upvotes: 5
Reputation: 583
There is a maintained (used in production by several companies) distributed XGBoost library as mentioned above (https://github.com/dmlc/xgboost), however to use it from PySpark is a bit tricky, someone made a working pyspark wrapper for version 0.72 of the library, with 0.8 support in progress.
See here https://medium.com/@bogdan.cojocar/pyspark-and-xgboost-integration-tested-on-the-kaggle-titanic-dataset-4e75a568bdb, and https://github.com/dmlc/xgboost/issues/1698 for the full discussion.
Make sure the xgboost jars are in your pyspark jar path.
Upvotes: 5
Reputation: 2477
There is no XGBoost classifier in Apache Spark ML (as of version 2.3). Available models are listed here : https://spark.apache.org/docs/2.3.0/ml-classification-regression.html
If you want to use XGBoost you should do it without pyspark (convert your spark dataframe to a pandas dataframe with .toPandas()
) or use another algorithm (https://spark.apache.org/docs/2.3.0/api/python/pyspark.ml.html#module-pyspark.ml.classification).
But if you really want to use XGBoost with pyspark, you'll have to dive into pyspark to implement a distributed XGBoost yourself. Here is an article where they do so : http://dmlc.ml/2016/10/26/a-full-integration-of-xgboost-and-spark.html
Upvotes: 5