使用管道从 S3 加载 Pyspark.ml 模型答案

【问题标题】：Load Pyspark.ml model from S3 using Pipeline使用管道从 S3 加载 Pyspark.ml 模型
【发布时间】：2021-05-11 23:29:50
【问题描述】：

我正在尝试将经过训练的模型保存到 S3 存储中，然后尝试通过 pyspark.ml 中的 Pipeline 包使用此模型进行加载和预测。这是我如何保存模型的示例。

#stage_1 to stage_4 are some basic trasnformation on data one-hot encoding e.t.c
# define stage 5: logistic regression model                          
 stage_5 = LogisticRegression(featuresCol='features',labelCol='label')

 # SETUP THE PIPELINE
 regression_pipeline = Pipeline(stages= [stage_1, stage_2, stage_3, stage_4, stage_5])

 # fit the pipeline for the trainind data
 model = regression_pipeline.fit(dataFrame1)

 model_path ="s3://s3-dummy_path-orch/dummy models/pipeline_testing_1.model"
 model.save(model_path)

我能够成功保存模型并在上述模型路径中创建了两个文件夹

阶段
元数据。

但是，当我尝试加载模型时，出现以下错误。

Traceback (most recent call last):
  File "/tmp/pythonScript_85ff2462_e087_4805_9f50_0c75fc4302e2958379757178872310.py", line 75, in <module>
    pipelineModel = Pipeline.load(model_path)
  File "/usr/lib/spark/python/lib/pyspark.zip/pyspark/ml/util.py", line 362, in load
  File "/usr/lib/spark/python/lib/pyspark.zip/pyspark/ml/pipeline.py", line 207, in load
  File "/usr/lib/spark/python/lib/pyspark.zip/pyspark/ml/util.py", line 300, in load
  File "/usr/lib/spark/python/lib/py4j-0.10.7-src.zip/py4j/java_gateway.py", line 1257, in __call__
  File "/usr/lib/spark/python/lib/pyspark.zip/pyspark/sql/utils.py", line 79, in deco
pyspark.sql.utils.IllegalArgumentException: 'requirement failed: Error loading metadata: Expected class name org.apache.spark.ml.Pipeline but found class name org.apache.spark.ml.PipelineModel'

我正在尝试如下加载模型：

from pyspark.ml import Pipeline

## same path used while #model.save in the above code snippet
model_path ="s3://s3-dummy_path-orch/dummy models/pipeline_testing_1.model" 

pipelineModel = Pipeline.load(model_path)

我该如何解决这个问题？

【问题讨论】：

标签： apache-spark pyspark pipeline apache-spark-ml

【解决方案1】：

如果您保存了管道模型，则应将其作为管道模型加载，而不是作为管道加载。不同之处在于管道模型适合数据帧，但管道不是。

from pyspark.ml import PipelineModel

pipelineModel = PipelineModel.load(model_path)

【讨论】：