如何序列化pyspark管道对象?

我正在尝试序列化PySpark Pipeline对象,以便以后可以保存和检索它.尝试使用
Python pickle库以及PySpark的PickleSerializer,dumps()调用本身就失败了.

使用本机pickle库时提供代码片段.

pipeline = Pipeline(stages=[tokenizer, hashingTF, lr])
with open ('myfile', 'wb') as f:
   pickle.dump(pipeline,f,2)
with open ('myfile', 'rb') as f:
   pipeline1 = pickle.load(f)

运行时出现以下错误:

py4j.protocol.Py4JError: An error occurred while calling o32.__getnewargs__. Trace:
py4j.Py4JException: Method __getnewargs__([]) does not exist
    at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:335)
    at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:344)
    at py4j.Gateway.invoke(Gateway.java:252)
    at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:133)
    at py4j.commands.CallCommand.execute(CallCommand.java:79)
    at py4j.GatewayConnection.run(GatewayConnection.java:209)
    at java.lang.Thread.run(Thread.java:785)

是否可以序列化PySpark Pipeline对象?

最佳答案 从技术上讲,您可以轻松挑选Pipeline对象:

from pyspark.ml.pipeline import Pipeline
import pickle

pickle.dumps(Pipeline(stages=[]))
## b'\x80\x03cpyspark.ml.pipeline\nPipeline\nq ...

你不能腌制的是Spark变形金刚和估算器,它们只是JVM对象的瘦包装器.如果你真的需要这个,你可以将它包装在一个函数中,例如:

def make_pipeline():
    return Pipeline(stages=[Tokenizer(inputCol="text", outputCol="words")])

pickle.dumps(make_pipeline)
## b'\x80\x03c__ ...

但由于它只是一段代码并且不存储任何持久性数据,因此它看起来并不特别有用.

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