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PySpark到PMML-“字段标签不存在”错误

  •  1
  • michalrudko  · 技术社区  · 7 年前

    我是PySpark的新手,所以这可能是一个基本问题。我正在尝试导出 代码至 使用 JPMML SparkML 图书馆 JPMML-SparkML

    from pyspark.ml import Pipeline
    from pyspark.ml.classification import DecisionTreeClassifier
    from pyspark.ml.feature import RFormula
    
    df = spark.read.csv("Iris.csv", header = True, inferSchema = True)
    formula = RFormula(formula = "Species ~ .")
    classifier = DecisionTreeClassifier()
    pipeline = Pipeline(stages = [formula, classifier])
    pipelineModel = pipeline.fit(df)
    

    我出错了 Field "label" does not exist 。运行 斯卡拉

    完整的错误消息:

    Traceback (most recent call last): File "/usr/lib/spark/spark-2.1.1-bin-hadoop2.7/python/pyspark/sql/utils.py", line 63, in deco return f(*a, **kw) File "/usr/lib/spark/spark-2.1.1-bin-hadoop2.7/python/lib/py4j-0.10.4-src.zip/py4j/protocol.py", line 319, in get_return_value format(target_id, ".", name), value) py4j.protocol.Py4JJavaError: An error occurred while calling o48.fit. :
     java.lang.IllegalArgumentException: Field "label" does not exist. at
     org.apache.spark.sql.types.StructType$$anonfun$apply$1.apply(StructType.scala:264) at
     org.apache.spark.sql.types.StructType$$anonfun$apply$1.apply(StructType.scala:264) at
     scala.collection.MapLike$class.getOrElse(MapLike.scala:128) at scala.collection.AbstractMap.getOrElse(Map.scala:59) at
     org.apache.spark.sql.types.StructType.apply(StructType.scala:263) at 
     org.apache.spark.ml.util.SchemaUtils$.checkNumericType(SchemaUtils.scala:71) at 
     org.apache.spark.ml.PredictorParams$class.validateAndTransformSchema(Predictor.scala:53) at
     org.apache.spark.ml.classification.Classifier.org$apache$spark$ml$classification$ClassifierParams$$super$validateAndTransformSchema(Cla
     ssifier.scala:58) at org.apache.spark.ml.classification.ClassifierParams$class.validateAndTransformSchema(Classifier.scala:42) at org.apache.spark.ml.classification.ProbabilisticClassifier.org$apache$spark$ml$classification$ProbabilisticClassifierParams$$super$vali
     dateAndTransformSchema(ProbabilisticClassifier.scala:53) at org.apache.spark.ml.classification.ProbabilisticClassifierParams$class.validateAndTransformSchema(ProbabilisticClassifier.scala:37) at
     org.apache.spark.ml.classification.ProbabilisticClassifier.validateAndTransformSchema(ProbabilisticClassifier.scala:53) at
     org.apache.spark.ml.Predictor.transformSchema(Predictor.scala:122) at org.apache.spark.ml.PipelineStage.transformSchema(Pipeline.scala:74) at org.apache.spark.ml.Predictor.fit(Predictor.scala:90) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at
     sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) at
     java.lang.reflect.Method.invoke(Method.java:497) at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:244) at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:357) at py4j.Gateway.invoke(Gateway.java:280) at
     py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:132) at py4j.commands.CallCommand.execute(CallCommand.java:79) at py4j.GatewayConnection.run(GatewayConnection.java:214) at java.lang.Thread.run(Thread.java:745)
    

    谢谢,迈克尔

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  •  5
  •   Philip    3 年前

    您需要提供要预测的列作为标签。您可以将dataframe中的列别名为“label”并使用分类器,也可以在分类器的构造函数中将该列作为labelCol参数提供。

    classifier = DecisionTreeClassifier(labelCol='some prediction field')