java - Spark - 使用 OpenCSV 解析文件的序列化问题

Question

我正在使用 Spark 处理 csv 文件。最近我用 opencsv 替换了手动 CSV 行解析。这是简化的代码

public class Main {

    public static void main(String[] args) {

        CSVParser parser = new CSVParserBuilder()
                .withSeparator(';')
                .build();

        SparkConf cfg = new SparkConf()
                .setMaster("local[4]")
                .setAppName("Testapp");
        JavaSparkContext sc = new JavaSparkContext(cfg);

        JavaRDD<String> textFile = sc.textFile("testdata.csv", 1);

        List<String> categories = textFile
                .map(line -> parser.parseLine(line)[10])
                .collect();
        System.out.println(categories);
    }
}

不幸的是，代码不起作用。它产生一个异常

Caused by: java.io.NotSerializableException: com.opencsv.CSVParser
Serialization stack:
    - object not serializable (class: com.opencsv.CSVParser, value: com.opencsv.CSVParser@1290c49)
    - element of array (index: 0)
    - array (class [Ljava.lang.Object;, size 1)
    - field (class: java.lang.invoke.SerializedLambda, name: capturedArgs, type: class [Ljava.lang.Object;)
    - object (class java.lang.invoke.SerializedLambda, SerializedLambda[capturingClass=class test.Main, functionalInterfaceMethod=org/apache/spark/api/java/function/Function.call:(Ljava/lang/Object;)Ljava/lang/Object;, implementation=invokeStatic test/Main.lambda$main$49bd2722$1:(Lcom/opencsv/CSVParser;Ljava/lang/String;)Ljava/lang/String;, instantiatedMethodType=(Ljava/lang/String;)Ljava/lang/String;, numCaptured=1])
    - writeReplace data (class: java.lang.invoke.SerializedLambda)
    - object (class test.Main$$Lambda$19/429639728, test.Main$$Lambda$19/429639728@72456279)
    - field (class: org.apache.spark.api.java.JavaPairRDD$$anonfun$toScalaFunction$1, name: fun$1, type: interface org.apache.spark.api.java.function.Function)
    - object (class org.apache.spark.api.java.JavaPairRDD$$anonfun$toScalaFunction$1, <function1>)
    at org.apache.spark.serializer.SerializationDebugger$.improveException(SerializationDebugger.scala:40)
    at org.apache.spark.serializer.JavaSerializationStream.writeObject(JavaSerializer.scala:46)
    at org.apache.spark.serializer.JavaSerializerInstance.serialize(JavaSerializer.scala:100)
    at org.apache.spark.util.ClosureCleaner$.ensureSerializable(ClosureCleaner.scala:400)
    ... 12 more

似乎 Spark 试图序列化 lambda 表达式，而 lamba 表达式以某种方式保持对 parser 的引用，这导致了上述错误。

问题是:是否有任何方法可以避免该异常并在传递给 Spark 的 lambda 表达式中使用不可序列化的库？我真的不想实现自己的 csv 解析器。

Answer 1

Spark 开箱即用地支持 CSV 文件

import org.apache.spark.sql.Row;
import org.apache.spark.sql.Dataset;

Dataset<Row> df = spark.read().format("csv")
                      .option("sep", ";")
                      .option("header", "true") //or "false" if no headers
                      .load("filename.csv");

编辑(对主要答案的推荐评论)

如果你真的需要它，你可以使用 df.javaRDD() 从 DataFrame 中获取 RDD 尽管最好使用 DataSet/DataFrame API(例如参见 here)