hadoop - 用pyspark将图像写为序列文件的值

Question

我正在使用pyspark编写序列文件，关键是图像文件名，值是由字节字符串表示的图像

from PIL import Image

def get_image(filename):
 s = StringIO()
 im=io.imread(filename)
 io.imsave(s, im)
 return [(filename, s)]

rdd  =  sc.parallelize(filenames)
rdd.flatMap(get_image).saveAsSequenceFile("/user/myname/output")

但pyspark引发异常，表明泡菜不支持该格式

Caused by: net.razorvine.pickle.InvalidOpcodeException: opcode not implemented: OBJ
    at net.razorvine.pickle.Unpickler.dispatch(Unpickler.java:224)
    at net.razorvine.pickle.Unpickler.load(Unpickler.java:85)
    at net.razorvine.pickle.Unpickler.loads(Unpickler.java:98)
    at org.apache.spark.api.python.SerDeUtil$$anonfun$pythonToJava$1$$anonfun$apply$1.apply(SerDeUtil.scala:151)
    at org.apache.spark.api.python.SerDeUtil$$anonfun$pythonToJava$1$$anonfun$apply$1.apply(SerDeUtil.scala:150)
    at scala.collection.Iterator$$anon$13.hasNext(Iterator.scala:371)
    at scala.collection.Iterator$$anon$10.hasNext(Iterator.scala:308)
    at scala.collection.Iterator$class.foreach(Iterator.scala:727)
    at scala.collection.AbstractIterator.foreach(Iterator.scala:1157)
    at scala.collection.generic.Growable$class.$plus$plus$eq(Growable.scala:48)
    at scala.collection.mutable.ArrayBuffer.$plus$plus$eq(ArrayBuffer.scala:103)
    at scala.collection.mutable.ArrayBuffer.$plus$plus$eq(ArrayBuffer.scala:47)
    at scala.collection.TraversableOnce$class.to(TraversableOnce.scala:273)
    at scala.collection.AbstractIterator.to(Iterator.scala:1157)
    at scala.collection.TraversableOnce$class.toBuffer(TraversableOnce.scala:265)
    at scala.collection.AbstractIterator.toBuffer(Iterator.scala:1157)
    at scala.collection.TraversableOnce$class.toArray(TraversableOnce.scala:252)
    at scala.collection.AbstractIterator.toArray(Iterator.scala:1157)
    at org.apache.spark.rdd.RDD$$anonfun$take$1$$anonfun$28.apply(RDD.scala:1298)
    at org.apache.spark.rdd.RDD$$anonfun$take$1$$anonfun$28.apply(RDD.scala:1298)
    at org.apache.spark.SparkContext$$anonfun$runJob$5.apply(SparkContext.scala:1850)
    at org.apache.spark.SparkContext$$anonfun$runJob$5.apply(SparkContext.scala:1850)
    at org.apache.spark.scheduler.ResultTask.runTask(ResultTask.scala:66)
    at org.apache.spark.scheduler.Task.run(Task.scala:88)
    at org.apache.spark.executor.Executor$TaskRunner.run(Executor.scala:214)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
    ... 1 more

Answer 1

当您尝试编码/解码python类/对象以进行序列化时，将使用用于腌制的OBJ操作码。就我而言，我无意将对象写入序列文件，因此对我而言，解决方法是修复该错误。

对于整个生态系统，问题在于spark uses Pyrolite 4.13直到version 4.17才引入OBJ编码/解码。至于该怎么办，我想您有几个选择:

通过请求请求或github问题说服 Spark 维护者使用更高版本的Pyrolite。

使用该版本的Pyrolite生成自己的Spark版本

不要将类/对象写入序列文件。