我有一个 DataFrame(已转换为 RDD)并想重新分区,以便每个键(第一列)都有自己的分区。这就是我所做的:
# Repartition to # key partitions and map each row to a partition given their key rank
my_rdd = df.rdd.partitionBy(len(keys), lambda row: int(row[0]))
但是,当我尝试将其映射回 DataFrame 或保存它时,出现此错误:
Caused by: org.apache.spark.api.python.PythonException: Traceback (most recent call last):
File "spark-1.5.1-bin-hadoop2.6/python/lib/pyspark.zip/pyspark/worker.py", line 111, in main
process()
File "spark-1.5.1-bin-hadoop2.6/python/lib/pyspark.zip/pyspark/worker.py", line 106, in process
serializer.dump_stream(func(split_index, iterator), outfile)
File "spark-1.5.1-bin-hadoop2.6/python/lib/pyspark.zip/pyspark/serializers.py", line 133, in dump_stream
for obj in iterator:
File "spark-1.5.1-bin-hadoop2.6/python/pyspark/rdd.py", line 1703, in add_shuffle_key
for k, v in iterator:
ValueError: too many values to unpack
at org.apache.spark.api.python.PythonRunner$$anon$1.read(PythonRDD.scala:166)
at org.apache.spark.api.python.PythonRunner$$anon$1.(PythonRDD.scala:207)
at org.apache.spark.api.python.PythonRunner.compute(PythonRDD.scala:125)
at org.apache.spark.api.python.PythonRDD.compute(PythonRDD.scala:70)
at org.apache.spark.rdd.RDD.computeOrReadCheckpoint(RDD.scala:297)
at org.apache.spark.rdd.RDD.iterator(RDD.scala:264)
at org.apache.spark.api.python.PairwiseRDD.compute(PythonRDD.scala:342)
at org.apache.spark.rdd.RDD.computeOrReadCheckpoint(RDD.scala:297)
at org.apache.spark.rdd.RDD.iterator(RDD.scala:264)
at org.apache.spark.scheduler.ShuffleMapTask.runTask(ShuffleMapTask.scala:73)
at org.apache.spark.scheduler.ShuffleMapTask.runTask(ShuffleMapTask.scala:41)
at org.apache.spark.scheduler.Task.run(Task.scala:88)
at org.apache.spark.executor.Executor$TaskRunner.run(Executor.scala:214)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
... 1 more
更多的测试表明即使这样也会导致同样的错误: my_rdd = df.rdd.partitionBy(x) # x = 可以是 5、100 等
你们有没有遇到过这种情况。如果有,您是如何解决的?
partitionBy
需要一个 PairwiseRDD
,它在 Python 中相当于 RDD
的长度为 2 的元组(列表),其中第一个元素是一个键第二个是一个值。
partitionFunc
获取 key 并将其映射到分区号。当您在 RDD[Row]
上使用它时,它会尝试将行解压缩为键和值,但失败了:
from pyspark.sql import Row
row = Row(1, 2, 3)
k, v = row
## Traceback (most recent call last):
## ...
## ValueError: too many values to unpack (expected 2)
即使您提供了正确的数据来做这样的事情:
my_rdd = (df.rdd.map(lambda row: (int(row[0]), row)).partitionBy(len(keys))
这真的没有意义。在 DataFrames
的情况下,分区不是特别有意义。参见 my answerĐẾN How to define partitioning of DataFrame?更多细节。
Tôi là một lập trình viên xuất sắc, rất giỏi!