python - 如何在 PySpark 中获得不同的字典 RDD？

Question

我有一个字典的 RDD，我想获得一个仅包含不同元素的 RDD。但是，当我尝试调用

rdd.distinct()

PySpark 给我以下错误

TypeError: unhashable type: 'dict'

    at org.apache.spark.api.python.PythonRunner$$anon$1.read(PythonRDD.scala:166)
    at org.apache.spark.api.python.PythonRunner$$anon$1.<init>(PythonRDD.scala:207)
    at org.apache.spark.api.python.PythonRunner.compute(PythonRDD.scala:125)
    at org.apache.spark.api.python.PythonRDD.compute(PythonRDD.scala:70)
    at org.apache.spark.rdd.RDD.computeOrReadCheckpoint(RDD.scala:306)
    at org.apache.spark.rdd.RDD.iterator(RDD.scala:270)
    at org.apache.spark.api.python.PairwiseRDD.compute(PythonRDD.scala:342)
    at org.apache.spark.rdd.RDD.computeOrReadCheckpoint(RDD.scala:306)
    at org.apache.spark.rdd.RDD.iterator(RDD.scala:270)
    at org.apache.spark.scheduler.ShuffleMapTask.runTask(ShuffleMapTask.scala:73)
    at org.apache.spark.scheduler.ShuffleMapTask.runTask(ShuffleMapTask.scala:41)
    at org.apache.spark.scheduler.Task.run(Task.scala:89)
    at org.apache.spark.executor.Executor$TaskRunner.run(Executor.scala:213)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
    at java.lang.Thread.run(Thread.java:745)
16/02/19 16:55:56 WARN TaskSetManager: Lost task 0.0 in stage 0.0 (TID 0, localhost): org.apache.spark.api.python.PythonException: Traceback (most recent call last):
  File "/usr/local/Cellar/apache-spark/1.6.0/libexec/python/lib/pyspark.zip/pyspark/worker.py", line 111, in main
    process()
  File "/usr/local/Cellar/apache-spark/1.6.0/libexec/python/lib/pyspark.zip/pyspark/worker.py", line 106, in process
    serializer.dump_stream(func(split_index, iterator), outfile)
  File "/usr/local/Cellar/apache-spark/1.6.0/libexec/python/lib/pyspark.zip/pyspark/rdd.py", line 2346, in pipeline_func
  File "/usr/local/Cellar/apache-spark/1.6.0/libexec/python/lib/pyspark.zip/pyspark/rdd.py", line 2346, in pipeline_func
  File "/usr/local/Cellar/apache-spark/1.6.0/libexec/python/lib/pyspark.zip/pyspark/rdd.py", line 317, in func
  File "/usr/local/Cellar/apache-spark/1.6.0/libexec/python/lib/pyspark.zip/pyspark/rdd.py", line 1776, in combineLocally
  File "/usr/local/Cellar/apache-spark/1.6.0/libexec/python/lib/pyspark.zip/pyspark/shuffle.py", line 238, in mergeValues
    d[k] = comb(d[k], v) if k in d else creator(v)
TypeError: unhashable type: 'dict'

我确实在字典中有一个键，可以将其用作不同的元素，但文档没有提供有关如何解决此问题的任何线索。

编辑:内容由字符串、字符串数组和数字字典组成

编辑2:字典示例...我希望具有相同“data_fingerprint”键的字典被视为相等:

{"id":"4eece341","data_fingerprint":"1707db7bddf011ad884d132bf80baf3c"}

谢谢

Answer 1

正如 @zero323 在他的评论中指出的那样，您必须决定如何比较字典，因为它们不可散列。一种方法是对键进行排序(因为它们不按任何特定顺序)，例如按字典顺序。然后创建一个以下形式的字符串:

def dict_to_string(dict):
    ...
    return 'key1|value1|key2|value2...|keyn|valuen'

如果您嵌套了不可散列的对象，则必须递归地执行此操作。

现在您可以将 RDD 转换为与字符串作为键(或其某种哈希值)配对

pairs = dictRDD.map(lambda d: (dict_to_string(d), d))

要得到你想要的东西，你只需要减少关键作为休闲

distinctDicts = pairs.reduceByKey(lambda val1, val2: val1).values()