Question

我正在尝试在PySpark应用程序中广播用户定义的变量，但我总是遇到以下错误：

 File "/usr/local/spark-2.1.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/worker.py", line 174, in main
    process()
  File "/usr/local/spark-2.1.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/worker.py", line 169, in process
    serializer.dump_stream(func(split_index, iterator), outfile)
  File "/usr/local/spark-2.1.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/serializers.py", line 268, in dump_stream
    vs = list(itertools.islice(iterator, batch))
  File "/home/.../sparkbroad.py", line 29, in <lambda>
    output = input_.map(lambda item: b.value.map(item))
  File "/usr/local/spark-2.1.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/broadcast.py", line 106, in value
    self._value = self.load(self._path)
  File "/usr/local/spark-2.1.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/broadcast.py", line 97, in load
    return pickle.load(f)

AttributeError: 'module' object has no attribute 'FooMap'

模块sparkbrad.py中的代码如下：

import random
import pyspark as spark

class FooMap(object):

    def __init__(self):
        keys = list(range(10))
        values = [2 * key for key in keys]
        self._map = dict(zip(keys, values))

    def map(self, value):
        if value not in self._map:
            return -1
        return self._map[value]


class FooMapJob(object):

    def __init__(self, inputs):
        self._inputs = inputs
        self._foomap = FooMap()

    def run(self):
        sc = spark.SparkContext('local', 'FooMap')
        input_ = sc.parallelize(self._inputs, 4)
        b = sc.broadcast(self._foomap)
        output = input_.map(lambda item: b.value.map(item))
        b.unpersist()
        result = list(output.toLocalIterator())
        sc.stop()
        return result


def main():
    inputs = [random.randint(0, 10) for _ in range(10)]
    job = FooMapJob(inputs)
    print(job.run())

if __name__ == '__main__':
    main()

我正在通过：

运行它

:~$ spark-submit --master local[4] --py-files sparkbroad.py sparkbroad.py

我添加了--py-files参数，但看起来它没有那么大的变化。不幸的是，我找不到任何在线处理复杂类广播（只是列表或词典）的例子。任何提示都表示赞赏。提前谢谢。

更新：将FooMap类放在一个单独的模块中，即使没有--py-files指令，一切似乎也能正常工作。

Answer 1

将FooMap类放在单独的模块中，一切正常。

在Spark中广播用户定义的类

1 个答案: