spark-submit和pyspark有什么区别？

Question

如果我启动pyspark然后运行此命令:

import my_script; spark = my_script.Sparker(sc); spark.collapse('./data/')

一切都很好.但是,如果我尝试通过命令行和spark-submit执行相同的操作,则会收到错误消息:

Command: /usr/local/spark/bin/spark-submit my_script.py collapse ./data/
  File "/usr/local/spark/python/pyspark/rdd.py", line 352, in func
    return f(iterator)
  File "/usr/local/spark/python/pyspark/rdd.py", line 1576, in combineLocally
    merger.mergeValues(iterator)
  File "/usr/local/spark/python/pyspark/shuffle.py", line 245, in mergeValues
    for k, v in iterator:
  File "/.../my_script.py", line 173, in _json_args_to_arr
    js = cls._json(line)
RuntimeError: uninitialized staticmethod object

my_script:

...
if __name__ == "__main__":
    args = sys.argv[1:]
    if args[0] == 'collapse':
        directory = args[1]
        from pyspark import SparkContext
        sc = SparkContext(appName="Collapse")
        spark = Sparker(sc)
        spark.collapse(directory)
        sc.stop()

为什么会这样？运行pyspark和运行spark-submit会导致这种分歧有什么区别？我怎样才能在spark-submit中完成这项工作？

编辑:我尝试通过这样做从bash shell运行它pyspark my_script.py collapse ./data/,我得到了同样的错误.一切正常的唯一时间是我在python shell中并导入脚本.

Answer 1

1. 如果您构建了一个spark应用程序,则需要使用spark-submit来运行该应用程序

   • 代码可以用python/scala编写

   • 模式可以是本地/集群

2. 如果您只想测试/运行几个单独的命令,则可以使用spark提供的shell

   • pyspark(用于python中的火花)
   • 火花壳(用于scala中的火花)