Pyspark --py文件不起作用

Question

我使用此文件建议http://spark.apache.org/docs/1.1.1/submitting-applications.html

spsark版本1.1.0

./spark/bin/spark-submit --py-files /home/hadoop/loganalysis/parser-src.zip \
/home/hadoop/loganalysis/ship-test.py 

和代码中的conf:

conf = (SparkConf()
        .setMaster("yarn-client")
        .setAppName("LogAnalysis")
        .set("spark.executor.memory", "1g")
        .set("spark.executor.cores", "4")
        .set("spark.executor.num", "2")
        .set("spark.driver.memory", "4g")
        .set("spark.kryoserializer.buffer.mb", "128"))

和slave节点抱怨ImportError

14/12/25 05:09:53 WARN scheduler.TaskSetManager: Lost task 0.0 in stage 0.0 (TID 0, ip-172-31-10-8.cn-north-1.compute.internal): org.apache.spark.api.python.PythonException: Traceback (most recent call last):
  File "/home/hadoop/spark/python/pyspark/worker.py", line 75, in main
    command = pickleSer._read_with_length(infile)
  File "/home/hadoop/spark/python/pyspark/serializers.py", line 150, in _read_with_length
    return self.loads(obj)
ImportError: No module named parser

和parser-src.zip在本地测试.

[hadoop@ip-172-31-10-231 ~]$ python
Python 2.7.8 (default, Nov  3 2014, 10:17:30) 
[GCC 4.8.2 20140120 (Red Hat 4.8.2-16)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import sys
>>> sys.path.insert(1, '/home/hadoop/loganalysis/parser-src.zip')
>>> from parser import parser
>>> parser.parse
<function parse at 0x7fa5ef4c9848>
>>> 

我正在尝试获取有关远程工作人员的信息.看看它是否复制了文件.sys.path看起来是什么......这很棘手.

更新:我用这个发现zip文件已被删除.和sys.path已设置.仍导入获取错误.

data = list(range(4))
disdata = sc.parallelize(data)
result = disdata.map(lambda x: "sys.path:  {0}\nDIR: {1}   \n FILES: {2} \n parser: {3}".format(sys.path, os.getcwd(), os.listdir('.'), str(parser)))
result.collect()
print(result.take(4))

我似乎不得不深入挖掘cloudpickle.这意味着我需要了解cloudpickle是如何工作的并且首先失败.

: An error occurred while calling o40.collect.
: org.apache.spark.SparkException: Job aborted due to stage failure: Task 4 in stage 0.0 failed 4 times, most recent failure: Lost task 4.3 in stage 0.0 (TID 23, ip-172-31-10-8.cn-north-1.compute.internal): org.apache.spark.api.python.PythonException: Traceback (most recent call last):
  File "/home/hadoop/spark/python/pyspark/worker.py", line 75, in main
    command = pickleSer._read_with_length(infile)
  File "/home/hadoop/spark/python/pyspark/serializers.py", line 150, in _read_with_length
    return self.loads(obj)
  File "/home/hadoop/spark/python/pyspark/cloudpickle.py", line 811, in subimport
    __import__(name)
ImportError: ('No module named parser', <function subimport at 0x7f219ffad7d0>, ('parser.parser',))

更新:

有人在火花0.8遇到同样的问题 http://apache-spark-user-list.1001560.n3.nabble.com/pyspark-Importing-other-py-files-in-PYTHONPATH-td2301.html

但他把他的lib放在python dist-packages和导入作品中.我试过但仍然导致导入错误.

更新:

OH.gush ..我认为问题是由于不了解zip文件和python导入行为引起的.我将parser.py传递给--py文件,它有效,抱怨另一个依赖.和zip只有.py文件[不包括.pyc]似乎也有效.

但我不太明白为什么.

Answer 1

尝试从方法本身而不是在驱动程序脚本的顶部导入自定义模块,例如:

def parse_record(record):
    import parser
    p = parser.parse(record)
    return p

而不是

import parser
def parse_record(record):
    p = parser.parse(record)
    return p

Cloud Pickle似乎无法识别何时导入自定义模块,因此它似乎尝试挑选顶级模块以及运行该方法所需的其他数据.根据我的经验,这意味着顶层模块似乎存在,但它们缺少可用的成员,并且嵌套模块无法按预期使用.一旦from A import *从方法(import A.B)内部导入或从内部导入,模块就按预期工作.

Answer 2

试试这个功能吧 SparkContext

sc.addPyFile(path)

根据这里的pyspark文件

为将来在此SparkContext上执行的所有任务添加.py或.zip依赖项.传递的路径可以是本地文件,HDFS(或其他Hadoop支持的文件系统)中的文件,也可以是HTTP,HTTPS或FTP URI.

尝试将您的python模块文件上传到公共云存储(例如AWS S3)并将URL传递给该方法.

这是一个更全面的阅读材料:http://www.cloudera.com/documentation/enterprise/5-5-x/topics/spark_python.html

Answer 3

听起来一个或多个节点配置不正确。集群上的所有节点是否都具有相同的 Python 版本/配置（即它们都安装了解析器模块）？

如果您不想一一检查，您可以编写一个脚本来检查它是否已安装/为您安装。该线程展示了几种实现此目的的方法。