我有以下文件夹结构
- libfolder
- lib1.py
- lib2.py
- main.py
main.py调用libfolder.lib1.py,然后调用libfolder.lib2.py和其他人.
这一切在本地机器上工作得很好,但在我将其部署到Dataproc后,我得到以下错误
File "/usr/lib/spark/python/lib/pyspark.zip/pyspark/serializers.py", line 455, in loads
return pickle.loads(obj, encoding=encoding)
ModuleNotFoundError: No module named 'libfolder'
我已将文件夹压缩到xyz.zip并运行以下命令:
spark-submit --py-files=xyz.zip main.py
序列化程序无法找到libfolder的位置.我打包文件夹的方式有问题吗?
这个问题类似于this one,但没有回答.
编辑:回应伊戈尔的问题
unzip -l为zip文件返回以下内容
- libfolder
- lib1.py
- lib2.py
- main.py
在main.py中使用此import语句调用lib1.py
from libfolder import lib1
最佳答案 这对我有用:
$cat main.py
from pyspark import SparkContext, SparkConf
from subpkg import sub
conf = SparkConf().setAppName("Shell Count")
sc = SparkContext(conf = conf)
text_file = sc.textFile("file:///etc/passwd")
counts = text_file.map(lambda line: sub.map(line)) \
.map(lambda shell: (shell, 1)) \
.reduceByKey(lambda a, b: sub.reduce(a, b))
counts.saveAsTextFile("hdfs:///count5.txt")
$cat subpkg/sub.py
def map(line):
return line.split(":")[6]
def reduce(a, b):
return a + b
$unzip -l /tmp/deps.zip
Archive: /tmp/deps.zip
Length Date Time Name
--------- ---------- ----- ----
0 2019-01-07 14:22 subpkg/
0 2019-01-07 13:51 subpkg/__init__.py
79 2019-01-07 14:13 subpkg/sub.py
--------- -------
79 3 files
$gcloud dataproc jobs submit pyspark --cluster test-cluster main.py --py-files deps.zip
Job [1f0f15108a4149c5942f49513ce04440] submitted.
Waiting for job output...
Hello world!
Job [1f0f15108a4149c5942f49513ce04440] finished successfully.