如何为--py-files依赖关系创建zip文件

Question

I Have a small project where I need to run Spark-submit with python code. here is the code link .

[XXXXX pysparktest]$ ls
README.md  spark_sample.py  utils

If I run this code with below spark-submit it is successful.

$ spark-submit spark_sample.py

But, I want to make utils package as zip file and pass zipfile to spark submit with help of --py-files

[XXXXXX pysparktest]$ ls
README.md  spark_sample.py  utils.zip

[XXXXXX pysparktest]$ spark-submit --py-files utils.zip spark_sample.py

When I tried ziping utils package directly and run through spark-submit I'm getting error

Traceback (most recent call last):
  File "/home/mapr/pysparktest/spark_sample.py", line 1, in <module>
    from utils import SparkOperations
ImportError: No module named utils

Can any one help me how to make the zip file and pass the zip file with --py-files parameter.