apache-spark - 如何修复 pyspark 中的 "No FileSystem for scheme: gs"？

Question

我正在尝试将 json 文件从 google 存储桶读取到本地 Spark 机器上的 pyspark 数据帧中。这是代码:

import pandas as pd
import numpy as np

from pyspark import SparkContext, SparkConf
from pyspark.sql import SparkSession, SQLContext

conf = SparkConf().setAll([('spark.executor.memory', '16g'),
                        ('spark.executor.cores','4'),
                         ('spark.cores.max','4')]).setMaster('local[*]')


spark = (SparkSession.
              builder.
              config(conf=conf).
              getOrCreate())


sc = spark.sparkContext

import glob
import bz2
import json
import pickle


bucket_path = "gs://<SOME_PATH>/"
client = storage.Client(project='<SOME_PROJECT>')
bucket = client.get_bucket ('<SOME_PATH>')
blobs = bucket.list_blobs()

theframes = []

for blob in blobs:
    print(blob.name)        
    testspark = spark.read.json(bucket_path + blob.name).cache()
    theframes.append(testspark) 

它正在从存储桶中读取文件(我可以看到 blob.name 的打印结果)，但随后崩溃如下:

 Traceback (most recent call last):
 File "test_code.py", line 66, in <module>
   testspark = spark.read.json(bucket_path + blob.name).cache()
 File "/home/anaconda3/envs/py37base/lib/python3.6/site-packages/pyspark/sql/readwriter.py", line 274, in json
return self._df(self._jreader.json(self._spark._sc._jvm.PythonUtils.toSeq(path)))
 File "/home/anaconda3/envs/py37base/lib/python3.6/site-packages/py4j/java_gateway.py", line 1257, in __call__
answer, self.gateway_client, self.target_id, self.name)
 File "/home/anaconda3/envs/py37base/lib/python3.6/site-packages/pyspark/sql/utils.py", line 63, in deco
return f(*a, **kw)
 File "/home/anaconda3/envs/py37base/lib/python3.6/site-packages/py4j/protocol.py", line 328, in get_return_value
format(target_id, ".", name), value)
py4j.protocol.Py4JJavaError: An error occurred while calling o51.json.
: java.io.IOException: No FileSystem for scheme: gs

我已经在 stackoverflow 上看到过这种类型的错误，但是当我有 pyspark 时，大多数解决方案似乎都在 Scala 中，和/或涉及到 core-site.xml 的困惑，我已经做了没有效果。

我正在使用 spark 2.4.1 和 python 3.6.7。

帮助将不胜感激!

Answer 1

需要一些配置参数才能将“gs”识别为分布式文件系统。
将此设置用于 google 云存储连接器 gcs-connector-hadoop2-latest.jar

spark = SparkSession \
        .builder \
        .config("spark.jars", "/path/to/gcs-connector-hadoop2-latest.jar") \
        .getOrCreate()

可以从 pyspark 设置的其他配置

spark._jsc.hadoopConfiguration().set('fs.gs.impl', 'com.google.cloud.hadoop.fs.gcs.GoogleHadoopFileSystem')
# This is required if you are using service account and set true, 
spark._jsc.hadoopConfiguration().set('fs.gs.auth.service.account.enable', 'true')
spark._jsc.hadoopConfiguration().set('google.cloud.auth.service.account.json.keyfile', "/path/to/keyfile")
# Following are required if you are using oAuth
spark._jsc.hadoopConfiguration().set('fs.gs.auth.client.id', 'YOUR_OAUTH_CLIENT_ID')
spark._jsc.hadoopConfiguration().set('fs.gs.auth.client.secret', 'OAUTH_SECRET')

或者，您可以在 core-site.xml 或 spark-defaults.conf 中设置这些配置。
命令行上的 Hadoop 配置
您也可以使用 spark.hadoop - 前缀配置属性以在 pyspark 时进行设置(或 spark-submit 一般)，例如

--conf spark.hadoop.fs.gs.impl=com.google.cloud.hadoop.fs.gcs.GoogleHadoopFileSystem