python - 方案 : s3 with pyspark 没有文件系统

Question

我正在尝试使用 Spark 从 S3 读取一个 txt 文件，但我收到了这个错误:

No FileSystem for scheme: s3

这是我的代码:

from pyspark import SparkContext, SparkConf
conf = SparkConf().setAppName("first")
sc = SparkContext(conf=conf)
data = sc.textFile("s3://"+AWS_ACCESS_KEY+":" + AWS_SECRET_KEY + "@/aaa/aaa/aaa.txt")

header = data.first()

这是完整的追溯:

An error occurred while calling o25.partitions.
: java.io.IOException: No FileSystem for scheme: s3
    at org.apache.hadoop.fs.FileSystem.getFileSystemClass(FileSystem.java:2660)
    at org.apache.hadoop.fs.FileSystem.createFileSystem(FileSystem.java:2667)
    at org.apache.hadoop.fs.FileSystem.access$200(FileSystem.java:94)
    at org.apache.hadoop.fs.FileSystem$Cache.getInternal(FileSystem.java:2703)
    at org.apache.hadoop.fs.FileSystem$Cache.get(FileSystem.java:2685)
    at org.apache.hadoop.fs.FileSystem.get(FileSystem.java:373)
    at org.apache.hadoop.fs.Path.getFileSystem(Path.java:295)
    at org.apache.hadoop.mapred.FileInputFormat.singleThreadedListStatus(FileInputFormat.java:258)
    at org.apache.hadoop.mapred.FileInputFormat.listStatus(FileInputFormat.java:229)
    at org.apache.hadoop.mapred.FileInputFormat.getSplits(FileInputFormat.java:315)
    at org.apache.spark.rdd.HadoopRDD.getPartitions(HadoopRDD.scala:194)
    at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:252)
    at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:250)
    at scala.Option.getOrElse(Option.scala:121)
    at org.apache.spark.rdd.RDD.partitions(RDD.scala:250)
    at org.apache.spark.rdd.MapPartitionsRDD.getPartitions(MapPartitionsRDD.scala:35)
    at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:252)
    at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:250)
    at scala.Option.getOrElse(Option.scala:121)
    at org.apache.spark.rdd.RDD.partitions(RDD.scala:250)
    at org.apache.spark.api.java.JavaRDDLike$class.partitions(JavaRDDLike.scala:61)
    at org.apache.spark.api.java.AbstractJavaRDDLike.partitions(JavaRDDLike.scala:45)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:498)
    at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:244)
    at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:357)
    at py4j.Gateway.invoke(Gateway.java:280)
    at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:132)
    at py4j.commands.CallCommand.execute(CallCommand.java:79)
    at py4j.GatewayConnection.run(GatewayConnection.java:214)
    at java.lang.Thread.run(Thread.java:748)

我该如何解决这个问题？

Answer 1

如果您使用的是本地计算机，则可以使用 boto3:

s3 = boto3.resource('s3')
# get a handle on the bucket that holds your file
bucket = s3.Bucket('yourBucket')
# get a handle on the object you want (i.e. your file)
obj = bucket.Object(key='yourFile.extension')
# get the object
response = obj.get()
# read the contents of the file and split it into a list of lines
lines = response[u'Body'].read().split('\n')

(不要忘记设置您的 AWS S3 凭证)。

如果您使用的是 AWS 虚拟机 (EC2)，另一个干净的解决方案是 granting S3 permissions to your EC2并使用此命令启动 pyspark:

pyspark --packages com.amazonaws:aws-java-sdk-pom:1.10.34,org.apache.hadoop:hadoop-aws:2.7.2

如果添加其他包，请确保格式为:'groupId:artifactId:version'，包之间用逗号分隔。

如果您使用的是 Jupyter Notebooks 中的 pyspark，这将有效:

import os
import pyspark
os.environ['PYSPARK_SUBMIT_ARGS'] = '--packages com.amazonaws:aws-java-sdk-pom:1.10.34,org.apache.hadoop:hadoop-aws:2.7.2 pyspark-shell'
from pyspark.sql import SQLContext
from pyspark import SparkContext
sc = SparkContext()
sqlContext = SQLContext(sc)
filePath = "s3a://yourBucket/yourFile.parquet"
df = sqlContext.read.parquet(filePath) # Parquet file read example