Question

我正在使用本地窗口并尝试使用python上的以下代码加载XML文件，我遇到此错误，有人知道如何解决它，

这是代码

df1 = sqlContext.read.format("xml").options(rowTag="IRS990EZ").load("https://irs-form-990.s3.amazonaws.com/201611339349202661_public.xml")

这是错误

Py4JJavaError                             Traceback (most recent call last)
<ipython-input-7-4832eb48a4aa> in <module>()
----> 1 df1 = sqlContext.read.format("xml").options(rowTag="IRS990EZ").load("https://irs-form-990.s3.amazonaws.com/201611339349202661_public.xml")

C:\SPARK_HOME\spark-2.2.0-bin-hadoop2.7\python\pyspark\sql\readwriter.py in load(self, path, format, schema, **options)
    157         self.options(**options)
    158         if isinstance(path, basestring):
--> 159             return self._df(self._jreader.load(path))
    160         elif path is not None:
    161             if type(path) != list:

C:\SPARK_HOME\spark-2.2.0-bin-hadoop2.7\python\lib\py4j-0.10.4-src.zip\py4j\java_gateway.py in __call__(self, *args)
   1131         answer = self.gateway_client.send_command(command)
   1132         return_value = get_return_value(
-> 1133             answer, self.gateway_client, self.target_id, self.name)
   1134 
   1135         for temp_arg in temp_args:

C:\SPARK_HOME\spark-2.2.0-bin-hadoop2.7\python\pyspark\sql\utils.py in deco(*a, **kw)
     61     def deco(*a, **kw):
     62         try:
---> 63             return f(*a, **kw)
     64         except py4j.protocol.Py4JJavaError as e:
     65             s = e.java_exception.toString()

C:\SPARK_HOME\spark-2.2.0-bin-hadoop2.7\python\lib\py4j-0.10.4-src.zip\py4j\protocol.py in get_return_value(answer, gateway_client, target_id, name)
    317                 raise Py4JJavaError(
    318                     "An error occurred while calling {0}{1}{2}.\n".
--> 319                     format(target_id, ".", name), value)
    320             else:
    321                 raise Py4JError(

Py4JJavaError: An error occurred while calling o38.load.
: java.io.IOException: No FileSystem for scheme: https
    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.mapreduce.lib.input.FileInputFormat.setInputPaths(FileInputFormat.java:500)
    at org.apache.hadoop.mapreduce.lib.input.FileInputFormat.setInputPaths(FileInputFormat.java:469)
    at org.apache.spark.SparkContext$$anonfun$newAPIHadoopFile$2.apply(SparkContext.scala:1160)
    at org.apache.spark.SparkContext$$anonfun$newAPIHadoopFile$2.apply(SparkContext.scala:1148)
    at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:151)
    at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:112)
    at org.apache.spark.SparkContext.withScope(SparkContext.scala:701)
    at org.apache.spark.SparkContext.newAPIHadoopFile(SparkContext.scala:1148)
    at com.databricks.spark.xml.util.XmlFile$.withCharset(XmlFile.scala:46)
    at com.databricks.spark.xml.DefaultSource$$anonfun$createRelation$1.apply(DefaultSource.scala:62)
    at com.databricks.spark.xml.DefaultSource$$anonfun$createRelation$1.apply(DefaultSource.scala:62)
    at com.databricks.spark.xml.XmlRelation$$anonfun$1.apply(XmlRelation.scala:47)
    at com.databricks.spark.xml.XmlRelation$$anonfun$1.apply(XmlRelation.scala:46)
    at scala.Option.getOrElse(Option.scala:121)
    at com.databricks.spark.xml.XmlRelation.<init>(XmlRelation.scala:45)
    at com.databricks.spark.xml.DefaultSource.createRelation(DefaultSource.scala:65)
    at com.databricks.spark.xml.DefaultSource.createRelation(DefaultSource.scala:43)
    at org.apache.spark.sql.execution.datasources.DataSource.resolveRelation(DataSource.scala:306)
    at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:178)
    at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:156)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
    at java.lang.reflect.Method.invoke(Unknown Source)
    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(Unknown Source)

Answer 1

某种方式pyspark无法加载http或https，我的一位同事找到了答案，所以这里是解决方案，

在创建spark上下文和sql上下文之前，我们需要加载这两行代码

import os
os.environ['PYSPARK_SUBMIT_ARGS'] = '--packages com.databricks:spark-xml_2.11:0.4.1 pyspark-shell'

从sc = pyspark.SparkContext.getOrCreate和sqlContext = SQLContext(sc)

创建sparkcontext和sqlcontext之后

使用sc.addFile(url)

将http或https网址添加到sc中

Data_XMLFile = sqlContext.read.format("xml").options(rowTag="anytaghere").load(pyspark.SparkFiles.get("*_public.xml")).coalesce(10).cache()

这个解决方案对我有用

Answer 2

错误消息说明了一切：您无法使用数据帧读取器＆amp; load访问网络上的文件（http或htpps）。我建议你先在本地下载文件。

有关可用来源的更多信息，请参阅pyspark.sql.DataFrameReader docs（通常，本地文件系统，HDFS和通过JDBC的数据库）。

与错误无关，请注意您似乎错误地使用了命令的format部分：假设您使用XML Data Source for Apache Spark package，正确的用法应为format('com.databricks.spark.xml')（请参阅example）。

Answer 3

我提交了一个类似但略有不同的错误：忘记文件路径的“s3：//”前缀。添加此前缀以形成“s3：// path / to / object”后，以下代码可以正常工作：

my_data = spark.read.format("com.databricks.spark.csv")\
               .option("header", "true")\
               .option("inferSchema", "true")\
               .option("delimiter", ",")\
               .load("s3://path/to/object")

PySpark java.io.IOException：没有用于方案的FileSystem：https

3 个答案: