使用 pyspark 读取 csv

Question

我是新手。我正在尝试使用 pyspark 读取 csv 文件。我提到了 PySpark How to read CSV into Dataframe, and manipulate it , Get CSV to Spark dataframe 等等。我尝试通过两种方式阅读它：

1个

from  pyspark.sql import SparkSession
from pyspark.sql import SQLContext
from pyspark.conf import SparkConf
sc = SparkContext.getOrCreate()
df = spark.read.csv('D:/Users/path/csv/test.csv')
df.show()

2个

import pyspark
sc = pyspark.SparkContext()
sql = SQLContext(sc)

df = (sql.read
         .format("com.databricks.spark.csv")
         .option("header", "true")
         .load("D:/Users/path/csv/test.csv"))
df.show()

这两个代码都不起作用。我收到以下错误：

 Py4JJavaError                             Traceback (most recent call last)
<ipython-input-28-c6263cc7dab9> in <module>()
      4
      5 sc = SparkContext.getOrCreate()
----> 6 df = spark.read.csv('D:/Users/path/csv/test.csv')
      7 df.show()
      8

~\opt\spark\spark-2.1.0-bin-hadoop2.7\python\pyspark\sql\readwriter.py in csv(self, path, schema, sep, encoding, quote, escape, comment, header, inferSchema, ignoreLeadingWhiteSpace, ignoreTrailingWhiteSpace, nullValue, nanValue, positiveInf, negativeInf, dateFormat, timestampFormat, maxColumns, maxCharsPerColumn, maxMalformedLogPerPartition, mode)
    378         if isinstance(path, basestring):
    379             path = [path]
--> 380         return self._df(self._jreader.csv(self._spark._sc._jvm.PythonUtils.toSeq(path)))
    381
    382     @since(1.5)

~\opt\spark\spark-2.1.0-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:

~\opt\spark\spark-2.1.0-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()

~\opt\spark\spark-2.1.0-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 o663.csv.
: java.util.ServiceConfigurationError: org.apache.spark.sql.sources.DataSourceRegister: Provider org.apache.spark.sql.hive.execution.HiveFileFormat not found
    at java.util.ServiceLoader.fail(ServiceLoader.java:239)
    at java.util.ServiceLoader.access$300(ServiceLoader.java:185)
    at java.util.ServiceLoader$LazyIterator.nextService(ServiceLoader.java:372)
    at java.util.ServiceLoader$LazyIterator.next(ServiceLoader.java:404)
    at java.util.ServiceLoader$1.next(ServiceLoader.java:480)
    at scala.collection.convert.Wrappers$JIteratorWrapper.next(Wrappers.scala:43)
    at scala.collection.Iterator$class.foreach(Iterator.scala:893)

我不明白为什么它会抛出一些配置单元异常 `Py4JJavaError: An error occurred while calling o663.csv.

java.util.ServiceConfigurationError: org.apache.spark.sql.sources.DataSourceRegister: Provider org.apache.spark.sql.hive.execution.HiveFileFormat not found。如何解决此错误HiveFileFormat not found.`

谁能指导我解决此错误？

原文由 user15051990 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

首先，系统需要将 Spark Session 识别为如下命令：

 from pyspark import SparkConf, SparkContext
sc = SparkContext()

之后，必须像这样将 SQL 库 引入系统：

 from pyspark.sql import SQLContext
sqlContext = SQLContext(sc)

最后您可以通过以下命令读取您的 CSV：

 df = sqlContext.read.format('com.databricks.spark.csv').options(header='true', inferschema='true').load('path/to/your/file.csv')

原文由 Mohammad Heydari 发布，翻译遵循 CC BY-SA 4.0 许可协议