Abh*_*ary 26 python apache-spark apache-spark-sql
我已经构建了Spark-csv,并且可以使用以下命令从pyspark shell中使用它
bin/spark-shell --packages com.databricks:spark-csv_2.10:1.0.3
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得到错误
>>> df_cat.save("k.csv","com.databricks.spark.csv")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Users/abhishekchoudhary/bigdata/cdh5.2.0/spark-1.3.1/python/pyspark/sql/dataframe.py", line 209, in save
self._jdf.save(source, jmode, joptions)
File "/Users/abhishekchoudhary/bigdata/cdh5.2.0/spark-1.3.1/python/lib/py4j-0.8.2.1-src.zip/py4j/java_gateway.py", line 538, in __call__
File "/Users/abhishekchoudhary/bigdata/cdh5.2.0/spark-1.3.1/python/lib/py4j-0.8.2.1-src.zip/py4j/protocol.py", line 300, in get_return_value
py4j.protocol.Py4JJavaError
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我应该将jar文件放在我的spark预构建设置中,以便我也可以spark-csv直接从python编辑器访问.
小智 26
当我使用spark-csv时,我也不得不下载commons-csvjar(不确定它是否仍然相关).两个罐子里面都有火花分布文件夹.
我下载了以下罐子:
wget http://search.maven.org/remotecontent?filepath=org/apache/commons/commons-csv/1.1/commons-csv-1.1.jar -O commons-csv-1.1.jar<br/>
wget http://search.maven.org/remotecontent?filepath=com/databricks/spark-csv_2.10/1.0.0/spark-csv_2.10-1.0.0.jar -O spark-csv_2.10-1.0.0.jar
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./bin/pyspark --jars "spark-csv_2.10-1.0.0.jar,commons-csv-1.1.jar"
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from pyspark.sql import SQLContext
sqlContext = SQLContext(sc)
df = sqlContext.load(source="com.databricks.spark.csv", path = "/path/to/you/file.csv")
df.show()
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另一个选择是将以下内容添加到spark-defaults.conf:
spark.jars.packages com.databricks:spark-csv_2.11:1.2.0
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Jim*_*mmy 16
而不是将jar放在任何特定的文件夹中,一个简单的解决方法是使用以下参数启动pyspark shell:
bin/pyspark --packages com.databricks:spark-csv_2.10:1.0.3
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这将自动加载所需的spark-csv罐子.
然后执行以下操作以阅读csv文件:
from pyspark.sql import SQLContext
sqlContext = SQLContext(sc)
df = sqlContext.read.format('com.databricks.spark.csv').options(header='true').load('file.csv')
df.show()
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