You can make a Hadoop input format which passes through the name of the
file. I generally find it easier to just hit Hadoop, get the file names,
and construct the RDDs though
El martes, 1 de septiembre de 2015, Matt K <matvey1...@gmail.com> escribió:
> Just want to add - I'm looking to partition the resulting Parquet files by
> customer-id, which is why I'm looking to extract the customer-id from the
> path.
>
> On Tue, Sep 1, 2015 at 7:00 PM, Matt K <matvey1...@gmail.com
> <javascript:_e(%7B%7D,'cvml','matvey1...@gmail.com');>> wrote:
>
>> Hi all,
>>
>> TL;DR - is there a way to extract the source path from an RDD via the
>> Scala API?
>>
>> I have sequence files on S3 that look something like this:
>> s3://data/customer=123/...
>> s3://data/customer=456/...
>>
>> I am using Spark Dataframes to convert these sequence files to Parquet.
>> As part of the processing, I actually need to know the customer-id. I'm
>> doing something like this:
>>
>> val rdd = sql.sparkContext.sequenceFile("s3://data/customer=*/*",
>> classOf[BytesWritable],
>> classOf[Text])
>>
>> val rowRdd = rdd.map(x => convertTextRowToTypedRdd(x._2, schema,
>> delimiter))
>>
>> val dataFrame = sql.createDataFrame(rowRdd, schema)
>>
>>
>> What I am trying to figure out is how to get the customer-id, which is
>> part of the path. I am not sure if there's a way to extract the source path
>> from the resulting HadoopRDD. Do I need to create one RDD per customer to
>> get around this?
>>
>>
>> Thanks,
>>
>> -Matt
>>
>
>
>
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