Anwar,
Will try this as it might do exactly what I need. I will follow your
pattern but use sc.textFile() for each file.
I am now thinking that I could start with an RDD of file paths and map it
into (path, content) pairs, provided I could read a file on the server.
Thank you,
Oleg
On 1 June 2014 18:41, Anwar Rizal <anriza...@gmail.com> wrote:
> I presume that you need to have access to the path of each file you are
> reading.
>
> I don't know whether there is a good way to do that for HDFS, I need to
> read the files myself, something like:
>
> def openWithPath(inputPath: String, sc:SparkContext) = {
> val fs = (new
> Path(inputPath)).getFileSystem(sc.hadoopConfiguration)
> val filesIt = fs.listFiles(path, false)
> val paths = new ListBuffer[URI]
> while (filesIt.hasNext) {
> paths += filesIt.next.getPath.toUri
> }
> val withPaths = paths.toList.map{ p =>
> sc.newAPIHadoopFile[LongWritable, Text,
> TextInputFormat](p.toString).map{ case (_,s) => (p, s.toString) }
> }
> withPaths.reduce{ _ ++ _ }
> }
> ...
>
> I would be interested if there is a better way to do the same thing ...
>
> Cheers,
> a:
>
>
> On Sun, Jun 1, 2014 at 6:00 PM, Nicholas Chammas <
> nicholas.cham...@gmail.com> wrote:
>
>> Could you provide an example of what you mean?
>>
>> I know it's possible to create an RDD from a path with wildcards, like in
>> the subject.
>>
>> For example, sc.textFile('s3n://bucket/2014-??-??/*.gz'). You can also
>> provide a comma delimited list of paths.
>>
>> Nick
>>
>> 2014년 6월 1일 일요일, Oleg Proudnikov<oleg.proudni...@gmail.com>님이 작성한 메시지:
>>
>> Hi All,
>>>
>>> Is it possible to create an RDD from a directory tree of the following
>>> form?
>>>
>>> RDD[(PATH, Seq[TEXT])]
>>>
>>> Thank you,
>>> Oleg
>>>
>>>
>
--
Kind regards,
Oleg
