Yes.
Kind regards,
Michał Michalski,
michal.michal...@boxever.com
On 24 April 2015 at 17:12, Jeetendra Gangele <gangele...@gmail.com> wrote:
> you used ZipWithUniqueID?
>
> On 24 April 2015 at 21:28, Michal Michalski <michal.michal...@boxever.com>
> wrote:
>
>> I somehow missed zipWithIndex (and Sean's email), thanks for hint. I mean
>> - I saw it before, but I just thought it's not doing what I want. I've
>> re-read the description now and it looks like it might be actually what I
>> need. Thanks.
>>
>> Kind regards,
>> Michał Michalski,
>> michal.michal...@boxever.com
>>
>> On 24 April 2015 at 16:26, Ganelin, Ilya <ilya.gane...@capitalone.com>
>> wrote:
>>
>>> To maintain the order you can use zipWithIndex as Sean Owen pointed
>>> out. This is the same as zipWithUniqueId except the assigned number is the
>>> index of the data in the RDD which I believe matches the order of data as
>>> it's stored on HDFS.
>>>
>>>
>>>
>>> Sent with Good (www.good.com)
>>>
>>>
>>> -----Original Message-----
>>> *From: *Michal Michalski [michal.michal...@boxever.com]
>>> *Sent: *Friday, April 24, 2015 11:18 AM Eastern Standard Time
>>> *To: *Ganelin, Ilya
>>> *Cc: *Spico Florin; user
>>> *Subject: *Re: Does HadoopRDD.zipWithIndex method preserve the order of
>>> the input data from Hadoop?
>>>
>>> I read it one by one as I need to maintain the order, but it doesn't
>>> mean that I process them one by one later. Input lines refer to different
>>> entities I update, so once I read them in order, I group them by the id of
>>> the entity I want to update, sort the updates on per-entity basis and
>>> process them further in parallel (including writing data to C* and Kafka at
>>> the very end). That's what I use Spark for - the first step I ask about is
>>> just a requirement related to the input format I get and need to support.
>>> Everything what happens after that is just a normal data processing job
>>> that you want to distribute.
>>>
>>> Kind regards,
>>> Michał Michalski,
>>> michal.michal...@boxever.com
>>>
>>> On 24 April 2015 at 16:10, Ganelin, Ilya <ilya.gane...@capitalone.com>
>>> wrote:
>>>
>>>> If you're reading a file one by line then you should simply use Java's
>>>> Hadoop FileSystem class to read the file with a BuffereInputStream. I don't
>>>> think you need an RDD here.
>>>>
>>>>
>>>>
>>>> Sent with Good (www.good.com)
>>>>
>>>>
>>>> -----Original Message-----
>>>> *From: *Michal Michalski [michal.michal...@boxever.com]
>>>> *Sent: *Friday, April 24, 2015 11:04 AM Eastern Standard Time
>>>> *To: *Ganelin, Ilya
>>>> *Cc: *Spico Florin; user
>>>> *Subject: *Re: Does HadoopRDD.zipWithIndex method preserve the order
>>>> of the input data from Hadoop?
>>>>
>>>> The problem I'm facing is that I need to process lines from input file
>>>> in the order they're stored in the file, as they define the order of
>>>> updates I need to apply on some data and these updates are not commutative
>>>> so that order matters. Unfortunately the input is purely order-based,
>>>> theres no timestamp per line etc. in the file and I'd prefer to avoid
>>>> preparing the file in advance by adding ordinals before / after each line.
>>>> From the approaches you suggested first two won't work as there's nothing I
>>>> could sort by. I'm not sure about the third one - I'm just not sure what
>>>> you meant there to be honest :-)
>>>>
>>>> Kind regards,
>>>> Michał Michalski,
>>>> michal.michal...@boxever.com
>>>>
>>>> On 24 April 2015 at 15:48, Ganelin, Ilya <ilya.gane...@capitalone.com>
>>>> wrote:
>>>>
>>>>> Michael - you need to sort your RDD. Check out the shuffle
>>>>> documentation on the Spark Programming Guide. It talks about this
>>>>> specifically. You can resolve this in a couple of ways - either by
>>>>> collecting your RDD and sorting it, using sortBy, or not worrying about
>>>>> the
>>>>> internal ordering. You can still extract elements in order by using a
>>>>> filter with the zip if e.g RDD.filter(s => s._2 < 50).sortBy(_._1)
>>>>>
>>>>>
>>>>>
>>>>> Sent with Good (www.good.com)
>>>>>
>>>>>
>>>>>
>>>>> -----Original Message-----
>>>>> *From: *Michal Michalski [michal.michal...@boxever.com]
>>>>> *Sent: *Friday, April 24, 2015 10:41 AM Eastern Standard Time
>>>>> *To: *Spico Florin
>>>>> *Cc: *user
>>>>> *Subject: *Re: Does HadoopRDD.zipWithIndex method preserve the order
>>>>> of the input data from Hadoop?
>>>>>
>>>>> Of course after you do it, you probably want to call
>>>>> repartition(somevalue) on your RDD to "get your paralellism back".
>>>>>
>>>>> Kind regards,
>>>>> Michał Michalski,
>>>>> michal.michal...@boxever.com
>>>>>
>>>>> On 24 April 2015 at 15:28, Michal Michalski <
>>>>> michal.michal...@boxever.com> wrote:
>>>>>
>>>>>> I did a quick test as I was curious about it too. I created a file
>>>>>> with numbers from 0 to 999, in order, line by line. Then I did:
>>>>>>
>>>>>> scala> val numbers = sc.textFile("./numbers.txt")
>>>>>> scala> val zipped = numbers.zipWithUniqueId
>>>>>> scala> zipped.foreach(i => println(i))
>>>>>>
>>>>>> Expected result if the order was preserved would be something like:
>>>>>> (0, 0), (1, 1) etc.
>>>>>> Unfortunately, the output looks like this:
>>>>>>
>>>>>> (126,1)
>>>>>> (223,2)
>>>>>> (320,3)
>>>>>> (1,0)
>>>>>> (127,11)
>>>>>> (2,10)
>>>>>> (...)
>>>>>>
>>>>>> The workaround I found that works for me for my specific use case
>>>>>> (relatively small input files) is setting explicitly the number of
>>>>>> partitions to 1 when reading a single *text* file:
>>>>>>
>>>>>> scala> val numbers_sp = sc.textFile("./numbers.txt", 1)
>>>>>>
>>>>>> Than the output is exactly as I would expect.
>>>>>>
>>>>>> I didn't dive into the code too much, but I took a very quick look at
>>>>>> it and figured out - correct me if I missed something, it's Friday
>>>>>> afternoon! ;-) - that this workaround will work fine for all the input
>>>>>> formats inheriting from org.apache.hadoop.mapred.FileInputFormat
>>>>>> including
>>>>>> TextInputFormat, of course - see the implementation of getSplits() method
>>>>>> there (
>>>>>> http://grepcode.com/file/repo1.maven.org/maven2/org.jvnet.hudson.hadoop/hadoop-core/0.19.1-hudson-2/org/apache/hadoop/mapred/FileInputFormat.java#FileInputFormat.getSplits%28org.apache.hadoop.mapred.JobConf%2Cint%29
>>>>>> ).
>>>>>> The numSplits variable passed there is exactly the same value as you
>>>>>> provide as a second argument to textFile, which is minPartitions.
>>>>>> However,
>>>>>> while *min* suggests that we can only define a minimal number of
>>>>>> partitions, while we have no control over the max, from what I can see in
>>>>>> the code, that value specifies the *exact* number of partitions per the
>>>>>> FileInputFormat.getSplits implementation. Of course it can differ for
>>>>>> other
>>>>>> input formats, but in this case it should work just fine.
>>>>>>
>>>>>>
>>>>>> Kind regards,
>>>>>> Michał Michalski,
>>>>>> michal.michal...@boxever.com
>>>>>>
>>>>>> On 24 April 2015 at 14:05, Spico Florin <spicoflo...@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> Hello!
>>>>>>> I know that HadoopRDD partitions are built based on the number of
>>>>>>> splits in HDFS. I'm wondering if these partitions preserve the initial
>>>>>>> order of data in file.
>>>>>>> As an example, if I have an HDFS (myTextFile) file that has these
>>>>>>> splits:
>>>>>>>
>>>>>>> split 0-> line 1, ..., line k
>>>>>>> split 1->line k+1,..., line k+n
>>>>>>> splt 2->line k+n, line k+n+m
>>>>>>>
>>>>>>> and the code
>>>>>>> val lines=sc.textFile("hdfs://mytextFile")
>>>>>>> lines.zipWithIndex()
>>>>>>>
>>>>>>> will the order of lines preserved?
>>>>>>> (line 1, zipIndex 1) , .. (line k, zipIndex k), and so one.
>>>>>>>
>>>>>>> I found this question on stackoverflow (
>>>>>>> http://stackoverflow.com/questions/26046410/how-can-i-obtain-an-element-position-in-sparks-rdd)
>>>>>>> whose answer intrigued me:
>>>>>>> "Essentially, RDD's zipWithIndex() method seems to do this, but it
>>>>>>> won't preserve the original ordering of the data the RDD was created
>>>>>>> from"
>>>>>>>
>>>>>>> Can you please confirm that is this the correct answer?
>>>>>>>
>>>>>>> Thanks.
>>>>>>> Florin
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
>>>>>
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