I have an RDD<Object> which I get from Hbase scan using newAPIHadoopRDD. I
am running here ZipWithIndex and its preserving the order. first object got
1 second got 2 third got 3 and so on nth object got n.
On 24 April 2015 at 20:56, 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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