Apologies janardhan, i always get confused on this
Ok. so you have a (key, val) RDD (val is irrelevant here)
then you can do this
val reduced = myRDD.reduceByKey((first, second) => first ++ second)
val sorted = reduced.sortBy(tpl => tpl._1)
hth
On Tue, Jul 26, 2016 at 3:31 AM, janardhan shetty <janardhan...@gmail.com>
wrote:
> groupBy is a shuffle operation and index is already lost in this process
> if I am not wrong and don't see *sortWith* operation on RDD.
>
> Any suggestions or help ?
>
> On Mon, Jul 25, 2016 at 12:58 AM, Marco Mistroni <mmistr...@gmail.com>
> wrote:
>
>> Hi
>> after you do a groupBy you should use a sortWith.
>> Basically , a groupBy reduces your structure to (anyone correct me if i m
>> wrong) a RDD[(key,val)], which you can see as a tuple.....so you could use
>> sortWith (or sortBy, cannot remember which one) (tpl=> tpl._1)
>> hth
>>
>> On Mon, Jul 25, 2016 at 1:21 AM, janardhan shetty <janardhan...@gmail.com
>> > wrote:
>>
>>> Thanks Marco. This solved the order problem. Had another question which
>>> is prefix to this.
>>>
>>> As you can see below ID2,ID1 and ID3 are in order and I need to maintain
>>> this index order as well. But when we do groupByKey
>>> operation(*rdd.distinct.groupByKey().mapValues(v
>>> => v.toArray*))
>>> everything is *jumbled*.
>>> Is there any way we can maintain this order as well ?
>>>
>>> scala> RDD.foreach(println)
>>> (ID2,18159)
>>> (ID1,18159)
>>> (ID3,18159)
>>>
>>> (ID2,18159)
>>> (ID1,18159)
>>> (ID3,18159)
>>>
>>> (ID2,36318)
>>> (ID1,36318)
>>> (ID3,36318)
>>>
>>> (ID2,54477)
>>> (ID1,54477)
>>> (ID3,54477)
>>>
>>> *Jumbled version : *
>>> Array(
>>> (ID1,Array(*18159*, 308703, 72636, 64544, 39244, 107937, *54477*,
>>> 145272, 100079, *36318*, 160992, 817, 89366, 150022, 19622, 44683,
>>> 58866, 162076, 45431, 100136)),
>>> (ID3,Array(100079, 19622, *18159*, 212064, 107937, 44683, 150022,
>>> 39244, 100136, 58866, 72636, 145272, 817, 89366, * 54477*, *36318*,
>>> 308703, 160992, 45431, 162076)),
>>> (ID2,Array(308703, * 54477*, 89366, 39244, 150022, 72636, 817, 58866,
>>> 44683, 19622, 160992, 107937, 100079, 100136, 145272, 64544, *18159*,
>>> 45431, *36318*, 162076))
>>> )
>>>
>>> *Expected output:*
>>> Array(
>>> (ID1,Array(*18159*,*36318*, *54477,...*)),
>>> (ID3,Array(*18159*,*36318*, *54477, ...*)),
>>> (ID2,Array(*18159*,*36318*, *54477, ...*))
>>> )
>>>
>>> As you can see after *groupbyKey* operation is complete item 18519 is
>>> in index 0 for ID1, index 2 for ID3 and index 16 for ID2 where as expected
>>> is index 0
>>>
>>>
>>> On Sun, Jul 24, 2016 at 12:43 PM, Marco Mistroni <mmistr...@gmail.com>
>>> wrote:
>>>
>>>> Hello
>>>> Uhm you have an array containing 3 tuples?
>>>> If all the arrays have same length, you can just zip all of them,
>>>> creatings a list of tuples
>>>> then you can scan the list 5 by 5...?
>>>>
>>>> so something like
>>>>
>>>> (Array(0)_2,Array(1)._2,Array(2)._2).zipped.toList
>>>>
>>>> this will give you a list of tuples of 3 elements containing each items
>>>> from ID1, ID2 and ID3 ... sample below
>>>> res: List((18159,100079,308703), (308703, 19622, 54477), (72636,18159,
>>>> 89366)..........)
>>>>
>>>> then you can use a recursive function to compare each element such as
>>>>
>>>> def iterate(lst:List[(Int, Int, Int)]):T = {
>>>> if (lst.isEmpty): /// return your comparison
>>>> else {
>>>> val splits = lst.splitAt(5)
>>>> // do sometjhing about it using splits._1
>>>> iterate(splits._2)
>>>> }
>>>>
>>>> will this help? or am i still missing something?
>>>>
>>>> kr
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> On 24 Jul 2016 5:52 pm, "janardhan shetty" <janardhan...@gmail.com>
>>>> wrote:
>>>>
>>>>> Array(
>>>>> (ID1,Array(18159, 308703, 72636, 64544, 39244, 107937, 54477, 145272,
>>>>> 100079, 36318, 160992, 817, 89366, 150022, 19622, 44683, 58866, 162076,
>>>>> 45431, 100136)),
>>>>> (ID3,Array(100079, 19622, 18159, 212064, 107937, 44683, 150022, 39244,
>>>>> 100136, 58866, 72636, 145272, 817, 89366, 54477, 36318, 308703, 160992,
>>>>> 45431, 162076)),
>>>>> (ID2,Array(308703, 54477, 89366, 39244, 150022, 72636, 817, 58866,
>>>>> 44683, 19622, 160992, 107937, 100079, 100136, 145272, 64544, 18159, 45431,
>>>>> 36318, 162076))
>>>>> )
>>>>>
>>>>> I need to compare first 5 elements of ID1 with first five element of
>>>>> ID3 next first 5 elements of ID1 to ID2. Similarly next 5 elements in
>>>>> that
>>>>> order until the end of number of elements.
>>>>> Let me know if this helps
>>>>>
>>>>>
>>>>> On Sun, Jul 24, 2016 at 7:45 AM, Marco Mistroni <mmistr...@gmail.com>
>>>>> wrote:
>>>>>
>>>>>> Apologies I misinterpreted.... could you post two use cases?
>>>>>> Kr
>>>>>>
>>>>>> On 24 Jul 2016 3:41 pm, "janardhan shetty" <janardhan...@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> Marco,
>>>>>>>
>>>>>>> Thanks for the response. It is indexed order and not ascending or
>>>>>>> descending order.
>>>>>>> On Jul 24, 2016 7:37 AM, "Marco Mistroni" <mmistr...@gmail.com>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Use map values to transform to an rdd where values are sorted?
>>>>>>>> Hth
>>>>>>>>
>>>>>>>> On 24 Jul 2016 6:23 am, "janardhan shetty" <janardhan...@gmail.com>
>>>>>>>> wrote:
>>>>>>>>
>>>>>>>>> I have a key,value pair rdd where value is an array of Ints. I
>>>>>>>>> need to maintain the order of the value in order to execute downstream
>>>>>>>>> modifications. How do we maintain the order of values?
>>>>>>>>> Ex:
>>>>>>>>> rdd = (id1,[5,2,3,15],
>>>>>>>>> Id2,[9,4,2,5]....)
>>>>>>>>>
>>>>>>>>> Followup question how do we compare between one element in rdd
>>>>>>>>> with all other elements ?
>>>>>>>>>
>>>>>>>>
>>>>>
>>>
>>
>
