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 ?
>>>>>>>
>>>>>>
>>>
>
