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