Hello ayan,

Same key will not exists in different lists.
Which means, If "1" exists in a list, then it will not be presented in
another list.

Thank you.

On Thu, Mar 30, 2017 at 3:56 PM, ayan guha <[email protected]> wrote:
> Is it possible for one key in 2 groups in rdd2?
>
> [1,2,3]
> [1,4,5]
>
> ?
>
> On Thu, 30 Mar 2017 at 12:23 pm, Mungeol Heo <[email protected]> wrote:
>>
>> Hello Yong,
>>
>> First of all, thank your attention.
>> Note that the values of elements, which have values at RDD/DF1, in the
>> same list will be always same.
>> Therefore, the "1" and "3", which from RDD/DF 1, will always have the
>> same value which is "a".
>>
>> The goal here is assigning same value to elements of the list which
>> does not exist in RDD/DF 1.
>> So, all the elements in the same list can have same value.
>>
>> Or, the final RDD/DF also can be like this,
>>
>> [1, 2, 3], a
>> [4, 5], b
>>
>> Thank you again.
>>
>> - Mungeol
>>
>>
>> On Wed, Mar 29, 2017 at 9:03 PM, Yong Zhang <[email protected]> wrote:
>> > What is the desired result for
>> >
>> >
>> > RDD/DF 1
>> >
>> > 1, a
>> > 3, c
>> > 5, b
>> >
>> > RDD/DF 2
>> >
>> > [1, 2, 3]
>> > [4, 5]
>> >
>> >
>> > Yong
>> >
>> > ________________________________
>> > From: Mungeol Heo <[email protected]>
>> > Sent: Wednesday, March 29, 2017 5:37 AM
>> > To: [email protected]
>> > Subject: Need help for RDD/DF transformation.
>> >
>> > Hello,
>> >
>> > Suppose, I have two RDD or data frame like addressed below.
>> >
>> > RDD/DF 1
>> >
>> > 1, a
>> > 3, a
>> > 5, b
>> >
>> > RDD/DF 2
>> >
>> > [1, 2, 3]
>> > [4, 5]
>> >
>> > I need to create a new RDD/DF like below from RDD/DF 1 and 2.
>> >
>> > 1, a
>> > 2, a
>> > 3, a
>> > 4, b
>> > 5, b
>> >
>> > Is there an efficient way to do this?
>> > Any help will be great.
>> >
>> > Thank you.
>> >
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>> >
>>
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> --
> Best Regards,
> Ayan Guha

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