t;>> In both the cases data is distributed uniformaly.
>>>> I do have following questions on the basis of above observation:
>>>>
>>>> 1. In case of rdd1, hash partitioning should calculate hashcode of key
>>>> (i.e. "aa" in
t;>> I do have following questions on the basis of above observation:
>>>
>>> 1. In case of rdd1, hash partitioning should calculate hashcode of key
>>> (i.e. "aa" in this case), so all records should go to single partition
>>> instead of uniform dist
calculate hashcode of key
>> (i.e. "aa" in this case), so all records should go to single partition
>> instead of uniform distribution?
>> 2. In case of rdd2, there is no key value pair so how hash partitoning
>> going to work i.e. what is the key to calculate hashcode?
&
instead of uniform distribution?
> 2. In case of rdd2, there is no key value pair so how hash partitoning
> going to work i.e. what is the key to calculate hashcode?
>
> I have followed @zero323 answer but not getting answer of these.
> https://stackoverflow.com/questions/314243
late hashcode?
I have followed @zero323 answer but not getting answer of these.
https://stackoverflow.com/questions/31424396/how-does-hashpartitioner-work
-----
__Vikash Pareek
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