RDD is collection of object And if these objects are mutable and changed
then the same will reflect in RDD.
For immutable objects it will not. Changing the mutable objects that are in
the RDD is not right practise.
The RDD is immutable in the sense that any transformation on the RDD will
result i
You can, but you shouldn't. Using backdoors to mutate the data in an RDD
is a good way to produce confusing and inconsistent results when, e.g., an
RDD's lineage needs to be recomputed or a Task is resubmitted on fetch
failure.
On Tue, Dec 29, 2015 at 11:24 AM, ai he wrote:
> Same thing.
>
> Sa
Same thing.
Say, your underlying structure is like Array(ArrayBuffer(1, 2),
ArrayBuffer(3, 4)).
Then you can add/remove data in ArrayBuffers and then the change will
be reflected in the rdd.
On Tue, Dec 29, 2015 at 11:19 AM, salexln wrote:
> I see, so in order the RDD to be completely immutab
I see, so in order the RDD to be completely immutable, its content should be
immutable as well.
And if the content is not immutable, we can change its content, but cannot
add / remove data?
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Hi salexln,
RDD's immutability depends on the underlying structure. I have the
following example.
--
scala> val m = Array.fill(2, 2)(0)
m: Array[Array[Int]] = Array(Array(0, 0), Array(0
Hi Yu,
thanks for the reply.
I only run it in my machine (in order to debug the algorithm) . What other
conditions did you have it mind?
My whole code is here: https://github.com/salexln/FinalProject_FCM
Best,
Alex
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Hi salexln,
Did you reproduce the same issue under any different condition?
I can't reproduce this issue, since I don't know the details of the
algorithm.
Please let me know more detailed condition or the repository?
Thanks,
Yu
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-- Yu Ishikawa
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