​No, ​
I'm not advising you to use .rdd, just saying it is possible. 
​Although I'd only use RDDs if you had a good reason to, given Datasets now, they are not gone or even deprecated.​

You do not need to order the whole data set to get the top eleme
​nt. That isn't what top does though. You might be interested to look at the source code. Nor is it what orderBy does if the optimizer is any good.

​Computing .rdd doesn't materialize an RDD. It involves some non-zero overhead in creating a plan, which should be minor compared to execution. So would any computation of "top N" on a Dataset, so I don't think this is relevant.

​orderBy + take is already the way to accomplish "Dataset.top". It works on Datasets, and therefore DataFrames too, for the reason you give. I'm not sure what you're asking there.

On Mon, Sep 5, 2016, 13:01 Jakub Dubovsky <spark.dubovsky.jakub@gmail.com> wrote:
Thanks Sean,

I was under impression that spark creators are trying to persuade user community not to use RDD api directly. Spark summit I attended was full of this. So I am a bit surprised that I hear use-rdd-api as an advice from you. But if this is a way then I have a second question. For conversion from dataset to rdd I would use Dataset.rdd lazy val. Since it is a lazy val it suggests there is some computation going on to create rdd as a copy. The question is how much computationally expansive is this conversion? If there is a significant overhead then it is clear why one would want to have top method directly on Dataset class.

Ordering whole dataset only to take first 10 or so top records is not really an acceptable option for us. Comparison function can be expansive and the size of dataset is (unsurprisingly) big.

To be honest I do not really understand what do you mean by b). Since DataFrame is now only an alias for Dataset[Row] what do you mean by "DataFrame-like counterpart"?


On Thu, Sep 1, 2016 at 2:31 PM, Sean Owen <sowen@cloudera.com> wrote:
You can always call .rdd.top(n) of course. Although it's slightly
clunky, you can also .orderBy($"value".desc).take(n). Maybe there's an
easier way.

I don't think if there's a strong reason other than it wasn't worth it
to write this and many other utility wrappers that a) already exist on
the underlying RDD API if you want them, and b) have a DataFrame-like
counterpart already that doesn't really need wrapping in a different

On Thu, Sep 1, 2016 at 12:53 PM, Jakub Dubovsky
<spark.dubovsky.jakub@gmail.com> wrote:
> Hey all,
> in RDD api there is very usefull method called top. It finds top n records
> in according to certain ordering without sorting all records. Very usefull!
> There is no top method nor similar functionality in Dataset api. Has anybody
> any clue why? Is there any specific reason for this?
> Any thoughts?
> thanks
> Jakub D.