I think there is a shuffle stage involved. And the future count job will depends on the first job’s shuffle stages’s output data directly as long as it is still available. Thus it will be much faster. Best Regards, Raymond Liu
From: tomsheep...@gmail.com [mailto:tomsheep...@gmail.com] Sent: Friday, June 27, 2014 10:08 AM To: user Subject: Re: About StorageLevel Thank u Andrew, that's very helpful. I still have some doubts on a simple trial: I opened a spark shell in local mode, and typed in val r=sc.parallelize(0 to 500000) val r2=r.keyBy(x=>x).groupByKey(10) and then I invoked the count action several times on it, r2.count (multiple times) The first job obviously takes more time than the latter ones. Is there some magic underneath? Regards, Kang Liu From: Andrew Or<mailto:and...@databricks.com> Date: 2014-06-27 02:25 To: user<mailto:user@spark.apache.org> Subject: Re: About StorageLevel Hi Kang, You raise a good point. Spark does not automatically cache all your RDDs. Why? Simply because the application may create many RDDs, and not all of them are to be reused. After all, there is only so much memory available to each executor, and caching an RDD adds some overhead especially if we have to kick out old blocks with LRU. As an example, say you run the following chain: sc.textFile(...).map(...).filter(...).flatMap(...).map(...).reduceByKey(...).count() You might be interested in reusing only the final result, but each step of the chain actually creates an RDD. If we automatically cache all RDDs, then we'll end up doing extra work for the RDDs we don't care about. The effect can be much worse if our RDDs are big and there are many of them, in which case there may be a lot of churn in the cache as we constantly evict RDDs we reuse. After all, the users know best what RDDs they are most interested in, so it makes sense to give them control over caching behavior. Best, Andrew 2014-06-26 5:36 GMT-07:00 tomsheep...@gmail.com<mailto:tomsheep...@gmail.com> <tomsheep...@gmail.com<mailto:tomsheep...@gmail.com>>: Hi all, I have a newbie question about StorageLevel of spark. I came up with these sentences in spark documents: If your RDDs fit comfortably with the default storage level (MEMORY_ONLY), leave them that way. This is the most CPU-efficient option, allowing operations on the RDDs to run as fast as possible. And Spark automatically monitors cache usage on each node and drops out old data partitions in a least-recently-used (LRU) fashion. If you would like to manually remove an RDD instead of waiting for it to fall out of the cache, use the RDD.unpersist() method. http://spark.apache.org/docs/latest/programming-guide.html#rdd-persistence But I found the default storageLevel is NONE in source code, and if I never call 'persist(somelevel)', that value will always be NONE. The 'iterator' method goes to final def iterator(split: Partition, context: TaskContext): Iterator[T] = { if (storageLevel != StorageLevel.NONE) { SparkEnv.get.cacheManager.getOrCompute(this, split, context, storageLevel) } else { computeOrReadCheckpoint(split, context) } } Is that to say, the rdds are cached in memory (or somewhere else) if and only if the 'persist' or 'cache' method is called explicitly, otherwise they will be re-computed every time even in an iterative situation? It made me confused becase I had a first impression that spark is super-fast because it prefers to store intermediate results in memory automatically. Forgive me if I asked a stupid question. Regards, Kang Liu
