Hi Pengcheng YIN,
RDD cache / persist calls do not trigger evaluation.
Unpersist call is blocking (it does have an async flavor but am not sure what
are the SLAs on behavior).
val rdd = sc.textFile().map()
rdd.persist() // This does not trigger actual storage
while(true){
val count = rdd.filter().count // this will trigger storage of RDD, so far
so good
if(count == 0)
break
newRdd = /* some codes that use `rdd` several times, and produce an new RDD
*/
rdd.unpersist() // This is immediate !!, if newRDD has not been evaluated +
stored yet, it's not good
rdd = newRdd.persist() // this will do nothing till next iteration of loop
(at count).
}
IMHO, last 3 lines can be replaced with -
newRdd = /* some codes that use `rdd` several times, and produce an new RDD
*/
ADDED --> newRdd.persist( ) // mark for storage
ADDED --> newRdd.filter( ... ).count // trigger storage
rdd.unpersist()
rdd = newRdd
Although, others can correct me if I am mistaken. You can also verify this with
small dataset.
Thanks,
Kapil
-----Original Message-----
From: Pengcheng YIN [mailto:[email protected]]
Sent: 04 January 2015 12:53
To: [email protected]
Subject: does calling cache()/persist() on a RDD trigger its immediate
evaluation?
Hi Pro,
I have a question regarding calling cache()/persist() on an RDD. All RDDs in
Spark are lazily evaluated, but does calling cache()/persist() on a RDD trigger
its immediate evaluation?
My spark app is something like this:
val rdd = sc.textFile().map()
rdd.persist()
while(true){
val count = rdd.filter().count
if(count == 0)
break
newRdd = /* some codes that use `rdd` several times, and produce an new RDD
*/
rdd.unpersist()
rdd = newRdd.persist()
}
In each iteration, I persist `rdd`, and unpersist it at the end of the
iteration, replace `rdd` with persisted `newRdd`. My concern is that, if RDD is
not evaluated and persisted when persist() is called, I need to change the
position of persist()/unpersist() called to make it more efficient.
Thanks,
Pengcheng
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