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From innowireless TaeYun Kim <>
Subject RE: Question about RDD cache, unpersist, materialization
Date Fri, 13 Jun 2014 01:09:20 GMT
(I¡¯ve clarified the statement (1) of my previous mail. See below.)


From: innowireless TaeYun Kim [] 
Sent: Friday, June 13, 2014 10:05 AM
Subject: RE: Question about RDD cache, unpersist, materialization


Currently I use rdd.count() for forceful computation, as Nick Pentreath suggested.


I think that it will be nice to have a method that forcefully computes a rdd, so that the
unnecessary rdds are safely unpersist()ed.


Let’s think a case that a rdd_a is a parent of both:

(1) a short-term rdd_s that depends only on rdd_a (maybe rdd_s is a transformation of rdd_a)

(2) and a long-term rdd_t that depends on the rdds that may be computed later. (may be for
some aggregation)

Usually rdd_s is computed early, and rdd_a is computed for it. But it has to remain in memory
until rdd_t is eventually computed. (Or recomputed when rdd_t is computed)

It would be nice if rdd_t could be computed when rdd_s is computed, so that rdd_a can be unpersist()ed,
since it will not be used anymore.

(Currently I use rdd_t.count() for that)


sc.prune() which was suggested in the related discussion you provided is not helpful for this
case, since rdd_a is ‘referenced’ by ‘lazy’ rdd_t.

(Get up rdd_t, and do it with rdd_s while rdd_a is here for rdd_s.)



From: Nicholas Chammas [] 
Sent: Friday, June 13, 2014 9:31 AM
To: user
Subject: Re: Question about RDD cache, unpersist, materialization


FYI: Here is a related discussion <>
 about this.


On Thu, Jun 12, 2014 at 8:10 PM, innowireless TaeYun Kim <>

Maybe It would be nice that unpersist() ¡®triggers¡¯ the computations of other rdds that
depends on it but not yet computed.
The pseudo code can be as follows:


    if (this rdd has not been persisted)
    for (all rdds that depends on this rdd but not yet computed)


From: Daniel Siegmann [] 
Sent: Friday, June 13, 2014 5:38 AM
Subject: Re: Question about RDD cache, unpersist, materialization


I've run into this issue. The goal of caching / persist seems to be to avoid recomputing an
RDD when its data will be needed multiple times. However, once the following RDDs are computed
the cache is no longer needed. The currently design provides no obvious way to detect when
the cache is no longer needed so it can be discarded.

In the case of cache in memory, it may be handled by partitions being dropped (in LRU order)
when memory fills up. I need to do some more experimentation to see if this really works well,
or if allowing memory to fill up causes performance issues or possibly OOM errors if data
isn't correctly freed.

In the case of persisting to disk, I'm not sure if there's a way to limit the disk space used
for caching. Does anyone know if there is such a configuration option? This is a pressing
issue for me - I have had jobs fail because nodes ran out of disk space.


On Wed, Jun 11, 2014 at 2:26 AM, Nick Pentreath <> wrote:

If you want to force materialization use .count()


Also if you can simply don't unpersist anything, unless you really need to free the memory

Sent from Mailbox <>  


On Wed, Jun 11, 2014 at 5:13 AM, innowireless TaeYun Kim <>

BTW, it is possible that rdd.first() does not compute the whole partitions. 
So, first() cannot be uses for the situation below. 

-----Original Message----- 
From: innowireless TaeYun Kim [] 
Sent: Wednesday, June 11, 2014 11:40 AM 
Subject: Question about RDD cache, unpersist, materialization 


What I (seems to) know about RDD persisting API is as follows: 
- cache() and persist() is not an action. It only does a marking. 
- unpersist() is also not an action. It only removes a marking. But if the 
rdd is already in memory, it is unloaded. 

And there seems no API to forcefully materialize the RDD without requiring a 
data by an action method, for example first(). 

So, I am faced with the following scenario. 

JavaRDD<T> rddUnion = sc.parallelize(new ArrayList<T>()); // create 
empty for merging 
for (int i = 0; i < 10; i++) 
JavaRDD<T2> rdd = sc.textFile(inputFileNames[i]); 
rdd.cache(); // Since it will be used twice, cache.[i]); // 
Transform and save, rdd materializes 
rddUnion = rddUnion.union(; // Do another 
transform to T and merge by union 
rdd.unpersist(); // Now it seems not needed. (But needed actually) 
// Here, rddUnion actually materializes, and needs all 10 rdds that 
already unpersisted. 
// So, rebuilding all 10 rdds will occur. 

If rddUnion can be materialized before the rdd.unpersist() line and 
cache()d, the rdds in the loop will not be needed on 

Now what is the best strategy? 
- Do not unpersist all 10 rdds in the loop. 
- Materialize rddUnion in the loop by calling 'light' action API, like 
- Give up and just rebuild/reload all 10 rdds when saving rddUnion. 

Is there some misunderstanding? 




Daniel Siegmann, Software Developer

Accelerating Machine Learning

E:  <> W:  <>


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