Re: One element per node

2015-09-18 Thread Reynold Xin
lto:r...@databricks.com] > *Sent:* Friday, September 18, 2015 4:37 PM > *To:* Ulanov, Alexander > *Cc:* Feynman Liang; dev@spark.apache.org > *Subject:* Re: One element per node > > > > Use a global atomic boolean and return nothing from that partition if the > boolean is

RE: One element per node

2015-09-18 Thread Ulanov, Alexander
partitions per node? From: Reynold Xin [mailto:r...@databricks.com] Sent: Friday, September 18, 2015 4:37 PM To: Ulanov, Alexander Cc: Feynman Liang; dev@spark.apache.org Subject: Re: One element per node Use a global atomic boolean and return nothing from that partition if the boolean is true

Re: One element per node

2015-09-18 Thread Feynman Liang
hat I have only one element per executor > (per worker, or per physical node)? > > > > *From:* Feynman Liang [mailto:fli...@databricks.com] > *Sent:* Friday, September 18, 2015 4:06 PM > *To:* Ulanov, Alexander > *Cc:* dev@spark.apache.org > *Subject:* Re: One element per

Re: One element per node

2015-09-18 Thread Reynold Xin
sical node)? *From:* Feynman Liang [mailto:fli...@databricks.com ] *Sent:* Friday, September 18, 2015 4:06 PM *To:* Ulanov, Alexander *Cc:* dev@spark.apache.org *Subject:* Re: One element per node rdd.mapPartitions(x => new Iterator(x.head)) On Fri, Sep 18, 2015 at 3:57 PM, Ulanov, A

RE: One element per node

2015-09-18 Thread Ulanov, Alexander
Thank you! How can I guarantee that I have only one element per executor (per worker, or per physical node)? From: Feynman Liang [mailto:fli...@databricks.com] Sent: Friday, September 18, 2015 4:06 PM To: Ulanov, Alexander Cc: dev@spark.apache.org Subject: Re: One element per node

Re: One element per node

2015-09-18 Thread Feynman Liang
rdd.mapPartitions(x => new Iterator(x.head)) On Fri, Sep 18, 2015 at 3:57 PM, Ulanov, Alexander wrote: > Dear Spark developers, > > > > Is it possible (and how to do it if possible) to pick one element per > physical node from an RDD? Let’s say the first element of any partition on > that node.