Hi, I have a source that provides few items since it gives file names to the mappers. The mapper opens the file and process records. As the files are
huge, one input line (a filename) gives a consequent work to the next stage. My topology looks like : addSource(myFileSource).rebalance().setParallelism(100).map(myFileMapper) If 100 mappers are created, about 85 end immediately and only a few process the files (for hours). I suspect an optimization making that there is
a minimum number of lines to pass to the next node or it is “shutdown” ; but in my case I do want the lines to be evenly distributed to each mapper. How to enforce that ? Greetings, Arnaud L'intégrité de ce message n'étant pas assurée sur internet, la société expéditrice ne peut être tenue responsable de son contenu ni de ses pièces jointes. Toute utilisation ou diffusion non autorisée est interdite. Si vous n'êtes pas destinataire de ce message, merci de le détruire et d'avertir l'expéditeur. The integrity of this message cannot be guaranteed on the Internet. The company that sent this message cannot therefore be held liable for its content nor attachments. Any unauthorized use or dissemination is prohibited. If you are not the intended recipient of this message, then please delete it and notify the sender. |
Hi,
If I understand you correctly, you want to have 100 mappers. Thus you need to apply the .setParallelism() after .map() > addSource(myFileSource).rebalance().map(myFileMapper).setParallelism(100) The order of commands you used, set the dop for the source to 100 (which might be ignored, if the provided source function "myFileSource" does not implements "ParallelSourceFunction" interface). The dop for the mapper should be the default value. Using .rebalance() is absolutely correct. It distributes the emitted tuples in a round robin fashion to all consumer tasks. -Matthias On 09/02/2015 05:41 PM, LINZ, Arnaud wrote: > Hi, > > > > I have a source that provides few items since it gives file names to the > mappers. The mapper opens the file and process records. As the files are > huge, one input line (a filename) gives a consequent work to the next stage. > > My topology looks like : > > addSource(myFileSource).rebalance().setParallelism(100).map(myFileMapper) > > If 100 mappers are created, about 85 end immediately and only a few > process the files (for hours). I suspect an optimization making that > there is a minimum number of lines to pass to the next node or it is > “shutdown” ; but in my case I do want the lines to be evenly distributed > to each mapper. > > How to enforce that ? > > > > Greetings, > > Arnaud > > > ------------------------------------------------------------------------ > > L'intégrité de ce message n'étant pas assurée sur internet, la société > expéditrice ne peut être tenue responsable de son contenu ni de ses > pièces jointes. Toute utilisation ou diffusion non autorisée est > interdite. Si vous n'êtes pas destinataire de ce message, merci de le > détruire et d'avertir l'expéditeur. > > The integrity of this message cannot be guaranteed on the Internet. The > company that sent this message cannot therefore be held liable for its > content nor attachments. Any unauthorized use or dissemination is > prohibited. If you are not the intended recipient of this message, then > please delete it and notify the sender. signature.asc (836 bytes) Download Attachment |
Hi,
You are right, but in fact it does not solve my problem, since I have 100 parallelism everywhere. Each of my 100 sources gives only a few lines (say 14 max), and only the first 14 next nodes will receive data. Same problem by replacing rebalance() with shuffle(). But I found a workaround: setting parallelism to 1 for the source (I don't need a 100 directory scanners anyway), it forces the rebalancing evenly between the mappers. Greetings, Arnaud -----Message d'origine----- De : Matthias J. Sax [mailto:[hidden email]] Envoyé : mercredi 2 septembre 2015 17:56 À : [hidden email] Objet : Re: How to force the parallelism on small streams? Hi, If I understand you correctly, you want to have 100 mappers. Thus you need to apply the .setParallelism() after .map() > addSource(myFileSource).rebalance().map(myFileMapper).setParallelism(1 > 00) The order of commands you used, set the dop for the source to 100 (which might be ignored, if the provided source function "myFileSource" does not implements "ParallelSourceFunction" interface). The dop for the mapper should be the default value. Using .rebalance() is absolutely correct. It distributes the emitted tuples in a round robin fashion to all consumer tasks. -Matthias On 09/02/2015 05:41 PM, LINZ, Arnaud wrote: > Hi, > > > > I have a source that provides few items since it gives file names to > the mappers. The mapper opens the file and process records. As the > files are huge, one input line (a filename) gives a consequent work to the next stage. > > My topology looks like : > > addSource(myFileSource).rebalance().setParallelism(100).map(myFileMapp > er) > > If 100 mappers are created, about 85 end immediately and only a few > process the files (for hours). I suspect an optimization making that > there is a minimum number of lines to pass to the next node or it is > “shutdown” ; but in my case I do want the lines to be evenly > distributed to each mapper. > > How to enforce that ? > > > > Greetings, > > Arnaud > > > ---------------------------------------------------------------------- > -- > > L'intégrité de ce message n'étant pas assurée sur internet, la société > expéditrice ne peut être tenue responsable de son contenu ni de ses > pièces jointes. Toute utilisation ou diffusion non autorisée est > interdite. Si vous n'êtes pas destinataire de ce message, merci de le > détruire et d'avertir l'expéditeur. > > The integrity of this message cannot be guaranteed on the Internet. > The company that sent this message cannot therefore be held liable for > its content nor attachments. Any unauthorized use or dissemination is > prohibited. If you are not the intended recipient of this message, > then please delete it and notify the sender. |
Thanks for clarifying. shuffle() is similar to rebalance() -- however,
it redistributes randomly and not in round robin fashion. However, the problem you describe sounds like a bug to me. I included dev list. Maybe anyone else can step in so we can identify it there is a bug or not. -Matthias On 09/02/2015 06:19 PM, LINZ, Arnaud wrote: > Hi, > > You are right, but in fact it does not solve my problem, since I have 100 parallelism everywhere. Each of my 100 sources gives only a few lines (say 14 max), and only the first 14 next nodes will receive data. > Same problem by replacing rebalance() with shuffle(). > > But I found a workaround: setting parallelism to 1 for the source (I don't need a 100 directory scanners anyway), it forces the rebalancing evenly between the mappers. > > Greetings, > Arnaud > > > -----Message d'origine----- > De : Matthias J. Sax [mailto:[hidden email]] > Envoyé : mercredi 2 septembre 2015 17:56 > À : [hidden email] > Objet : Re: How to force the parallelism on small streams? > > Hi, > > If I understand you correctly, you want to have 100 mappers. Thus you need to apply the .setParallelism() after .map() > >> addSource(myFileSource).rebalance().map(myFileMapper).setParallelism(1 >> 00) > > The order of commands you used, set the dop for the source to 100 (which might be ignored, if the provided source function "myFileSource" does not implements "ParallelSourceFunction" interface). The dop for the mapper should be the default value. > > Using .rebalance() is absolutely correct. It distributes the emitted tuples in a round robin fashion to all consumer tasks. > > -Matthias > > On 09/02/2015 05:41 PM, LINZ, Arnaud wrote: >> Hi, >> >> >> >> I have a source that provides few items since it gives file names to >> the mappers. The mapper opens the file and process records. As the >> files are huge, one input line (a filename) gives a consequent work to the next stage. >> >> My topology looks like : >> >> addSource(myFileSource).rebalance().setParallelism(100).map(myFileMapp >> er) >> >> If 100 mappers are created, about 85 end immediately and only a few >> process the files (for hours). I suspect an optimization making that >> there is a minimum number of lines to pass to the next node or it is >> “shutdown” ; but in my case I do want the lines to be evenly >> distributed to each mapper. >> >> How to enforce that ? >> >> >> >> Greetings, >> >> Arnaud >> >> >> ---------------------------------------------------------------------- >> -- >> >> L'intégrité de ce message n'étant pas assurée sur internet, la société >> expéditrice ne peut être tenue responsable de son contenu ni de ses >> pièces jointes. Toute utilisation ou diffusion non autorisée est >> interdite. Si vous n'êtes pas destinataire de ce message, merci de le >> détruire et d'avertir l'expéditeur. >> >> The integrity of this message cannot be guaranteed on the Internet. >> The company that sent this message cannot therefore be held liable for >> its content nor attachments. Any unauthorized use or dissemination is >> prohibited. If you are not the intended recipient of this message, >> then please delete it and notify the sender. > signature.asc (836 bytes) Download Attachment |
Hi, I don't think it's a bug. If there are 100 sources that each emit only 14 elements then only the first 14 mappers will ever receive data. The round-robin distribution is not global, since the sources operate independently from each other. Cheers, Aljoscha On Wed, 2 Sep 2015 at 20:00 Matthias J. Sax <[hidden email]> wrote: Thanks for clarifying. shuffle() is similar to rebalance() -- however, |
If it would be only 14 elements, you are obviously right. However, if I
understood Arnaud correctly, the problem is, that there are more than 14 elements: > Each of my 100 sources gives only a few lines (say 14 max) That would be about 140 lines in total. Using non-parallel source, he is able to distribute the elements to all 100 mappers. I assume that about 40 mappers receive 2 lines, and 60 receive 1 line. @Arnaud: is this correct? -Matthias On 09/03/2015 03:04 PM, Aljoscha Krettek wrote: > Hi, > I don't think it's a bug. If there are 100 sources that each emit only > 14 elements then only the first 14 mappers will ever receive data. The > round-robin distribution is not global, since the sources operate > independently from each other. > > Cheers, > Aljoscha > > On Wed, 2 Sep 2015 at 20:00 Matthias J. Sax <[hidden email] > <mailto:[hidden email]>> wrote: > > Thanks for clarifying. shuffle() is similar to rebalance() -- however, > it redistributes randomly and not in round robin fashion. > > However, the problem you describe sounds like a bug to me. I included > dev list. Maybe anyone else can step in so we can identify it there is a > bug or not. > > -Matthias > > > On 09/02/2015 06:19 PM, LINZ, Arnaud wrote: > > Hi, > > > > You are right, but in fact it does not solve my problem, since I > have 100 parallelism everywhere. Each of my 100 sources gives only a > few lines (say 14 max), and only the first 14 next nodes will > receive data. > > Same problem by replacing rebalance() with shuffle(). > > > > But I found a workaround: setting parallelism to 1 for the source > (I don't need a 100 directory scanners anyway), it forces the > rebalancing evenly between the mappers. > > > > Greetings, > > Arnaud > > > > > > -----Message d'origine----- > > De : Matthias J. Sax [mailto:[hidden email] > <mailto:[hidden email]>] > > Envoyé : mercredi 2 septembre 2015 17:56 > > À : [hidden email] <mailto:[hidden email]> > > Objet : Re: How to force the parallelism on small streams? > > > > Hi, > > > > If I understand you correctly, you want to have 100 mappers. Thus > you need to apply the .setParallelism() after .map() > > > >> > addSource(myFileSource).rebalance().map(myFileMapper).setParallelism(1 > >> 00) > > > > The order of commands you used, set the dop for the source to 100 > (which might be ignored, if the provided source function > "myFileSource" does not implements "ParallelSourceFunction" > interface). The dop for the mapper should be the default value. > > > > Using .rebalance() is absolutely correct. It distributes the > emitted tuples in a round robin fashion to all consumer tasks. > > > > -Matthias > > > > On 09/02/2015 05:41 PM, LINZ, Arnaud wrote: > >> Hi, > >> > >> > >> > >> I have a source that provides few items since it gives file names to > >> the mappers. The mapper opens the file and process records. As the > >> files are huge, one input line (a filename) gives a consequent > work to the next stage. > >> > >> My topology looks like : > >> > >> > addSource(myFileSource).rebalance().setParallelism(100).map(myFileMapp > >> er) > >> > >> If 100 mappers are created, about 85 end immediately and only a few > >> process the files (for hours). I suspect an optimization making that > >> there is a minimum number of lines to pass to the next node or it is > >> “shutdown” ; but in my case I do want the lines to be evenly > >> distributed to each mapper. > >> > >> How to enforce that ? > >> > >> > >> > >> Greetings, > >> > >> Arnaud > >> > >> > >> > ---------------------------------------------------------------------- > >> -- > >> > >> L'intégrité de ce message n'étant pas assurée sur internet, la > société > >> expéditrice ne peut être tenue responsable de son contenu ni de ses > >> pièces jointes. Toute utilisation ou diffusion non autorisée est > >> interdite. Si vous n'êtes pas destinataire de ce message, merci de le > >> détruire et d'avertir l'expéditeur. > >> > >> The integrity of this message cannot be guaranteed on the Internet. > >> The company that sent this message cannot therefore be held > liable for > >> its content nor attachments. Any unauthorized use or dissemination is > >> prohibited. If you are not the intended recipient of this message, > >> then please delete it and notify the sender. > > > signature.asc (836 bytes) Download Attachment |
Btw, it is working with a parallelism 1 source, because only a single source partitions (round-robin or random) the data. Several sources do not assign work to the same few mappers.2015-09-03 15:22 GMT+02:00 Matthias J. Sax <[hidden email]>: If it would be only 14 elements, you are obviously right. However, if I |
In reply to this post by Matthias J. Sax-2
In case of rebalance(), all sources start the
round-robin partitioning at index 0. Since each source emits only very
few elements, only the first 15 mappers receive any input. It
would be better to let each source start the round-robin partitioning at
a different index, something like startIdx = (numReceivers /
numSenders) * myIdx.However, the ShufflePartitioner is only initialized once at the client side (if I see that correctly) and then the same instance is deserialized by all operators, i.e., all use random number generators with the same seed. Btw, it is working with a parallelism 1 source, because only a single source partitions (round-robin or random) the data. Several sources do not assign work to the same few mappers.2015-09-03 15:22 GMT+02:00 Matthias J. Sax <[hidden email]>: If it would be only 14 elements, you are obviously right. However, if I |
In reply to this post by Aljoscha Krettek
The purpose of rebalance() should be to rebalance the partitions of a data streams as evenly as possible, right? If all senders start sending data to the same receiver and there is less data in each partition than receivers, partitions are not evenly rebalanced. That is exactly the problem Arnaud ran into. 2015-09-03 15:53 GMT+02:00 Matthias J. Sax <[hidden email]>: For rebalance() this makes sense. I don't think anything must be |
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