Hi Till and Fabian,
I had to come back to this problem because we're putting out a maintenance release soon. I
think I overcomplicated the issue here. I don't need equal partitions. All that I need is
to ensure that we have continuous partitions based in the 0th partion. Ie. if there if there
are 4 partitions and 2 are empty, partition 0 will have data and partition 1 will have data.
@till, I see what you did in ALS, with a Custom partitioner, Is there a way that I can write
a custom partitioner to make sure that we have data in the 0th and 1st partition? I don't
see much documentation for custom partitioners.
Thanks.
Andy
________________________________________
From: Till Rohrmann <trohrmann@apache.org>
Sent: Tuesday, April 26, 2016 9:39:41 AM
To: dev@flink.apache.org
Subject: Re: Partition problem
If you don’t know the size of your matrix, then you cannot partition it
into continuous chunks of rows. The problem is that partitionByRange
samples the data set to generate a distribution and, thus, two matrices
will rarely be partitioned identically. Also if you want to provide a data
distribution, then you have to know how many rows you have. Thus, you would
first have to count the number of rows and broadcast this information to
the subsequent partitioning step (don’t use collect for that).
Alternatively, if it’s ok that the grouping just has to be consistent and
not continuous with respect to the row index, then you could use the row
index modulo the number of blocks, you want to create, to calculate a
partition id. Using a custom partitioner which partitions the data
according to this partition ID and then applying you flat map operation
should do the trick. You could take a look at the ALS (line 458)
implementation, where I did something similar.
Cheers,
Till
On Mon, Apr 25, 2016 at 8:18 PM, Andrew Palumbo <ap.dev@outlook.com> wrote:
> sorry  just noticed below should read:
>
> val rowsA = (0 until inCoreA.nrow).map(i => (i, inCoreA(i, ::)))
> drmA = env.fromCollection(rowsA).partitionByRange(0)
>
> val rowsB = (0 until inCoreB.nrow).map(i => (i, inCoreB(i, ::)))
> drmB = env.fromCollection(rowsB).partitionByRange(0)
>
>
> also:
>
> The Blockified representation is a `DataSet[(Array(K), Matrix)]`.
>
> Thanks
>
> ________________________________________
> From: Andrew Palumbo <ap.dev@outlook.com>
> Sent: Monday, April 25, 2016 1:58 PM
> To: dev@flink.apache.org
> Subject: Re: Partition problem
>
> Thank you Fabian and Till for answering,
>
> I think that my explanation of the problem was a bit over simplified (I
> am trying to implement an operator that will pass our tests, and didn't
> want to throw too much code at you). I realize that this is an odd case, a
> 2x2 matrix in a distributed context, but it is for a specific Unit test
> that we enforce.
>
> So we have two different Distributed Matrix Representations: RowWise and
> Blockified. The RowWise representation is a `DataSet[(K, Vector)]` where
> K is e.g., an Int Key and Vector is a row of the Matrix. The Blockified
> representation is a `DataSet[Array(K), Matrix]`. In the Gist that I
> posted, I was working with a Blockified Distributed dataset. Since it was
> a 2x2 matrix that was Blockified into 4 partitions, the nonempty
> partitions actually contain a 1x2 Matrix (rather than a (Vector) "row" as i
> think It reads I will update that to be more clear.
>
> @Fabian In this case, I am using ExecutionEnvironment.FromCollection to
> create the original RowWise Matrix DataSet. (There are other cases in
> which we read from HDFS directly). But for this problem I am doing
> something like:
>
> val inCoreA = dense((1, 2), (3, 4))
> val inCoreB = dense((3, 5), (4, 6))
>
> val rowsA = (0 until m.nrow).map(i => (i, inCoreA(i, ::)))
> drmA = env.fromCollection(rows).partitionByRange(0)
>
> val rowsB = (0 until m.nrow).map(i => (i, inCoreB(i, ::)))
> drmA = env.fromCollection(rows).partitionByRange(0)
>
> >If you need the split ID in your program, you can implement an
> InputFormat,
> >which wraps another IF and assigns the ID of the current InputSplit to the
> >read data, i.e., converts the DataType from T to Tuple2[Int, T].
>
> I'm not sure if the partitioning at this point matters (of the rowwise
> Matrices)? (In next map these into Blockified Matrices)
>
> @Till I think that you're right in that my assumption of Identical
> partitioning is a problem.
>
> The above Matrices are then mapped into Blockified Matrices currently
> using the method something as follows:
>
> val blocksA = drmA.mapPartition {
> values =>
> val (keys, vectors) = values.toIterable.unzip
>
> if (vectors.nonEmpty) {
> val vector = vectors.head
> val matrix: Matrix = if (vector.isDense) {
> val matrix = new DenseMatrix(vectors.size, ncolLocal)
> vectors.zipWithIndex.foreach { case (vec, idx) => matrix(idx,
> ::) := vec }
> matrix
> } else {
> new SparseRowMatrix(vectors.size, ncolLocal, vectors.toArray)
> }
> Seq((keys.toArray(classTag), matrix))
> } else {
> Seq()
> }
> }
>
> And the same for Matrix B.
>
> Which is where the partition index assignment begins in the gist:
> https://gist.github.com/andrewpalumbo/1768dac6d2f5fdf963abacabd859aaf3
>
>
> > I think the problem is that you assume that both matrices have the same
> > partitioning. If you guarantee that this is the case, then you can use
> the
> > subtask index as the block index.
>
> Yes I think this is the problem I'd assumed that when mapping into
> partitions, the 0 partiton would be used first and then the 1 partition and
> so on... I understand what your saying now though re: lazy assignment via
> task Id. So essentially the partition that the data ends up in is
> arbitrary based on the task ID that happens to be assigning it.
>
> But in the general case this is not true,
> > and then you have to calculate the blocks by first assigning a block
> index
> > (e.g. rows with 09 index are assigned to block 0, rows with 1019
> assigned
> > to block 1, etc.) and then create the blocks by reducing on this block
> > index. That's because the distribution of the individual rows in the
> > cluster is not necessarily the same between two matrices.
>
> I suppose this would have to be done when Blockifying in the method
> above. The rowwise matrix may be 2x2 or 20000000 x 2 and directly read
> from HDFS. I'm not sure how to assign divide the data and partition it
> myself when mapping a rowwise matrix into blocks. Eg. can I know the size
> of the DataSet before the computation is triggered by env.execute()? If I
> guess what you are saying is to hand partition the data in the above
> `.asBlockified()` method.
>
>
> As well is it not still possible that i may end up with the same problem
> when the # of matrix blocks is < the degree of parallelism?
>
>
>
> In the end what I really need to do is be able to join the two Bockified
> DataSets (of any size) in the correct order.. so maybe there is an other
> way to do this?
>
>
> Thanks again for your time.
>
> Andy
> ________________________________________
> From: Fabian Hueske <fhueske@gmail.com>
> Sent: Monday, April 25, 2016 6:09 AM
> To: dev@flink.apache.org
> Subject: Re: Partition problem
>
> Hi Andrew,
>
> I might be wrong, but I think this problem is caused by an assumption of
> how Flink reads input data.
> In Flink, each InputSplit is not read by a new task and a split does not
> correspond to a partition. This is different from how Hadoop MR and Spark
> handle InputSplits.
>
> Instead, Flink creates as many DataSource tasks as specified by the task
> parallelism and lazily assigns InputSplits to its subtasks. Idle DataSource
> subtasks request InputSplits from the JobManager and the assignment happens
> firstcomefirstserve.
> Hence, the subtask ID (or partition ID) of an InputSplit is not
> deterministic and a DataSource might read more than one or also no split at
> all (such as in your case).
>
> If you need the split ID in your program, you can implement an InputFormat,
> which wraps another IF and assigns the ID of the current InputSplit to the
> read data, i.e., converts the DataType from T to Tuple2[Int, T].
>
> Hope this helps,
> Fabian
>
>
> 20160425 11:27 GMT+02:00 Till Rohrmann <trohrmann@apache.org>:
>
> > Hi Andrew,
> >
> > I think the problem is that you assume that both matrices have the same
> > partitioning. If you guarantee that this is the case, then you can use
> the
> > subtask index as the block index. But in the general case this is not
> true,
> > and then you have to calculate the blocks by first assigning a block
> index
> > (e.g. rows with 09 index are assigned to block 0, rows with 1019
> assigned
> > to block 1, etc.) and then create the blocks by reducing on this block
> > index. That's because the distribution of the individual rows in the
> > cluster is not necessarily the same between two matrices.
> >
> > Cheers,
> > Till
> >
> > On Mon, Apr 25, 2016 at 1:40 AM, Andrew Palumbo <ap.dev@outlook.com>
> > wrote:
> >
> > > Hi All,
> > >
> > >
> > > I've run into a problem with empty partitions when the number of
> elements
> > > in a DataSet is less than the Degree of Parallelism. I've created a
> gist
> > > here to describe it:
> > >
> > >
> > > https://gist.github.com/andrewpalumbo/1768dac6d2f5fdf963abacabd859aaf3
> > >
> > >
> > > I have two 2x2 matrices, Matrix A and Matrix B and an execution
> > > environment where the degree of parallelism is 4. Both matrices are
> > > blockified in 2 different DataSet s . In this case (the case of a 2x2
> > > matrices with 4 partitions) this means that each row goes into a
> > partition
> > > leaving 2 empty partitions. In Matrix A, the rows go into partitions 0,
> > 1.
> > > However the rows of Matrix B end up in partitions 1, 2. I assign the
> > > ordinal index of the blockified matrix's partition to its block, and
> then
> > > join on that index.
> > >
> > >
> > > However in this case, with differently partitioned matrices of the same
> > > geometry, the intersection of the blockified matrices' indices is 1,
> and
> > > partitions 0 and 2 are dropped.
> > >
> > >
> > > I've tried explicitly defining the dop for Matrix B using the count of
> > > nonempty partitions in Matrix A, however this changes the order of the
> > > DataSet, placing partition 2 into partition 0.
> > >
> > >
> > > Is there a way to make sure that these datasets are partitioned in the
> > > same way?
> > >
> > >
> > > Thank you,
> > >
> > >
> > > Andy
> > >
> > >
> > >
> >
>
