Yes this would be helpful, otherwise the Q part of the decomposition is useless. One can use that to solve the system by transposing it and multiplying with b and solving for x (Ax = b) where A = R and b = Qt*b since the Upper triangular matrix is correctly available (R)
On Fri, Nov 11, 2016 at 3:56 AM Sean Owen <so...@cloudera.com> wrote: > @Xiangrui / @Joseph, do you think it would be reasonable to have > CoordinateMatrix sort the rows it creates to make an IndexedRowMatrix? in > order to make the ultimate output of toRowMatrix less surprising when it's > not ordered? > > > On Tue, Nov 8, 2016 at 3:29 PM Sean Owen <so...@cloudera.com> wrote: > > I think the problem here is that IndexedRowMatrix.toRowMatrix does *not* > result in a RowMatrix with rows in order of their indices, necessarily: > > > // Drop its row indices. > RowMatrix rowMat = indexedRowMatrix.toRowMatrix(); > > What you get is a matrix where the rows are arranged in whatever order > they were passed to IndexedRowMatrix. RowMatrix says it's for rows where > the ordering doesn't matter, but then it's maybe surprising it has a QR > decomposition method, because clearly the result depends on the order of > rows in the input. (CC Yuhao Yang for a comment?) > > You could say, well, why doesn't IndexedRowMatrix.toRowMatrix return at > least something with sorted rows? that would not be hard. It also won't > return "missing" rows (all zeroes), so it would not in any event result in > a RowMatrix whose implicit rows and ordering represented the same matrix. > That, at least, strikes me as something to be better documented. > > Maybe it would be nicer still to at least sort the rows, given the > existence of use cases like yours. For example, at least > CoordinateMatrix.toIndexedRowMatrix could sort? that is less surprising. > > In any event you should be able to make it work by manually getting the > RDD[IndexedRow] out of IndexedRowMatrix, sorting by index, then mapping it > to Vectors and making a RowMatrix from it. > > > > On Tue, Nov 8, 2016 at 2:41 PM Iman Mohtashemi <iman.mohtash...@gmail.com> > wrote: > > Hi Sean, > Here you go: > > sparsematrix.txt = > > row, col ,val > 0,0,.42 > 0,1,.28 > 0,2,.89 > 1,0,.83 > 1,1,.34 > 1,2,.42 > 2,0,.23 > 3,0,.42 > 3,1,.98 > 3,2,.88 > 4,0,.23 > 4,1,.36 > 4,2,.97 > > The vector is just the third column of the matrix which should give the > trivial solution of [0,0,1] > > This translates to this which is correct > There are zeros in the matrix (Not really sparse but just an example) > 0.42 0.28 0.89 > 0.83 0.34 0.42 > 0.23 0.0 0.0 > 0.42 0.98 0.88 > 0.23 0.36 0.97 > > > Here is what I get for the Q and R > > Q: -0.21470961288429483 0.23590615093828807 0.6784910613691661 > -0.3920784235278427 -0.06171221388256143 0.5847874866876442 > -0.7748216464954987 -0.4003560542230838 -0.29392323671555354 > -0.3920784235278427 0.8517909521421976 -0.31435038559403217 > -0.21470961288429483 -0.23389547730301666 -0.11165321782745863 > R: -1.0712142642814275 -0.8347536340918976 -1.227672225670157 > 0.0 0.7662808691141717 0.7553315911660984 > 0.0 0.0 0.7785210939368136 > > When running this in matlab the numbers are the same but row 1 is the last > row and the last row is interchanged with row 3 > > > > On Mon, Nov 7, 2016 at 11:35 PM Sean Owen <so...@cloudera.com> wrote: > > Rather than post a large section of code, please post a small example of > the input matrix and its decomposition, to illustrate what you're saying is > out of order. > > On Tue, Nov 8, 2016 at 3:50 AM im281 <iman.mohtash...@gmail.com> wrote: > > I am getting the correct rows but they are out of order. Is this a bug or > am > I doing something wrong? > > >
