Or you can use the factory method `Vectors.sparse`: val sv = Vectors.sparse(numProducts, productIds.map(x => (x, 1.0)))
where numProducts should be the largest product id plus one. Best, Xiangrui On Mon, Sep 15, 2014 at 12:46 PM, Chris Gore <cdg...@cdgore.com> wrote: > Hi Sameer, > > MLLib uses Breeze’s vector format under the hood. You can use that. > http://www.scalanlp.org/api/breeze/index.html#breeze.linalg.SparseVector > > For example: > > import breeze.linalg.{DenseVector => BDV, SparseVector => BSV, Vector => BV} > > val numClasses = classes.distinct.count.toInt > > val userWithClassesAsSparseVector = rows.map(x => (x.userID, new > BSV[Double](x.classIDs.sortWith(_ < _), > Seq.fill(x.classIDs.length)(1.0).toArray, > numClasses).asInstanceOf[BV[Double]])) > > Chris > > On Sep 15, 2014, at 11:28 AM, Sameer Tilak <ssti...@live.com> wrote: > > Hi All, > I have transformed the data into following format: First column is user id, > and then all the other columns are class ids. For a user only class ids that > appear in this row have value 1 and others are 0. I need to crease a sparse > vector from this. Does the API for creating a sparse vector that can > directly support this format? > > User id Product class ids > > 2622572 145447 1620 13421 28565 285556 293 4553 67261 130 3646 1671 18806 > 183576 3286 51715 57671 57476 > > --------------------------------------------------------------------- To unsubscribe, e-mail: user-unsubscr...@spark.apache.org For additional commands, e-mail: user-h...@spark.apache.org
