Hm, the root is a leaf? it's possible but that means there are no splits. If it's a toy example, could be. This was just off the top of my head looking at the code, so could be missing something, but a non-trivial tree should start with an internalnode.
On Thu, Jun 11, 2020 at 11:01 PM AaronLee <y...@wish.com.invalid> wrote: > Thanks srowen. I also checked > https://www.programcreek.com/scala/org.apache.spark.ml.tree.InternalNode. > Splits are available via "InternalNode" ".split" attribute. But > "dtm.rootNode" belongs to "LeafNode". > > ``` > scala> dtm.rootNode > res9: org.apache.spark.ml.tree.Node = LeafNode(prediction = 0.0, impurity = > 0.3153051824490453) > > scala> dftm.rootNode. > impurity prediction > > scala> dftm.rootNode.getClass.getSimpleName > res13: String = LeafNode > > scala> import org.apache.spark.ml.tree.{InternalNode, LeafNode, Node} > import org.apache.spark.ml.tree.{InternalNode, LeafNode, Node} > > scala> val intnode = dftm.rootNode.asInstanceOf[InternalNode] > java.lang.ClassCastException: org.apache.spark.ml.tree.LeafNode cannot be > cast to org.apache.spark.ml.tree.InternalNode > ... 51 elided > > ``` > > > > -- > Sent from: http://apache-spark-user-list.1001560.n3.nabble.com/ > > --------------------------------------------------------------------- > To unsubscribe e-mail: user-unsubscr...@spark.apache.org > >
