问题
I am using Spark Scala to calculate cosine similarity between the Dataframe rows.
Dataframe schema is below:
root
|-- itemId: string (nullable = true)
|-- features: vector (nullable = true)
Sample of the dataframe below
+-------+--------------------+
| itemId| features|
+-------+--------------------+
| ab |[4.7143,0.0,5.785...|
| cd |[5.5,0.0,6.4286,4...|
| ef |[4.7143,1.4286,6....|
........
+-------+--------------------+
Code to compute the cosine similarities:
val irm = new IndexedRowMatrix(myDataframe.rdd.zipWithIndex().map {
case (row, index) => IndexedRow(row.getAs[Vector]("features"), index)
}).toCoordinateMatrix.transpose.toRowMatrix.columnSimilarities
In the irm matrix, I have (i, j, score) where i, j are the indexes of item i, and j of my original dataframe. What I would like is to get (itemIdA, itemIdB, score) where itemIdA and itemIdB are the ids of index i and j respectively, by joining this irm with the initial dataframe or if there is any better option?
回答1:
Create a row index before converting the dataframe to a matrix and create a mapping between the index and the id. After the computation, use the created Map
to convert the column index (previously row index but changed with the transpose
) to the id.
val rdd = myDataframe.as[(String, org.apache.spark.mllib.linalg.Vector)].rdd.zipWithIndex()
val indexMap = rdd.map{case ((id, vec), index) => (index, id)}.collectAsMap()
Calculate the cosine similarities as before using the :
val irm = new IndexedRowMatrix(rdd.map{case ((id, vec), index) => IndexedRow(index, vec)})
.toCoordinateMatrix().transpose().toRowMatrix().columnSimilarities()
Convert column indices back to the ids:
irm.entries.map(e => (indexMap(e.i), indexMap(e.j), e.value))
This should give you what you are looking for.
来源:https://stackoverflow.com/questions/51163248/how-to-get-item-id-from-cosine-similarity-matrix