I have a saved h2o model in mojo format, and now I am trying to load it and use it to make predictions on a new dataset (df
) as part of a spark app written in scala. Ideally, I wish to append a new row to the existing DataFrame containing the class probability based on this model.
I can see how to apply a mojo to an individual row already in a RowData format (as per answer here), but I am not sure how to map over an existing DataFrame so that it is in the right format to make predictions using the mojo model. I have worked with DataFrames a fair bit, but never with the underlying RDDs.
Also, should this model be serialised / broadcast so that predictions can be done in parallel on a cluster, or will it be available to all executors as part of the map?
I have gotten this far:
# load mojo model and create easy predict model wrapper
val mojo = MojoModel.load("loca/path/to/mojo/mojo.zip")
val easyModel = new EasyPredictModelWrapper(mojo)
# map over spark DataFrame, converty to rdd, and make predictions on each row:
df.rdd.map { row =>
val prediction = easyModel.predictBinomial(row).classProbabilities
println(prediction)
}
But my row
variable is not in the right format for this to work. Any suggestions on what to try next?
EDIT: my DataFrame consists of 70 predictive feature columns which are a mixture of integers and category/factor columns. A very simple sample DataFrame:
val df = Seq(
(0, 3, "cat1"),
(1, 2, "cat2"),
(2, 6, "cat1")
).toDF("id", "age", "category")
Use this function to prepare RowData object needed for H2O:
def rowToRowData(df: DataFrame, row: Row): RowData = {
val rowAsMap = row.getValuesMap[Any](df.schema.fieldNames)
val rowData = rowAsMap.foldLeft(new RowData()) { case (rd, (k,v)) =>
if (v != null) { rd.put(k, v.toString) }
rd
}
rowData
}
I have a complete answer here: https://stackoverflow.com/a/47898040/9120484 You can call map on df directly instead of on rdd.
来源:https://stackoverflow.com/questions/47828510/how-to-map-over-dataframe-in-spark-to-extract-rowdata-and-make-predictions-using