How to create correct data frame for classification in Spark ML - scala

Question

I am trying to run random forest classification by using Spark ML api but I am having issues with creating right data frame input into pipeline.
Here is sample data:
age,hours_per_week,education,sex,salaryRange
38,40,"hs-grad","male","A"
28,40,"bachelors","female","A"
52,45,"hs-grad","male","B"
31,50,"masters","female","B"
42,40,"bachelors","male","B"
age and hours_per_week are integers while other features including label salaryRange are categorical (String)
Loading this csv file (lets call it sample.csv) can be done by Spark csv library like this:
val data = sqlContext.csvFile("/home/dusan/sample.csv")
By default all columns are imported as string so we need to change "age" and "hours_per_week" to Int:
val toInt = udf[Int, String]( _.toInt)
val dataFixed = data.withColumn("age", toInt(data("age"))).withColumn("hours_per_week",toInt(data("hours_per_week")))
Just to check how schema looks now:
scala> dataFixed.printSchema
root
|-- age: integer (nullable = true)
|-- hours_per_week: integer (nullable = true)
|-- education: string (nullable = true)
|-- sex: string (nullable = true)
|-- salaryRange: string (nullable = true)
Then lets set the cross validator and pipeline:
val rf = new RandomForestClassifier()
val pipeline = new Pipeline().setStages(Array(rf))
val cv = new CrossValidator().setNumFolds(10).setEstimator(pipeline).setEvaluator(new BinaryClassificationEvaluator)
Error shows up when running this line:
val cmModel = cv.fit(dataFixed)
java.lang.IllegalArgumentException: Field "features" does not exist.
It is possible to set label column and feature column in RandomForestClassifier ,however I have 4 columns as predictors (features) not only one.
How I should organize my data frame so it has label and features columns organized correctly?
For your convenience here is full code :
import org.apache.spark.SparkConf
import org.apache.spark.SparkContext
import org.apache.spark.ml.classification.RandomForestClassifier
import org.apache.spark.ml.evaluation.BinaryClassificationEvaluator
import org.apache.spark.ml.tuning.CrossValidator
import org.apache.spark.ml.Pipeline
import org.apache.spark.sql.DataFrame
import org.apache.spark.sql.functions._
import org.apache.spark.mllib.linalg.{Vector, Vectors}
object SampleClassification {
def main(args: Array[String]): Unit = {
//set spark context
val conf = new SparkConf().setAppName("Simple Application").setMaster("local");
val sc = new SparkContext(conf)
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
import sqlContext.implicits._
import com.databricks.spark.csv._
//load data by using databricks "Spark CSV Library"
val data = sqlContext.csvFile("/home/dusan/sample.csv")
//by default all columns are imported as string so we need to change "age" and "hours_per_week" to Int
val toInt = udf[Int, String]( _.toInt)
val dataFixed = data.withColumn("age", toInt(data("age"))).withColumn("hours_per_week",toInt(data("hours_per_week")))
val rf = new RandomForestClassifier()
val pipeline = new Pipeline().setStages(Array(rf))
val cv = new CrossValidator().setNumFolds(10).setEstimator(pipeline).setEvaluator(new BinaryClassificationEvaluator)
// this fails with error
//java.lang.IllegalArgumentException: Field "features" does not exist.
val cmModel = cv.fit(dataFixed)
}
}
Thanks for help!

Answer 1

As of Spark 1.4, you can use Transformer org.apache.spark.ml.feature.VectorAssembler.
Just provide column names you want to be features.
val assembler = new VectorAssembler()
.setInputCols(Array("col1", "col2", "col3"))
.setOutputCol("features")
and add it to your pipeline.

Answer 2

You simply need to make sure that you have a "features" column in your dataframe that is of type VectorUDF as show below:
scala> val df2 = dataFixed.withColumnRenamed("age", "features")
df2: org.apache.spark.sql.DataFrame = [features: int, hours_per_week: int, education: string, sex: string, salaryRange: string]
scala> val cmModel = cv.fit(df2)
java.lang.IllegalArgumentException: requirement failed: Column features must be of type org.apache.spark.mllib.linalg.VectorUDT#1eef but was actually IntegerType.
at scala.Predef$.require(Predef.scala:233)
at org.apache.spark.ml.util.SchemaUtils$.checkColumnType(SchemaUtils.scala:37)
at org.apache.spark.ml.PredictorParams$class.validateAndTransformSchema(Predictor.scala:50)
at org.apache.spark.ml.Predictor.validateAndTransformSchema(Predictor.scala:71)
at org.apache.spark.ml.Predictor.transformSchema(Predictor.scala:118)
at org.apache.spark.ml.Pipeline$$anonfun$transformSchema$4.apply(Pipeline.scala:164)
at org.apache.spark.ml.Pipeline$$anonfun$transformSchema$4.apply(Pipeline.scala:164)
at scala.collection.IndexedSeqOptimized$class.foldl(IndexedSeqOptimized.scala:51)
at scala.collection.IndexedSeqOptimized$class.foldLeft(IndexedSeqOptimized.scala:60)
at scala.collection.mutable.ArrayOps$ofRef.foldLeft(ArrayOps.scala:108)
at org.apache.spark.ml.Pipeline.transformSchema(Pipeline.scala:164)
at org.apache.spark.ml.tuning.CrossValidator.transformSchema(CrossValidator.scala:142)
at org.apache.spark.ml.PipelineStage.transformSchema(Pipeline.scala:59)
at org.apache.spark.ml.tuning.CrossValidator.fit(CrossValidator.scala:107)
at $iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC.<init>(<console>:67)
at $iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC.<init>(<console>:72)
at $iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC.<init>(<console>:74)
at $iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC.<init>(<console>:76)
EDIT1
Essentially there need to be two fields in your data frame "features" for feature vector and "label" for instance labels. Instance must be of type Double.
To create a "features" fields with Vector type first create a udf as show below:
val toVec4 = udf[Vector, Int, Int, String, String] { (a,b,c,d) =>
val e3 = c match {
case "hs-grad" => 0
case "bachelors" => 1
case "masters" => 2
}
val e4 = d match {case "male" => 0 case "female" => 1}
Vectors.dense(a, b, e3, e4)
}
Now to also encode the "label" field, create another udf as shown below:
val encodeLabel = udf[Double, String]( _ match { case "A" => 0.0 case "B" => 1.0} )
Now we transform original dataframe using these two udf:
val df = dataFixed.withColumn(
"features",
toVec4(
dataFixed("age"),
dataFixed("hours_per_week"),
dataFixed("education"),
dataFixed("sex")
)
).withColumn("label", encodeLabel(dataFixed("salaryRange"))).select("features", "label")
Note that there can be extra columns / fields present in the dataframe, but in this case I have selected only features and label:
scala> df.show()
+-------------------+-----+
| features|label|
+-------------------+-----+
|[38.0,40.0,0.0,0.0]| 0.0|
|[28.0,40.0,1.0,1.0]| 0.0|
|[52.0,45.0,0.0,0.0]| 1.0|
|[31.0,50.0,2.0,1.0]| 1.0|
|[42.0,40.0,1.0,0.0]| 1.0|
+-------------------+-----+
Now its upto you to set correct parameters for your learning algorithm to make it work.

Answer 3

According to spark documentation on mllib - random trees, seems to me that you should define the features map that you are using and the points should be a labeledpoint.
This will tell the algorithm which column should be used as prediction and which ones are the features.
https://spark.apache.org/docs/latest/mllib-decision-tree.html