Since the OneHotEncoder
/OneHotEncoderEstimator
does not accept empty string for name, or you'll get the following error :
java.lang.IllegalArgumentException: requirement failed: Cannot have an empty string for name.
at scala.Predef$.require(Predef.scala:233)
at org.apache.spark.ml.attribute.Attribute$$anonfun$5.apply(attributes.scala:33)
at org.apache.spark.ml.attribute.Attribute$$anonfun$5.apply(attributes.scala:32)
[...]
This is how I will do it : (There is other way to do it, rf. @Anthony 's answer)
I'll create an UDF
to process the empty category :
import org.apache.spark.sql.functions._
def processMissingCategory = udf[String, String] { s => if (s == "") "NA" else s }
Then, I'll apply the UDF on the column :
val df = sqlContext.createDataFrame(Seq(
(0, "a"),
(1, "b"),
(2, "c"),
(3, ""), //<- original example has "a" here
(4, "a"),
(5, "c")
)).toDF("id", "category")
.withColumn("category",processMissingCategory('category))
df.show
// +---+--------+
// | id|category|
// +---+--------+
// | 0| a|
// | 1| b|
// | 2| c|
// | 3| NA|
// | 4| a|
// | 5| c|
// +---+--------+
Now, you can go back to your transformations
val indexer = new StringIndexer().setInputCol("category").setOutputCol("categoryIndex").fit(df)
val indexed = indexer.transform(df)
indexed.show
// +---+--------+-------------+
// | id|category|categoryIndex|
// +---+--------+-------------+
// | 0| a| 0.0|
// | 1| b| 2.0|
// | 2| c| 1.0|
// | 3| NA| 3.0|
// | 4| a| 0.0|
// | 5| c| 1.0|
// +---+--------+-------------+
// Spark <2.3
// val encoder = new OneHotEncoder().setInputCol("categoryIndex").setOutputCol("categoryVec")
// Spark +2.3
val encoder = new OneHotEncoderEstimator().setInputCols(Array("categoryIndex")).setOutputCols(Array("category2Vec"))
val encoded = encoder.transform(indexed)
encoded.show
// +---+--------+-------------+-------------+
// | id|category|categoryIndex| categoryVec|
// +---+--------+-------------+-------------+
// | 0| a| 0.0|(3,[0],[1.0])|
// | 1| b| 2.0|(3,[2],[1.0])|
// | 2| c| 1.0|(3,[1],[1.0])|
// | 3| NA| 3.0| (3,[],[])|
// | 4| a| 0.0|(3,[0],[1.0])|
// | 5| c| 1.0|(3,[1],[1.0])|
// +---+--------+-------------+-------------+
EDIT:
@Anthony 's solution in Scala :
df.na.replace("category", Map( "" -> "NA")).show
// +---+--------+
// | id|category|
// +---+--------+
// | 0| a|
// | 1| b|
// | 2| c|
// | 3| NA|
// | 4| a|
// | 5| c|
// +---+--------+
I hope this helps!
.na
really needed here? Wouldn't justdf.replace('', 'EMPTY', 'category')
work? – Janinajanine