Fit a dataframe into randomForest pyspark

Question

I have a DataFrame that looks like this:

+--------------------+------------------+
|            features|           labels |
+------------------         


        

Answer 1

If you are using Spark 2.x, I believe that this is what you need :

from pyspark.sql.functions import udf
from pyspark.mllib.linalg import Vectors
from pyspark.ml.linalg import VectorUDT
from pyspark.ml.feature import StringIndexer

df = spark.createDataFrame([("[-0.38475, 0.568]", "label1"), ("[0.645734, 0.699]", "label2")], ("features", "label"))

def parse(s):
  try:
    return Vectors.parse(s).asML()
  except:
    return None

parse_ = udf(parse, VectorUDT())

parsed = df.withColumn("features", parse_("features"))

indexer = StringIndexer(inputCol="label", outputCol="label_indexed")

indexer.fit(parsed).transform(parsed).show()
## +----------------+------+-------------+
## |        features| label|label_indexed|
## +----------------+------+-------------+
## |[-0.38475,0.568]|label1|          0.0|
## |[0.645734,0.699]|label2|          1.0|
## +----------------+------+-------------+

With Spark 1.6, it isn't much different :

from pyspark.sql.functions import udf
from pyspark.ml.feature import StringIndexer
from pyspark.mllib.linalg import Vectors, VectorUDT

df = sqlContext.createDataFrame([("[-0.38475, 0.568]", "label1"), ("[0.645734, 0.699]", "label2")], ("features", "label"))

parse_ = udf(Vectors.parse, VectorUDT())

parsed = df.withColumn("features", parse_("features"))

indexer = StringIndexer(inputCol="label", outputCol="label_indexed")

indexer.fit(parsed).transform(parsed).show()
## +----------------+------+-------------+
## |        features| label|label_indexed|
## +----------------+------+-------------+
## |[-0.38475,0.568]|label1|          0.0|
## |[0.645734,0.699]|label2|          1.0|
## +----------------+------+-------------+

Vectors has a parse function that can help you achieve what you are trying to do.