CODEWITHSUNDEEP
pythonapache-sparkapache-spark-sqlapache-spark-ml
Tushar Kanade
Tushar Kanade

Reputation: 67

Create labledpoints from Spark Dataframe & how to pass list of names to VectorAssembler

I have a further questions from here https://stackoverflow.com/a/32557330/5235052 I am trying to build labledPoints from a dataframe, where I have the features and label in columns. The features are all boolean with 1/0.

Here is a sample row from the dataframe: 

|             0|       0|        0|            0|       0|            0|     1|        0|     0|           0|       0|       0|       0|           0|        0|         0|      0|            0|       0|           0|          0|         0|         0|              0|        0|        0|        0|         0|          0|    1|    0|    1|    0|    0|       0|           0|    0|     0|     0|     0|         0|         1|

#Using the code from above answer, 
#create a list of feature names from the column names of the dataframe
df_columns = []
for  c in df.columns:
    if c == 'is_item_return': continue
    df_columns.append(c)

#using VectorAssembler for transformation, am using only first 4 columns names
assembler = VectorAssembler()
assembler.setInputCols(df_columns[0:5])
assembler.setOutputCol('features')

transformed = assembler.transform(df)

   #mapping also from above link
   from pyspark.mllib.regression import LabeledPoint
   from pyspark.sql.functions import col

new_df = transformed.select(col('is_item_return'), col("features")).map(lambda row: LabeledPoint(row.is_item_return, row.features))

When I inspect the contents of the RDD, I get the right label, but the feature vector is wrong. 

(0.0,(5,[],[]))

Could someone help me understanding, how to pass the column names of an existing dataframe as feature names to the VectorAssembler?

Upvotes: 1

Views: 2538

Answers (1)

zero323
zero323

Reputation: 330153

There is nothing wrong here. What you get is a string representation of the SparseVector which exactly reflects your input:

  • you take first five columns (assembler.setInputCols(df_columns[0:5])) and the output vector is of length 5
  • since first columns of example input don't contain non-zero entries indices and values arrays are empty

To illustrate this lets use Scala which provides useful toSparse / toDense methods:

import org.apache.spark.mllib.linalg.Vectors

val v = Vectors.dense(Array(0.0, 0.0, 0.0, 0.0, 0.0))
v.toSparse.toString
// String = (5,[],[])

v.toSparse.toDense.toString
// String = [0.0,0.0,0.0,0.0,0.0]

So with PySpark:

from pyspark.ml.feature import VectorAssembler

df = sc.parallelize([
    tuple([0.0] * 5),
    tuple([1.0] * 5), 
    (1.0, 0.0, 1.0, 0.0, 1.0),
    (0.0, 1.0, 0.0, 1.0, 0.0)
]).toDF()

features = (VectorAssembler(inputCols=df.columns, outputCol="features")
    .transform(df)
    .select("features"))

features.show(4, False)

## +---------------------+
## |features             |
## +---------------------+
## |(5,[],[])            |
## |[1.0,1.0,1.0,1.0,1.0]|
## |[1.0,0.0,1.0,0.0,1.0]|
## |(5,[1,3],[1.0,1.0])  |
## +---------------------+

It also show that assembler is choosing different representation depending on number of non-zero entries.

features.flatMap(lambda x: x).map(type).collect()

## [pyspark.mllib.linalg.SparseVector,
##  pyspark.mllib.linalg.DenseVector,
##  pyspark.mllib.linalg.DenseVector,
##  pyspark.mllib.linalg.SparseVector]

Upvotes: 4

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