Spark Convert Data Frame Column to dense Vector for StandardScaler() "Column must be of type org.apache.spark.ml.linalg.VectorUDT"

Question

I am very new to Spark and I am trying to apply StandardScaler() to a column in a DataFrame.

+---------------+
|      DF_column|
+---------------+
| 0.114285714286|
| 0.115702479339|
| 0.267893660532|
|0.0730337078652|
| 0.124309392265|
| 0.365714285714|
| 0.111747851003|
| 0.279538904899|
| 0.134670487106|
| 0.523287671233|
| 0.404011461318|
|          0.375|
| 0.125517241379|
|0.0143266475645|
| 0.313684210526|
| 0.381088825215|
| 0.411428571429|
| 0.327683615819|
| 0.153409090909|
| 0.344827586207|
+---------------+

The problem is that applying it like this, gives me an error:

requirement failed: Input column DF_column must be a vector column.

I tried using UDF but still doesn't work.

scaler = StandardScaler(inputCol='DF_column', 
    outputCol="scaledFeatures",withStd=True, withMean=False)

I did the example of the LIBSVM but that is easy a the TXT file is loading features as Vectors.

Answer 1

If you have a column of scalars then StandardScaler is a serious overkill. You can scale directly:

from pyspark.sql.functions import col, stddev_samp

df.withColumn("scaled",
  col("DF_column") / df.agg(stddev_samp("DF_column")).first()[0])

but if you really want to use scaler than assemble a vector first:

from pyspark.ml.feature import VectorAssembler
from pyspark.ml.feature import StandardScaler

assembler = VectorAssembler(
  inputCols=["DF_column"], outputCol="features"
)

assembled = assembler.transform(df)

scaler = StandardScaler(
  inputCol="features", outputCol="scaledFeatures",
  withStd=True, withMean=False
).fit(assembled)

scaler.transform(assembled)