Unpacking a list to select multiple columns from a spark data frame

Question

I have a spark data frame df. Is there a way of sub selecting a few columns using a list of these columns?

scala> df.columns
res0: Array[String] = Array("a", "b", "c", "d")

I know I can do something like df.select("b", "c"). But suppose I have a list containing a few column names val cols = List("b", "c"), is there a way to pass this to df.select? df.select(cols) throws an error. Something like df.select(*cols) as in python

Answer 1

Use df.select(cols.head, cols.tail: _*)

Let me know if it works :)

Explanation from @Ben:

The key is the method signature of select:

select(col: String, cols: String*)

The cols:String* entry takes a variable number of arguments. :_* unpacks arguments so that they can be handled by this argument. Very similar to unpacking in python with *args. See here and here for other examples.

Answer 2

You can typecast String to spark column like this:

import org.apache.spark.sql.functions._
df.select(cols.map(col): _*)

Answer 3

Another option that I've just learnt.

import org.apache.spark.sql.functions.col
val columns = Seq[String]("col1", "col2", "col3")
val colNames = columns.map(name => col(name))
val df = df.select(colNames:_*)

Answer 4

First convert the String Array to a List of Spark dataset Column type as below

String[] strColNameArray = new String[]{"a", "b", "c", "d"};

List<Column> colNames = new ArrayList<>();

for(String strColName : strColNameArray){
    colNames.add(new Column(strColName));
}

then convert the List using JavaConversions functions within the select statement as below. You need the following import statement.

import scala.collection.JavaConversions;

Dataset<Row> selectedDF = df.select(JavaConversions.asScalaBuffer(colNames ));

Answer 5

You can pass arguments of type Column* to select:

val df = spark.read.json("example.json")
val cols: List[String] = List("a", "b")
//convert string to Column
val col: List[Column] = cols.map(df(_))
df.select(col:_*)

Answer 6

you can do like this

String[] originCols = ds.columns();
ds.selectExpr(originCols)

spark selectExp source code

     /**
   * Selects a set of SQL expressions. This is a variant of `select` that accepts
   * SQL expressions.
   *
   * {{{
   *   // The following are equivalent:
   *   ds.selectExpr("colA", "colB as newName", "abs(colC)")
   *   ds.select(expr("colA"), expr("colB as newName"), expr("abs(colC)"))
   * }}}
   *
   * @group untypedrel
   * @since 2.0.0
   */
  @scala.annotation.varargs
  def selectExpr(exprs: String*): DataFrame = {
    select(exprs.map { expr =>
      Column(sparkSession.sessionState.sqlParser.parseExpression(expr))
    }: _*)
  }

Answer 7

Yes , You can make use of .select in scala.

Use .head and .tail to select the whole values mentioned in the List()

Example

val cols = List("b", "c")
df.select(cols.head,cols.tail: _*)

Explanation

Answer 8

Prepare a list where all the requirement features are listed then use spark inbuilt function using *, reference given below.

lst = ["col1", "col2", "col3"]
result = df.select(*lst)

Some time we get an error of:" Analysis Exception: cannot resolve 'col1' given input columns" try to convert features to string type as mentioned below:

from pyspark.sql.functions import lit
from pyspark.sql.types import StringType
for i in lst:
   if i not in df.columns:
      df = df.withColumn(i, lit(None).cast(StringType()))

And finally you will get the dataset with required features.