how to select all columns that starts with a common label

Question

I have a dataframe in Spark 1.6 and want to select just some columns out of it. The column names are like:

colA, colB, colC, colD, colE, colF-0, colF-1, colF-2
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Answer 1

First grab the column names with df.columns, then filter down to just the column names you want .filter(_.startsWith("colF")). This gives you an array of Strings. But the select takes select(String, String*). Luckily select for columns is select(Column*), so finally convert the Strings into Columns with .map(df(_)), and finally turn the Array of Columns into a var arg with : _*.

df.select(df.columns.filter(_.startsWith("colF")).map(df(_)) : _*).show

This filter could be made more complex (same as Pandas). It is however a rather ugly solution (IMO):

df.select(df.columns.filter(x => (x.equals("colA") || x.startsWith("colF"))).map(df(_)) : _*).show 

If the list of other columns is fixed you could also merge a fixed array of columns names with filtered array.

df.select((Array("colA", "colB") ++ df.columns.filter(_.startsWith("colF"))).map(df(_)) : _*).show

Answer 2

I wrote a function that does that. Read the comments to see how it works.

  /**
    * Given a sequence of prefixes, select suitable columns from [[DataFrame]]
    * @param columnPrefixes Sequence of prefixes
    * @param dF Incoming [[DataFrame]]
    * @return [[DataFrame]] with prefixed columns selected
    */
  def selectPrefixedColumns(columnPrefixes: Seq[String], dF: DataFrame): DataFrame = {
    // Find out if given column name matches any of the provided prefixes
    def colNameStartsWith: String => Boolean = (colName: String) =>
        columnsPrefix.map(prefix => colName.startsWith(prefix)).reduce(_ || _)
    // Filter columns list by checking against given prefixes sequence
    val columns = dF.columns.filter(colNameStartsWith)
    // Select filtered columns list
    dF.select(columns.head, columns.tail:_*)
  }