PySpark - rename more than one column using withColumnRenamed

I want to change names of two columns using spark withColumnRenamed function. Of course, I can write:

data = sqlContext.createDataFrame([(1,2), (3,4)], ['x1', 'x2'])
data = (data
       .withColumnRenamed('x1','x3')
       .withColumnRenamed('x2', 'x4'))

but I want to do this in one step (having list/tuple of new names). Unfortunately, neither this:

data = data.withColumnRenamed(['x1', 'x2'], ['x3', 'x4'])

nor this:

data = data.withColumnRenamed(('x1', 'x2'), ('x3', 'x4'))

is working. Is it possible to do this that way?

Solution 1:

It is not possible to use a single withColumnRenamed call.

• You can use DataFrame.toDF method*

  data.toDF('x3', 'x4')
  

  or

  new_names = ['x3', 'x4']
  data.toDF(*new_names)
  

• It is also possible to rename with simple select:

  from pyspark.sql.functions import col
  
  mapping = dict(zip(['x1', 'x2'], ['x3', 'x4']))
  data.select([col(c).alias(mapping.get(c, c)) for c in data.columns])
  

Similarly in Scala you can:

• Rename all columns:

  val newNames = Seq("x3", "x4")
  
  data.toDF(newNames: _*)
  

• Rename from mapping with select:

  val  mapping = Map("x1" -> "x3", "x2" -> "x4")
  
  df.select(
    df.columns.map(c => df(c).alias(mapping.get(c).getOrElse(c))): _*
  )
  

  or foldLeft + withColumnRenamed

  mapping.foldLeft(data){
    case (data, (oldName, newName)) => data.withColumnRenamed(oldName, newName) 
  }
  

* Not to be confused with RDD.toDF which is not a variadic functions, and takes column names as a list,

Solution 2:

I couldn't find an easy pyspark solution either, so just built my own one, similar to pandas' df.rename(columns={'old_name_1':'new_name_1', 'old_name_2':'new_name_2'}).

import pyspark.sql.functions as F

def rename_columns(df, columns):
    if isinstance(columns, dict):
        return df.select(*[F.col(col_name).alias(columns.get(col_name, col_name)) for col_name in df.columns])
    else:
        raise ValueError("'columns' should be a dict, like {'old_name_1':'new_name_1', 'old_name_2':'new_name_2'}")

So your solution will look like data = rename_columns(data, {'x1': 'x3', 'x2': 'x4'})

If you want to chain your method calls, Spark 3.0 brought in pyspark.sql.DataFrame.transform, which you can use in the following way:

my_df.transform(lambda df: rename_columns(df, {'old_name_1':'new_name_1', 'old_name_2':'new_name_2'}))

It saves me some lines of code, hope it will help you too.

Solution 3:

why do you want to perform it in a single line if you print the execution plan it is actually done in single line only

data = spark.createDataFrame([(1,2), (3,4)], ['x1', 'x2'])
data = (data
   .withColumnRenamed('x1','x3')
   .withColumnRenamed('x2', 'x4'))
data.explain()

OUTPUT

== Physical Plan ==
*(1) Project [x1#1548L AS x3#1552L, x2#1549L AS x4#1555L]
+- Scan ExistingRDD[x1#1548L,x2#1549L]

if you want to do it with a tuple of list you can use a simple map function

data = spark.createDataFrame([(1,2), (3,4)], ['x1', 'x2'])
new_names = [("x1","x3"),("x2","x4")]
data = data.select(list(
       map(lambda old,new:F.col(old).alias(new),*zip(*new_names))
       ))

data.explain()

still has same plan

OUTPUT

== Physical Plan ==
*(1) Project [x1#1650L AS x3#1654L, x2#1651L AS x4#1655L]
+- Scan ExistingRDD[x1#1650L,x2#1651L]

Solution 4:

This should work if you want to rename multiple columns using the same column name with a prefix

df.select([f.col(c).alias(PREFIX + c) for c in df.columns])