How to add a constant column in a Spark DataFrame?

I want to add a column in a DataFrame with some arbitrary value (that is the same for each row). I get an error when I use withColumn as follows:

    dt.withColumn('new_column', 10).head(5)

    ---------------------------------------------------------------------------
    AttributeError                            Traceback (most recent call last)
    <ipython-input-50-a6d0257ca2be> in <module>()
          1 dt = (messages
          2     .select(messages.fromuserid, messages.messagetype, floor(messages.datetime/(1000*60*5)).alias("dt")))
    ----> 3 dt.withColumn('new_column', 10).head(5)

    /Users/evanzamir/spark-1.4.1/python/pyspark/sql/dataframe.pyc in withColumn(self, colName, col)
       1166         [Row(age=2, name=u'Alice', age2=4), Row(age=5, name=u'Bob', age2=7)]
       1167         """
    -> 1168         return self.select('*', col.alias(colName))
       1169 
       1170     @ignore_unicode_prefix

    AttributeError: 'int' object has no attribute 'alias'

It seems that I can trick the function into working as I want by adding and subtracting one of the other columns (so they add to zero) and then adding the number I want (10 in this case):

    dt.withColumn('new_column', dt.messagetype - dt.messagetype + 10).head(5)

    [Row(fromuserid=425, messagetype=1, dt=4809600.0, new_column=10),
     Row(fromuserid=47019141, messagetype=1, dt=4809600.0, new_column=10),
     Row(fromuserid=49746356, messagetype=1, dt=4809600.0, new_column=10),
     Row(fromuserid=93506471, messagetype=1, dt=4809600.0, new_column=10),
     Row(fromuserid=80488242, messagetype=1, dt=4809600.0, new_column=10)]

This is supremely hacky, right? I assume there is a more legit way to do this?

Spark 2.2+

Spark 2.2 introduces typedLit to support Seq, Map, and Tuples (SPARK-19254) and following calls should be supported (Scala):

    import org.apache.spark.sql.functions.typedLit

    df.withColumn("some_array", typedLit(Seq(1, 2, 3)))
    df.withColumn("some_struct", typedLit(("foo", 1, .0.3)))
    df.withColumn("some_map", typedLit(Map("key1" -> 1, "key2" -> 2)))

Spark 1.3+ (lit), 1.4+ (array, struct), 2.0+ (map):

The second argument for DataFrame.withColumn should be a Column so you have to use a literal:

    from pyspark.sql.functions import lit

    df.withColumn('new_column', lit(10))

If you need complex columns you can build these using blocks like array:

    from pyspark.sql.functions import array, create_map, struct

    df.withColumn("some_array", array(lit(1), lit(2), lit(3)))
    df.withColumn("some_struct", struct(lit("foo"), lit(1), lit(.3)))
    df.withColumn("some_map", create_map(lit("key1"), lit(1), lit("key2"), lit(2)))

Exactly the same methods can be used in Scala.

    import org.apache.spark.sql.functions.{array, lit, map, struct}

    df.withColumn("new_column", lit(10))
    df.withColumn("map", map(lit("key1"), lit(1), lit("key2"), lit(2)))

To provide names for structs use either alias on each field:

    df.withColumn(
        "some_struct",
        struct(lit("foo").alias("x"), lit(1).alias("y"), lit(0.3).alias("z"))
     )

or cast on the whole object

    df.withColumn(
        "some_struct", 
        struct(lit("foo"), lit(1), lit(0.3)).cast("struct<x: string, y: integer, z: double>")
     )

It is also possible, although slower, to use an UDF.

Note :

The same constructs can be used to pass constant arguments to UDFs or SQL functions.

From: stackoverflow.com/q/32788322

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