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How to get keys and values from MapType column in SparkSQL DataFrame

I have data in a parquet file which has 2 fields: object_id: String and alpha: Map<>.

It is read into a data frame in sparkSQL and the schema looks like this:

scala> alphaDF.printSchema()
root
 |-- object_id: string (nullable = true)
 |-- ALPHA: map (nullable = true)
 |    |-- key: string
 |    |-- value: struct (valueContainsNull = true)

I am using Spark 2.0 and I am trying to create a new data frame in which columns need to be object_id plus keys of the ALPHA map as in object_id, key1, key2, key2, ...

I was first trying to see if I could at least access the map like this:

scala> alphaDF.map(a => a(0)).collect()
<console>:32: error: Unable to find encoder for type stored in a Dataset.
Primitive types (Int, String, etc) and Product types (case classes) are 
supported by importing spark.implicits._  Support for serializing other
types will be added in future releases.
   alphaDF.map(a => a(0)).collect()

but unfortunately I can't seem to be able to figure out how to access the keys of the map.

Can someone please show me a way to get the object_id plus map keys as column names and map values as respective values in a new dataframe?

like image 984
lloydh Avatar asked Nov 15 '16 05:11

lloydh


Video Answer


2 Answers

Spark >= 2.3

You can simplify the process using map_keys function:

import org.apache.spark.sql.functions.map_keys

There is also map_values function, but it won't be directly useful here.

Spark < 2.3

General method can be expressed in a few steps. First required imports:

import org.apache.spark.sql.functions.udf
import org.apache.spark.sql.Row

and example data:

val ds = Seq(
  (1, Map("foo" -> (1, "a"), "bar" -> (2, "b"))),
  (2, Map("foo" -> (3, "c"))),
  (3, Map("bar" -> (4, "d")))
).toDF("id", "alpha")

To extract keys we can use UDF (Spark < 2.3)

val map_keys = udf[Seq[String], Map[String, Row]](_.keys.toSeq)

or built-in functions

import org.apache.spark.sql.functions.map_keys

val keysDF = df.select(map_keys($"alpha"))

Find distinct ones:

val distinctKeys = keysDF.as[Seq[String]].flatMap(identity).distinct
  .collect.sorted

You can also generalize keys extraction with explode:

import org.apache.spark.sql.functions.explode

val distinctKeys = df
  // Flatten the column into key, value columns
 .select(explode($"alpha"))
 .select($"key")
 .as[String].distinct
 .collect.sorted

And select:

ds.select($"id" +: distinctKeys.map(x => $"alpha".getItem(x).alias(x)): _*)
like image 123
zero323 Avatar answered Oct 17 '22 08:10

zero323


And if you are in PySpark, I just find an easy implementation:

from pyspark.sql.functions import map_keys

alphaDF.select(map_keys("ALPHA").alias("keys")).show()

You can check details in here

like image 31
Hailin FU Avatar answered Oct 17 '22 07:10

Hailin FU