Spark HiveContext get the same format as hive client select - apache-spark

Question

When a Hive table has values like maps or arrays, if you select it in the Hive client they are shown as JSON, e.g.: {"a":1,"b":1} or [1,2,2].
When you select those in Spark, they are map/array objects in the DataFrame. If you stringify each row they are Map("a" -> 1, "b" -> 1) or WrappedArray(1, 2, 2).
I want to have the same format as the Hive client when using Spark's HiveContext.
How can I do this?

Answer 1

Spark has its own functions to convert complex objects into their JSON representation.
Here is the documentation for the org.apache.spark.sql.functions package, which also comes with the to_json function that does the following:
Converts a column containing a StructType, ArrayType of StructTypes, a MapType or ArrayType of MapTypes into a JSON string with the specified schema. Throws an exception, in the case of an unsupported type.
Here is a short example as ran on the spark-shell:
scala> val df = spark.createDataFrame(
| Seq(("hello", Map("a" -> 1)), ("world", Map("b" -> 2)))
| ).toDF("name", "map")
df: org.apache.spark.sql.DataFrame = [name: string, map: map<string,int>]
scala> df.show
+-----+-----------+
| name| map|
+-----+-----------+
|hello|Map(a -> 1)|
|world|Map(b -> 2)|
+-----+-----------+
scala> df.select($"name", to_json(struct($"map")) as "json").show
+-----+---------------+
| name| json|
+-----+---------------+
|hello|{"map":{"a":1}}|
|world|{"map":{"b":2}}|
+-----+---------------+
Here is a similar example, with arrays instead of maps:
scala> val df = spark.createDataFrame(
| Seq(("hello", Seq("a", "b")), ("world", Seq("c", "d")))
| ).toDF("name", "array")
df: org.apache.spark.sql.DataFrame = [name: string, array: array<string>]
scala> df.select($"name", to_json(struct($"array")) as "json").show
+-----+-------------------+
| name| json|
+-----+-------------------+
|hello|{"array":["a","b"]}|
|world|{"array":["c","d"]}|
+-----+-------------------+