CODEWITHSUNDEEP
jsonscalaapache-sparkapache-spark-sql
navige
navige

Reputation: 2517

Spark Row to JSON

I would like to create a JSON from a Spark v.1.6 (using scala) dataframe. I know that there is the simple solution of doing df.toJSON.

However, my problem looks a bit different. Consider for instance a dataframe with the following columns:

|  A  |     B     |  C1  |  C2  |    C3   |
-------------------------------------------
|  1  | test      |  ab  |  22  |  TRUE   |
|  2  | mytest    |  gh  |  17  |  FALSE  |

I would like to have at the end a dataframe with 

|  A  |     B     |                        C                   |
----------------------------------------------------------------
|  1  | test      | { "c1" : "ab", "c2" : 22, "c3" : TRUE }    |
|  2  | mytest    | { "c1" : "gh", "c2" : 17, "c3" : FALSE }   |

where C is a JSON containing C1, C2, C3. Unfortunately, I at compile time I do not know what the dataframe looks like (except the columns A and B that are always "fixed").

As for the reason why I need this: I am using Protobuf for sending around the results. Unfortunately, my dataframe sometimes has more columns than expected and I would still send those via Protobuf, but I do not want to specify all columns in the definition.

How can I achieve this?

Upvotes: 21

Views: 47832

Answers (4)

Cyanny
Cyanny

Reputation: 530

I use this command to solve the to_json problem:

output_df = (df.select(to_json(struct(col("*"))).alias("content")))

Upvotes: 14

Michael Armbrust
Michael Armbrust

Reputation: 1565

Spark 2.1 should have native support for this use case (see #15354).

import org.apache.spark.sql.functions.to_json
df.select(to_json(struct($"c1", $"c2", $"c3")))

Upvotes: 28

David Griffin
David Griffin

Reputation: 13927

Here, no JSON parser, and it adapts to your schema:

import org.apache.spark.sql.functions.{col, concat, concat_ws, lit}

df.select(
  col(df.columns(0)),
  col(df.columns(1)),
  concat(
    lit("{"), 
    concat_ws(",",df.dtypes.slice(2, df.dtypes.length).map(dt => {
      val c = dt._1;
      val t = dt._2;
      concat(
        lit("\"" + c + "\":" + (if (t == "StringType") "\""; else "")  ),
        col(c),
        lit(if(t=="StringType") "\""; else "") 
      )
    }):_*), 
    lit("}")
  ) as "C"
).collect()

Upvotes: 7

zero323
zero323

Reputation: 330413

First lets convert C's to a struct:

val dfStruct = df.select($"A", $"B", struct($"C1", $"C2", $"C3").alias("C"))

This is structure can be converted to JSONL using toJSON as before:

dfStruct.toJSON.collect
// Array[String] = Array(
//   {"A":1,"B":"test","C":{"C1":"ab","C2":22,"C3":true}}, 
//   {"A":2,"B":"mytest","C":{"C1":"gh","C2":17,"C3":false}})

I am not aware of any built-in method that can convert a single column but you can either convert it individually and join or use your favorite JSON parser in an UDF. 

case class C(C1: String, C2: Int, C3: Boolean)

object CJsonizer {
  import org.json4s._
  import org.json4s.JsonDSL._
  import org.json4s.jackson.Serialization
  import org.json4s.jackson.Serialization.write

  implicit val formats = Serialization.formats(org.json4s.NoTypeHints)

  def toJSON(c1: String, c2: Int, c3: Boolean) = write(C(c1, c2, c3))
}


val cToJSON = udf((c1: String, c2: Int, c3: Boolean) => 
  CJsonizer.toJSON(c1, c2, c3))

df.withColumn("c_json", cToJSON($"C1", $"C2", $"C3"))

Upvotes: 5

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