CODEWITHSUNDEEP
jsonscalaapache-spark
stepandel
stepandel

Reputation: 133

How to convert nested JSON to map object in scala

I have the following JSON objects:

{
    "user_id": "123",
    "data": {
        "city": "New York"
    },
    "timestamp": "1563188698.31",
    "session_id": "6a793439-6535-4162-b333-647a6761636b"
}
{
    "user_id": "123",
    "data": {
        "name": "some_name",
        "age": "23",
        "occupation": "teacher"
    },
    "timestamp": "1563188698.31",
    "session_id": "6a793439-6535-4162-b333-647a6761636b"
}

I'm using val df = sqlContext.read.json("json") to read the file to dataframe

Which combines all data attributes into data struct like so:

root
 |-- data: struct (nullable = true)
 |    |-- age: string (nullable = true)
 |    |-- city: string (nullable = true)
 |    |-- name: string (nullable = true)
 |    |-- occupation: string (nullable = true)
 |-- session_id: string (nullable = true)
 |-- timestamp: string (nullable = true)
 |-- user_id: string (nullable = true)

Is it possible to transform data field to MAP[String, String] Data type? And so it only has the same attributes as original json?

Upvotes: 3

Views: 5853

Answers (2)

abiratsis
abiratsis

Reputation: 7316

Yes you can achieve that by exporting a Map[String, String] from the JSON data as shown next:

import org.apache.spark.sql.types.{MapType, StringType}
import org.apache.spark.sql.functions.{to_json, from_json}

val jsonStr = """{
    "user_id": "123",
    "data": {
        "name": "some_name",
        "age": "23",
        "occupation": "teacher"
    },
    "timestamp": "1563188698.31",
    "session_id": "6a793439-6535-4162-b333-647a6761636b"
}"""

val df = spark.read.json(Seq(jsonStr).toDS)

val mappingSchema = MapType(StringType, StringType)

df.select(from_json(to_json($"data"), mappingSchema).as("map_data"))

//Output
// +-----------------------------------------------------+
// |map_data                                             |
// +-----------------------------------------------------+
// |[age -> 23, name -> some_name, occupation -> teacher]|
// +-----------------------------------------------------+

First we extract the content of the data field into a string with to_json($"data"), then we parse and extract the Map with from_json(to_json($"data"), schema).

Upvotes: 7

msrv499
msrv499

Reputation: 339

Not sure what you mean to convert it to a Map of (String, String), But see if below can help.

val dataDF = spark.read.option("multiline","true").json("madhu/user.json").select("data").toDF

dataDF
.withColumn("age", $"data"("age")).withColumn("city", $"data"("city"))
.withColumn("name", $"data"("name"))
.withColumn("occupation", $"data"("occupation"))
.drop("data")
.show

Upvotes: 1

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