create a spark dataframe from a nested json file in scala

Question

I have a json file that looks like this

{
"group" : {},
"lang" : [ 
    [ 1, "scala", "functional" ], 
    [ 2, "java","object" ], 
    [ 3, "py","interpreted" ]
]
}

I tried to create a dataframe using

val path = "some/path/to/jsonFile.json"
val df = sqlContext.read.json(path)
df.show()

when I run this I get

df: org.apache.spark.sql.DataFrame = [_corrupt_record: string]

How do we create a df based on contents of "lang" key? I do not care about group{} all I need is, pull data out of "lang" and apply case class like this

case class ProgLang (id: Int, lang: String, type: String )

I have read this post Reading JSON with Apache Spark - `corrupt_record` and understand that each record needs to be on a newline but in my case I cannot change the file structure

Ramesh Maharjan · Accepted Answer · 2017-07-19T02:30:23.507

7

The json format is wrong. The the json api of sqlContext is reading it as corrupt record. Correct form is

{"group":{},"lang":[[1,"scala","functional"],[2,"java","object"],[3,"py","interpreted"]]}

and supposing you have it in a file ("/home/test.json"), then you can use following method to get the dataframe you want

import org.apache.spark.sql.functions._
import sqlContext.implicits._

val df = sqlContext.read.json("/home/test.json")

val df2 = df.withColumn("lang", explode($"lang"))
    .withColumn("id", $"lang"(0))
    .withColumn("langs", $"lang"(1))
    .withColumn("type", $"lang"(2))
    .drop("lang")
    .withColumnRenamed("langs", "lang")
    .show(false)

You should have

+---+-----+-----------+
|id |lang |type       |
+---+-----+-----------+
|1  |scala|functional |
|2  |java |object     |
|3  |py   |interpreted|
+---+-----+-----------+

Updated

If you don't want to change your input json format as mentioned in your comment below, you can use wholeTextFiles to read the json file and parse it as below

import sqlContext.implicits._
import org.apache.spark.sql.functions._

val readJSON = sc.wholeTextFiles("/home/test.json")
  .map(x => x._2)
  .map(data => data.replaceAll("\n", ""))

val df = sqlContext.read.json(readJSON)

val df2 = df.withColumn("lang", explode($"lang"))
  .withColumn("id", $"lang"(0).cast(IntegerType))
  .withColumn("langs", $"lang"(1))
  .withColumn("type", $"lang"(2))
  .drop("lang")
  .withColumnRenamed("langs", "lang")

df2.show(false)
df2.printSchema

It should give you dataframe as above and schema as

root
 |-- id: integer (nullable = true)
 |-- lang: string (nullable = true)
 |-- type: string (nullable = true)

edited Jul 19 '17 at 02:30

answered Jul 19 '17 at 00:19

Ramesh Maharjan

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Ramesh, I should have been more clear, I cannot change my json file format and it contains just a single json doc i.e. data shared earlier and it has newlines whereas in your case its all in a single line. – devtest13 Jul 19 '17 at 01:13
is it in the same format as you pasted in your question? how another record is added in that format then? – Ramesh Maharjan Jul 19 '17 at 01:22
yes, format remains same, within .json file it is the "lang" : [ 4, "ruby","interpreted" ] that gets new data – devtest13 Jul 19 '17 at 01:39
I think I am getting closer, I now have a df with group and lang schema root |-- lang: array (nullable = true) | |-- element: array (containsNull = true) | | |-- element: string (containsNull = true) here's the code `val readJSON = sc.wholeTextFiles("'home/data.json").map(x => x._2).map(data => data.replaceAll("""[\n]+""", " ")) val df = sqlContext.read.json(readJSON) df.printSchema` – devtest13 Jul 19 '17 at 02:07
yeah I was thinking to use wholeTextFiles as well. You got it right . After that you can just use my solution above. :) the part of creating df2 :) – Ramesh Maharjan Jul 19 '17 at 02:24
I have updated my answer :) please check – Ramesh Maharjan Jul 19 '17 at 02:30

score 2 · Answer 2 · answered Jul 19 '17 at 08:43

As of Spark 2.2 you can use multiLine option to deal with the case of multi-line JSONs.

scala> spark.read.option("multiLine", true).json("jsonFile.json").printSchema
root
 |-- lang: array (nullable = true)
 |    |-- element: array (containsNull = true)
 |    |    |-- element: string (containsNull = true)

Before Spark 2.2 see How to access sub-entities in JSON file? or Read multiline JSON in Apache Spark.

create a spark dataframe from a nested json file in scala

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