how to convert json string to dataframe on spark

Question

I want to convert string variable below to dataframe on spark.

val jsonStr = "{ "metadata": { "key": 84896, "value": 54 }}"

I know how to create dataframe from json file.

sqlContext.read.json("file.json")

but I don't know how to create dataframe from string variable.

How can I convert json String variable to dataframe.

Answer 1

For Spark 2.2+ :

import spark.implicits._
val jsonStr = """{ "metadata": { "key": 84896, "value": 54 }}"""
val df = spark.read.json(Seq(jsonStr).toDS)

For Spark 2.1.x :

val events = sc.parallelize("""{"action":"create","timestamp":"2016-01-07T00:01:17Z"}""" :: Nil)    
val df = sqlContext.read.json(events)

Hint: this is using sqlContext.read.json(jsonRDD: RDD[Stirng]) overload. There is also sqlContext.read.json(path: String) where it reads a Json file directly.

For older versions :

val jsonStr = """{ "metadata": { "key": 84896, "value": 54 }}"""
val rdd = sc.parallelize(Seq(jsonStr))
val df = sqlContext.read.json(rdd)

Answer 2

Since the function for reading JSON from an RDD got deprecated in Spark 2.2, this would be another option:

val jsonStr = """{ "metadata": { "key": 84896, "value": 54 }}"""
import spark.implicits._ // spark is your SparkSession object
val df = spark.read.json(Seq(jsonStr).toDS)

Answer 3

simple_json = '{"results":[{"a":1,"b":2,"c":"name"},{"a":2,"b":5,"c":"foo"}]}'
rddjson = sc.parallelize([simple_json])
df = sqlContext.read.json(rddjson)

The reference to the answer is https://stackoverflow.com/a/49399359/2187751

Answer 4

To convert list of json Strings into DataFrame in Spark 2.2 =>

val spark = SparkSession
          .builder()
          .master("local")
          .appName("Test")
          .getOrCreate()

var strList = List.empty[String]
var jsonString1 = """{"ID" : "111","NAME":"Arkay","LOC":"Pune"}"""
var jsonString2 = """{"ID" : "222","NAME":"DineshS","LOC":"PCMC"}"""
strList = strList :+ jsonString1
strList = strList :+ jsonString2

val rddData = spark.sparkContext.parallelize(strList)
resultDF = spark.read.json(rddData)
resultDF.show()

Result:

+---+----+-------+
| ID| LOC|   NAME|
+---+----+-------+
|111|Pune|  Arkay|
|222|PCMC|DineshS|
+---+----+-------+

Answer 5

Here is an example how to convert Json string to Dataframe in Java (Spark 2.2+):

String str1 = "{\"_id\":\"123\",\"ITEM\":\"Item 1\",\"CUSTOMER\":\"Billy\",\"AMOUNT\":285.2}";
String str2 = "{\"_id\":\"124\",\"ITEM\":\"Item 2\",\"CUSTOMER\":\"Sam\",\"AMOUNT\":245.85}";
List<String> jsonList = new ArrayList<>();
jsonList.add(str1);
jsonList.add(str2);
SparkContext sparkContext = new SparkContext(new SparkConf()
        .setAppName("myApp").setMaster("local"));
JavaSparkContext javaSparkContext = new JavaSparkContext(sparkContext);
SQLContext sqlContext = new SQLContext(sparkContext);
JavaRDD<String> javaRdd = javaSparkContext.parallelize(jsonList);
Dataset<Row> data = sqlContext.read().json(javaRdd);
data.show();

Here is the result:

+------+--------+------+---+
|AMOUNT|CUSTOMER|  ITEM|_id|
+------+--------+------+---+
| 285.2|   Billy|Item 1|123|
|245.85|     Sam|Item 2|124|
+------+--------+------+---+

Answer 6

you can now directly read json from Dataset[String]:https://spark.apache.org/docs/latest/sql-data-sources-json.html

val otherPeopleDataset = spark.createDataset(
  """{"name":"Yin","address":{"city":"Columbus","state":"Ohio"}}""" :: Nil)
val otherPeople = spark.read.json(otherPeopleDataset)
otherPeople.show()
// +---------------+----+
// |        address|name|
// +---------------+----+
// |[Columbus,Ohio]| Yin|
// +---------------+----+

Answer 7

There will be some error in some case like Illegal Patter component : XXX so for that you need to add .option with timestamp in spark.read so updated code will be.

val spark = SparkSession
          .builder()
          .master("local")
          .appName("Test")
          .getOrCreate()
import spark.implicits._
val jsonStr = """{ "metadata": { "key": 84896, "value": 54 }}"""
val df = spark.read.option("timestampFormat", "yyyy/MM/dd HH:mm:ss ZZ").json(Seq(jsonStr).toDS)
df.show()