CODEWITHSUNDEEP
jsonscalaapache-spark
LearningSlowly
LearningSlowly

Reputation: 9451

Reading JSON with Apache Spark - `corrupt_record`

I have a json file, nodes that looks like this:

[{"toid":"osgb4000000031043205","point":[508180.748,195333.973],"index":1}
,{"toid":"osgb4000000031043206","point":[508163.122,195316.627],"index":2}
,{"toid":"osgb4000000031043207","point":[508172.075,195325.719],"index":3}
,{"toid":"osgb4000000031043208","point":[508513,196023],"index":4}]

I am able to read and manipulate this record with Python.

I am trying to read this file in scala through the spark-shell.

From this tutorial, I can see that it is possible to read json via sqlContext.read.json

val vfile = sqlContext.read.json("path/to/file/nodes.json")

However, this results in a corrupt_record error:

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

Can anyone shed some light on this error? I can read and use the file with other applications and I am confident it is not corrupt and sound json.

Upvotes: 31

Views: 51313

Answers (4)

dk14
dk14

Reputation: 22374

Spark cannot read JSON-array to a record on top-level, so you have to pass:

{"toid":"osgb4000000031043205","point":[508180.748,195333.973],"index":1} 
{"toid":"osgb4000000031043206","point":[508163.122,195316.627],"index":2} 
{"toid":"osgb4000000031043207","point":[508172.075,195325.719],"index":3} 
{"toid":"osgb4000000031043208","point":[508513,196023],"index":4}

As it's described in the tutorial you're referring to:

Let's begin by loading a JSON file, where each line is a JSON object

The reasoning is quite simple. Spark expects you to pass a file with a lot of JSON-entities (entity per line), so it could distribute their processing (per entity, roughly saying).

To put more light on it, here is a quote form the official doc

Note that the file that is offered as a json file is not a typical JSON file. Each line must contain a separate, self-contained valid JSON object. As a consequence, a regular multi-line JSON file will most often fail.

This format is called JSONL. Basically it's an alternative to CSV.

Upvotes: 31

SandeepGodara
SandeepGodara

Reputation: 1528

As Spark expects "JSON Line format" not a typical JSON format, we can tell spark to read typical JSON by specifying:

val df = spark.read.option("multiline", "true").json("<file>")

Upvotes: 43

Datageek
Datageek

Reputation: 26739

To read the multi-line JSON as a DataFrame:

val spark = SparkSession.builder().getOrCreate()

val df = spark.read.json(spark.sparkContext.wholeTextFiles("file.json").values)

Reading large files in this manner is not recommended, from the wholeTextFiles docs

Small files are preferred, large file is also allowable, but may cause bad performance.

Upvotes: 12

Robert Gabriel
Robert Gabriel

Reputation: 1190

I run into the same problem. I used sparkContext and sparkSql on the same configuration:

val conf = new SparkConf()
  .setMaster("local[1]")
  .setAppName("Simple Application")


val sc = new SparkContext(conf)

val spark = SparkSession
  .builder()
  .config(conf)
  .getOrCreate()

Then, using the spark context I read the whole json (JSON - path to file) file:

 val jsonRDD = sc.wholeTextFiles(JSON).map(x => x._2)

You can create a schema for future selects, filters...

val schema = StructType( List(
  StructField("toid", StringType, nullable = true),
  StructField("point", ArrayType(DoubleType), nullable = true),
  StructField("index", DoubleType, nullable = true)
))

Create a DataFrame using spark sql:

var df: DataFrame = spark.read.schema(schema).json(jsonRDD).toDF()

For testing use show and printSchema:

df.show()
df.printSchema()

sbt build file:

name := "spark-single"

version := "1.0"

scalaVersion := "2.11.7"

libraryDependencies += "org.apache.spark" %% "spark-core" % "2.0.2"
libraryDependencies +="org.apache.spark" %% "spark-sql" % "2.0.2"

Upvotes: 1

Related Questions