How to read a Json file with a specific format with Spark Scala?

Question

I'm trying to read a Json file which is like :

[ 
{"IFAM":"EQR","KTM":1430006400000,"COL":21,"DATA":[{"MLrate":"30","Nrout":"0","up":null,"Crate":"2"} 
,{"MLrate":"31","Nrout":"0","up":null,"Crate":"2"} 
,{"MLrate":"30","Nrout":"5","up":null,"Crate":"2"} 
,{"MLrate":"34","Nrout":"0","up":null,"Crate":"4"} 
,{"MLrate":"33","Nrout":"0","up":null,"Crate":"2"} 
,{"MLrate":"30","Nrout":"8","up":null,"Crate":"2"} 
]} 
,{"IFAM":"EQR","KTM":1430006400000,"COL":22,"DATA":[{"MLrate":"30","Nrout":"0","up":null,"Crate":"2"} 
,{"MLrate":"30","Nrout":"0","up":null,"Crate":"0"} 
,{"MLrate":"35","Nrout":"1","up":null,"Crate":"5"} 
,{"MLrate":"30","Nrout":"6","up":null,"Crate":"2"} 
,{"MLrate":"30","Nrout":"0","up":null,"Crate":"2"} 
,{"MLrate":"38","Nrout":"8","up":null,"Crate":"1"} 
]} 
,...
]

I've tried the command:

    val df = sqlContext.read.json("namefile") 
    df.show()

But this does not work : my columns are not recognized...

is your single json data is spread into multiple lines or single line ? — vijay kumar
– vijay kumar, Commented Jul 29, 2015 at 13:55
single line but when I open it with notePad this is easy to read and on several lines. — SparkUser
– SparkUser, Commented Jul 29, 2015 at 14:01

zero323 · Accepted Answer · 2015-07-29 15:42:19Z

6

If you want to use read.json you need a single JSON document per line. If your file contains a valid JSON array with documents it simply won't work as expected. For example if we take your example data input file should be formatted like this:

{"IFAM":"EQR","KTM":1430006400000,"COL":21,"DATA":[{"MLrate":"30","Nrout":"0","up":null,"Crate":"2"}, {"MLrate":"31","Nrout":"0","up":null,"Crate":"2"}, {"MLrate":"30","Nrout":"5","up":null,"Crate":"2"} ,{"MLrate":"34","Nrout":"0","up":null,"Crate":"4"} ,{"MLrate":"33","Nrout":"0","up":null,"Crate":"2"} ,{"MLrate":"30","Nrout":"8","up":null,"Crate":"2"} ]}
{"IFAM":"EQR","KTM":1430006400000,"COL":22,"DATA":[{"MLrate":"30","Nrout":"0","up":null,"Crate":"2"} ,{"MLrate":"30","Nrout":"0","up":null,"Crate":"0"} ,{"MLrate":"35","Nrout":"1","up":null,"Crate":"5"} ,{"MLrate":"30","Nrout":"6","up":null,"Crate":"2"} ,{"MLrate":"30","Nrout":"0","up":null,"Crate":"2"} ,{"MLrate":"38","Nrout":"8","up":null,"Crate":"1"} ]}

If you use read.json on above structure you'll see it is parsed as expected:

scala> sqlContext.read.json("namefile").printSchema
root
 |-- COL: long (nullable = true)
 |-- DATA: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- Crate: string (nullable = true)
 |    |    |-- MLrate: string (nullable = true)
 |    |    |-- Nrout: string (nullable = true)
 |    |    |-- up: string (nullable = true)
 |-- IFAM: string (nullable = true)
 |-- KTM: long (nullable = true)

answered Jul 29, 2015 at 15:42

zero323

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1 Comment

Jyd Over a year ago

@zero323 I want o do something similar, but I just want to load the data from the json file, to an rdd and I dont want to go through the spark-sql just a simple transformation. How can I achieve this? Thanks!

brunocrt · Accepted Answer · 2018-01-17 19:48:40Z

If you don't want to format your JSON file (line by line) you could create a schema using StructType and MapType using SparkSQL functions

import org.apache.spark.sql.DataFrame 
import org.apache.spark.sql.functions._ 
import org.apache.spark.sql.types._

// Convenience function for turning JSON strings into DataFrames
def jsonToDataFrame(json: String, schema: StructType = null): 
DataFrame = {
    val reader = spark.read
    Option(schema).foreach(reader.schema)   
    reader.json(sc.parallelize(Array(json))) 
}

// Using a struct
val schema = new StructType().add("a", new StructType().add("b", IntegerType))

// call the function passing the sample JSON data and the schema as parameter
val json_df = jsonToDataFrame("""
   {
     "a": {
        "b": 1
     }
   } """, schema)

// now you can access your json fields
val b_value = json_df.select("a.b")
b_value.show()

See this reference documentation for more examples and details https://docs.databricks.com/spark/latest/spark-sql/complex-types.html#transform-complex-data-types-scala

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