How to get multiple line json File Into Single rec

rdd=sc.textFile(json or xml)
rdd.collect()

[u'{', u'    "glossary": {', u'        "title": "example glossary",', u'\t\t"GlossDiv": {', u'            "title": "S",', u'\t\t\t"GlossList": {', u'                "GlossEntry": {', u'                    "ID": "SGML",', u'\t\t\t\t\t"SortAs": "SGML",', u'\t\t\t\t\t"GlossTerm": "Standard Generalized Markup Language",', u'\t\t\t\t\t"Acronym": "SGML",', u'\t\t\t\t\t"Abbrev": "ISO 8879:1986",', u'\t\t\t\t\t"GlossDef": {', u'                        "para": "A meta-markup language, used to create markup languages such as DocBook.",', u'\t\t\t\t\t\t"GlossSeeAlso": ["GML", "XML"]', u'                    },', u'\t\t\t\t\t"GlossSee": "markup"', u'                }', u'            }', u'        }', u'    }', u'}', u'']

But My output should be every think in one line

{"glossary": {"title": "example glossary","GlossDiv": {"title": "S","GlossList":.....}}

回答1:

I'd recommend using Spark SQL JSON and then saving calling toJson (see https://spark.apache.org/docs/latest/sql-programming-guide.html#json-datasets )

val input = sqlContext.jsonFile(path)
val output = input...
output.toJSON.saveAsTextFile(outputath)

However if your json records can't be parsed by Spark SQL because of the multi-line issue or some other issue, we can take one of the examples from the Learning Spark book (slightly biased as a co-author of course) and and modify it to use wholeTextFiles.

case class Person(name: String, lovesPandas: Boolean)
// Read the input and throw away the file names
val input = sc.wholeTextFiles(inputFile).map(_._2)

// Parse it into a specific case class. We use mapPartitions beacuse:
// (a) ObjectMapper is not serializable so we either create a singleton object encapsulating ObjectMapper
//     on the driver and have to send data back to the driver to go through the singleton object.
//     Alternatively we can let each node create its own ObjectMapper but that's expensive in a map
// (b) To solve for creating an ObjectMapper on each node without being too expensive we create one per
//     partition with mapPartitions. Solves serialization and object creation performance hit.
val result = input.mapPartitions(records => {
    // mapper object created on each executor node
    val mapper = new ObjectMapper with ScalaObjectMapper
    mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
    mapper.registerModule(DefaultScalaModule)
    // We use flatMap to handle errors
    // by returning an empty list (None) if we encounter an issue and a
    // list with one element if everything is ok (Some(_)).
    records.flatMap(record => {
      try {
        Some(mapper.readValue(record, classOf[ioRecord]))
      } catch {
        case e: Exception => None
      }
    })
  }, true)
result.filter(_.lovesPandas).map(mapper.writeValueAsString(_))
  .saveAsTextFile(outputFile)
}

And in Python:

from pyspark import SparkContext
import json
import sys

if __name__ == "__main__":
    if len(sys.argv) != 4:
        print "Error usage: LoadJson [sparkmaster] [inputfile] [outputfile]"
        sys.exit(-1)
    master = sys.argv[1]
    inputFile = sys.argv[2]
    outputFile = sys.argv[3]
    sc = SparkContext(master, "LoadJson")
    input = sc.wholeTextFiles(inputFile).map(_._2)
    data = input.flatMap(lambda x: json.loads(x))
    data.filter(lambda x: 'lovesPandas' in x and x['lovesPandas']).map(
        lambda x: json.dumps(x)).saveAsTextFile(outputFile)
    sc.stop()
    print "Done!"

回答2:

Use sc.wholeTextFiles() instead.

回答3:

Also have a look at sqlContext.jsonFile: https://spark.apache.org/docs/1.3.1/sql-programming-guide.html#json-datasets