Question

我想将下面的字符串变量转换为spark上的数据帧。

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

我知道如何从json文件创建数据框。

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

但我不知道如何从字符串变量创建数据框。

如何将json String变量转换为dataframe。

Answer 1

Spark 2.2+：

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

Spark 2.1.x：

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

提示：这是使用sqlContext.read.json(jsonRDD: RDD[Stirng])重载。还有sqlContext.read.json(path: String)，它直接读取Json文件。

older versions：

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

Answer 2

由于在RDD中读取JSON的函数已被弃用 Spark 2.2，这将是另一种选择：

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

要将JSON字符串列表转换为Spark 2.2中的DataFrame =>

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()

结果：

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

Answer 4

这里是一个如何在Java（Spark 2.2+）中将Json字符串转换为Dataframe的示例：

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();

这是结果：

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

Answer 5

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)

答案的参考是https://stackoverflow.com/a/49399359/2187751

Answer 6

您现在可以直接从Dataset [String]中读取json：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

在某些情况下会出现一些错误，例如“非法模式”组件：XXX，因此您需要在spark.read中添加带有时间戳的.option，以便更新代码。

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()

如何将json字符串转换为spark

7 个答案: