火花嵌套的结构型转换为映射

Question

我在Scala中使用Spark 1.6。

我用对象创建了Elasticsearch中的索引。对象" params"是作为映射[字符串，地图[String，String]]创建的。示例：

val params : Map[String, Map[String, String]] = ("p1" -> ("p1_detail" -> "table1"), "p2" -> (("p2_detail" -> "table2"), ("p2_filter" -> "filter2")), "p3" -> ("p3_detail" -> "table3"))

为我提供了如下图的记录：

{
        "_index": "x",
        "_type": "1",
        "_id": "xxxxxxxxxxxx",
        "_score": 1,
        "_timestamp": 1506537199650,
        "_source": {
           "a": "toto",
           "b": "tata",
           "c": "description",
           "params": {
              "p1": {
                 "p1_detail": "table1"
              },
              "p2": {
                 "p2_detail": "table2",
                 "p2_filter": "filter2"
              },
              "p3": {
                 "p3_detail": "table3"
              }
           }
        }
     },

然后，我正在尝试读取Elasticsearch索引以更新值。

Spark用以下模式读取索引：

|-- a: string (nullable = true)
|-- b: string (nullable = true)
|-- c: string (nullable = true)
|-- params: struct (nullable = true)
|    |-- p1: struct (nullable = true)
|    |    |-- p1_detail: string (nullable = true)
|    |-- p2: struct (nullable = true)
|    |    |-- p2_detail: string (nullable = true)
|    |    |-- p2_filter: string (nullable = true)
|    |-- p3: struct (nullable = true)
|    |    |-- p3_detail: string (nullable = true)

我的问题是该对象被读为结构。为了管理和轻松更新我想拥有地图的字段，因为我对structtype并不熟悉。

我试图将对象作为地图作为地图，但我有以下错误：

 User class threw exception: org.apache.spark.sql.AnalysisException: cannot resolve 'UDF(params)' due to data type mismatch: argument 1 requires map<string,map<string,string>> type, however, 'params' is of struct<p1:struct<p1_detail:string>,p2:struct<p2_detail:string,p2_filter:string>,p3:struct<p3_detail:string>> type.;

UDF代码段：

val getSubField : Map[String, Map[String, String]] => String = (params : Map[String, Map[String, String]]) => { val return_string = (params ("p1") getOrElse("p1_detail", null.asInstanceOf[String]) return_string }

我的问题：我们如何将此结构转换为地图？我已经看过文档中可用的tomap方法，但由于我是Scala初学者，找不到如何使用它（对隐式参数不太熟悉）。

。

预先感谢

Answer 1

我最终解决了以下内容：

def convertRowToMap[T](row: Row): Map[String, T] = {
  row.schema.fieldNames
    .filter(field => !row.isNullAt(row.fieldIndex(field)))
    .map(field => field -> row.getAs[T](field))
    .toMap
}
/* udf that converts Row to Map */
val rowToMap: Row => Map[String, Map[String, String]] = (row: Row) => {
  val mapTemp = convertRowToMap[Row](row)
  
  val mapToReturn = mapTemp.map { case (k, v) => k -> convertRowToMap[String](v) }
  
  mapToReturn   
}
val udfrowToMap = udf(rowToMap)

Answer 2

您无法将param类型指定为structType对象，而是将类型指定为行。

//Schema of parameter
def schema:StructType = (new StructType).add("p1", (new StructType).add("p1_detail", StringType))
      .add("p2", (new StructType).add("p2_detail", StringType).add("p2_filter",StringType))
      .add("p3", (new StructType).add("p3_detail", StringType))
 //Not allowed
 val extractVal: schema => collection.Map[Nothing, Nothing] = _.getMap(0)

解决方案：

// UDF example to process struct column
val extractVal: (Row) => collection.Map[Nothing, Nothing] = _.getMap(0)
// You would implement something similar
   val getSubField : Map[String, Map[String, String]] => String =
  (params : Row) =>
  {
    val p1 = params.getAs[Row]("p1")
    .........
    return null;
  }

我希望这会有所帮助！