Spark scala - 将 StructType 嵌套转换为 Map [英] Spark scala - Nested StructType conversion to Map

问题描述

我在 Scala 中使用 Spark 1.6.

I am using Spark 1.6 in scala.

我在 ElasticSearch 中用一个对象创建了一个索引.对象params"被创建为 Map[String, Map[String, String]].示例:

I created an index in ElasticSearch with an object. The object "params" was created as a Map[String, Map[String, String]]. Example:

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

这给了我如下所示的记录:

That gives me records that look like the following:

{
        "_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"
              }
           }
        }
     },

然后我尝试读取 Elasticsearch 索引以更新值.

Then I am trying to read the Elasticsearch index in order to update the values.

Spark 使用以下架构读取索引:

Spark reads the index with the following schema:

|-- 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)

我的问题是对象被作为结构读取.为了管理和轻松更新字段，我想要一个 Map，因为我对 StructType 不太熟悉.

My problem is that the object is read as a struct. In order to manage and easily update the fields I want to have a Map as I am not very familiar with StructType.

我尝试将 UDF 中的对象作为 Map 获取，但出现以下错误:

I tried to get the object in a UDF as a Map but I have the following error:

 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 代码片段:

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?我已经阅读了文档中提供的 toMap 方法，但找不到如何使用它(对隐式参数不是很熟悉)，因为我是 Scala 初学者.

My question: How can we convert this Struct to a Map? I already read saw the toMap method available in the documentation but can not find how to use it (not very familiar with implicit parameters) as I am a scala beginner.

提前致谢，

推荐答案

我最终解决如下:

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)

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