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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