Question

我是Spark和Scala的新手，所以请原谅。我有一个文本文件，格式如下：

for t in stride(from: 0, through: duration, by: frameDelta) {
    draw(in: renderBuffer, depthTexture: depthBuffer, time: t) { (texture) in
        recorder.writeFrame(forTexture: texture, time: t)
    }
}

我已经可以使用sc.textFile命令创建RDD，并且可以使用以下命令处理每个部分：

328;ADMIN HEARNG;[street#939 W El Camino,city#Chicago,state#IL]

但是，如您所见，第3个元素是嵌套键/值对，到目前为止，我无法使用它。我正在寻找的是一种将上述数据转换为如下所示的RDD的方法：

val department_record = department_rdd.map(record => record.split(";"))

感谢您的帮助。

Answer 1

您可以将,处的地址字段拆分为一个数组，剥去括号，然后在#处再次拆分以提取所需的地址分量，如下所示：

val department_rdd = sc.parallelize(Seq(
  "328;ADMIN HEARNG;[street#939 W El Camino,city#Chicago,state#IL]",
  "400;ADMIN HEARNG;[street#800 First Street,city#San Francisco,state#CA]"
))

val department_record = department_rdd.
  map(_.split(";")).
  map{ case Array(id, name, address) =>
    val addressArr = address.split(",").
      map(_.replaceAll("^\\[|\\]$", "").split("#"))
    (id, name, addressArr(0)(1), addressArr(1)(1), addressArr(2)(1))
  }

department_record.collect
// res1: Array[(String, String, String, String, String)] = Array(
//   (328,ADMIN HEARNG,939 W El Camino,Chicago,IL),
//   (400,ADMIN HEARNG,800 First Street,San Francisco,CA)
// )

如果要转换为DataFrame，只需应用toDF()：

department_record.toDF("id", "name", "street", "city", "state").show
// +---+------------+----------------+-------------+-----+
// | id|        name|          street|         city|state|
// +---+------------+----------------+-------------+-----+
// |328|ADMIN HEARNG| 939 W El Camino|      Chicago|   IL|
// |400|ADMIN HEARNG|800 First Street|San Francisco|   CA|
// +---+------------+----------------+-------------+-----+

Answer 2

DF解决方案：

scala> val df = Seq(("328;ADMIN HEARNG;[street#939 W El Camino,city#Chicago,state#IL]"),
     |   ("400;ADMIN HEARNG;[street#800 First Street,city#San Francisco,state#CA]")).toDF("dept")
df: org.apache.spark.sql.DataFrame = [dept: string]

scala> val df2 =df.withColumn("arr",split('dept,";")).withColumn("address",split(regexp_replace('arr(2),"\\[|\\]",""),"#"))
df2: org.apache.spark.sql.DataFrame = [dept: string, arr: array<string> ... 1 more field]

scala> df2.select('arr(0) as "id",'arr(1) as "name",split('address(1),",")(0) as "street",split('address(2),",")(0) as "city",'address(3) as "state").show
+---+------------+----------------+-------------+-----+
| id|        name|          street|         city|state|
+---+------------+----------------+-------------+-----+
|328|ADMIN HEARNG| 939 W El Camino|      Chicago|   IL|
|400|ADMIN HEARNG|800 First Street|San Francisco|   CA|
+---+------------+----------------+-------------+-----+


scala>

如何在Spark / Scala数据导入中处理嵌套的键值对

2 个答案: