我有一个嵌套案例类的集合。我的工作是使用这些案例类生成数据集,并将输出写入镶木地板。
我非常恼火地发现我必须手动进行大量工作才能加载这些数据并将其转换回案例类,以便在后续工作中使用它。无论如何,这就是我现在正在尝试做的事情。
我的案例类是这样的:
case class Person(userId: String, tech: Option[Tech])
case class Tech(browsers: Seq[Browser], platforms: Seq[Platform])
case class Browser(family: String, version: Int)
所以我正在加载我的镶木地板数据。我可以将
tech
数据作为 Row
获取:
val df = sqlContext.load("part-r-00716.gz.parquet")
val x = df.head
val tech = x.getStruct(x.fieldIndex("tech"))
但现在我找不到如何实际迭代浏览器。如果我尝试
val browsers = tech.getStruct(tech.fieldIndex("browsers"))
我会得到一个例外:
java.lang.ClassCastException: scala.collection.mutable.WrappedArray$ofRef cannot be cast to org.apache.spark.sql.Row
如何使用 Spark 1.5.2 迭代嵌套浏览器数据?
更新 事实上,我的案例类包含可选值,所以
Browser
实际上是:
case class Browser(family: String,
major: Option[String] = None,
minor: Option[String] = None,
patch: Option[String] = None,
language: String,
timesSeen: Long = 1,
firstSeenAt: Long,
lastSeenAt: Long)
我也有类似的
Os
:
case class Os(family: String,
major: Option[String] = None,
minor: Option[String] = None,
patch: Option[String] = None,
patchMinor: Option[String],
override val timesSeen: Long = 1,
override val firstSeenAt: Long,
override val lastSeenAt: Long)
所以
Tech
真的是:
case class Technographic(browsers: Seq[Browser],
devices: Seq[Device],
oss: Seq[Os])
现在,考虑到某些值是可选的,我需要一个解决方案来正确地重建我的案例类。当前的解决方案不支持
None
值,因此例如给定输入数据:
Tech(browsers=Seq(
Browser(family=Some("IE"), major=Some(7), language=Some("en"), timesSeen=3),
Browser(family=None, major=None, language=Some("en-us"), timesSeen=1),
Browser(family=Some("Firefox), major=None, language=None, timesSeen=1)
)
)
我需要它来加载数据,如下所示:
family=IE, major=7, language=en, timesSeen=3,
family=None, major=None, language=en-us, timesSeen=1,
family=Firefox, major=None, language=None, timesSeen=1
因为当前的解决方案不支持
None
值,所以实际上每个列表项有任意数量的值,即:
browsers.family = ["IE", "Firefox"]
browsers.major = [7]
browsers.language = ["en", "en-us"]
timesSeen = [3, 1, 1]
如您所见,无法将最终数据(由 Spark 返回)转换为生成它的案例类。
我该如何解决这种疯狂的问题?
一些例子
// Select two columns
df.select("userId", "tech.browsers").show()
// Select the nested values only
df.select("tech.browsers").show(truncate = false)
+-------------------------+
|browsers |
+-------------------------+
|[[Firefox,4], [Chrome,2]]|
|[[Firefox,4], [Chrome,2]]|
|[[IE,25]] |
|[] |
|null |
+-------------------------+
// Extract the family (nested value)
// This way you can iterate over the persons, and get their browsers
// Family values are nested
df.select("tech.browsers.family").show()
+-----------------+
| family|
+-----------------+
|[Firefox, Chrome]|
|[Firefox, Chrome]|
| [IE]|
| []|
| null|
+-----------------+
// Normalize the family: One row for each family
// Then you can iterate over all families
// Family values are un-nested, empty values/null/None are handled by explode()
df.select(explode(col("tech.browsers.family")).alias("family")).show()
+-------+
| family|
+-------+
|Firefox|
| Chrome|
|Firefox|
| Chrome|
| IE|
+-------+
基于最后一个示例:
val families = df.select(explode(col("tech.browsers.family")))
.map(r => r.getString(0)).distinct().collect().toList
println(families)
给出“正常”本地 Scala 列表中唯一的浏览器列表:
列表(IE、Firefox、Chrome)