根据提供的列表过滤数组列

本文介绍了根据提供的列表过滤数组列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

限时送ChatGPT账号..

我在数据框中有以下类型:

I have the following types in a dataframe:

 root
 |-- id: string (nullable = true)
 |-- items: array (nullable = true)
 |    |-- element: string (containsNull = true)

输入:

val rawData = Seq(("id1",Array("item1","item2","item3","item4")),("id2",Array("item1","item2","item3")))
val data = spark.createDataFrame(rawData)

和项目列表:

 val filter_list = List("item1", "item2")

我想过滤掉不在 filter_list 中的项目，类似于 array_contains 的功能，但它不适用于提供的字符串列表，仅单个值.

I would like to filter out items that are non in the filter_list, similar to how array_contains would function, but its not working on a provided list of strings, only a single value.

所以输出看起来像这样:

so the output would look like this:

val rawData = Seq(("id1",Array("item1","item2")),("id2",Array("item1","item2")))
val data = spark.createDataFrame(rawData)

我尝试使用以下 UDF 解决此问题，但我可能在 Scala 和 Spark 之间混合了类型:

I tried solving this with the following UDF, but I probably mix types between Scala and Spark:

def filterItems(flist: List[String]) = udf {
  (recs: List[String]) => recs.filter(item => flist.contains(item))
}

我使用的是 Spark 2.2

I'm using Spark 2.2

谢谢！

推荐答案

你的代码几乎是对的.您所要做的就是将 List 替换为 Seq

You code is almost right. All you have to do is replace List with Seq

def filterItems(flist: List[String]) = udf {
  (recs: Seq[String]) => recs.filter(item => flist.contains(item))
}

从 List[String] => 更改签名也是有意义的.UserDefinedFunction 到 SeqString] =>UserDefinedFunction，但不是必需的.

It would also make sense to change signature from List[String] => UserDefinedFunction to SeqString] => UserDefinedFunction, but it is not required.

参考 SQL 编程指南 - 数据类型.

这篇关于根据提供的列表过滤数组列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！