来自 Array 列的所有组合的 Spark Dataframe [英] Spark Dataframe from all combinations of Array column

问题描述

假设我有一个包含两列 elements_1 和 elements_2 的 Spark DataFrame d1，其中包含大小为 k 的整数集 和 value_1、value_2 包含一个整数值.例如，使用 k = 3:

Assume I have a Spark DataFrame d1 with two columns, elements_1 and elements_2, that contain sets of integers of size k, and value_1, value_2 that contain a integer value. For example, with k = 3:

d1 = 
+------------+------------+
| elements_1 | elements_2 |
+-------------------------+
| (1, 4, 3)  |  (3, 4, 5) |
| (2, 1, 3)  |  (1, 0, 2) |
| (4, 3, 1)  |  (3, 5, 6) |
+-------------------------+

我需要创建一个新列 combinations ，其中包含每对集合 elements_1 和 elements_2 的集合列表来自它们元素的所有可能组合.这些集合必须具有以下属性:

I need to create a new column combinations made that contains, for each pair of sets elements_1 and elements_2, a list of the sets from all possible combinations of their elements. These sets must have the following properties:

1. 它们的大小必须是 k+1
2. 它们必须包含 elements_1 中的集合或 elements_2
中的集合

1. Their size must be k+1
2. They must contain either the set in elements_1 or the set in elements_2

例如，从 (1, 2, 3) 和 (3, 4, 5) 我们得到 [(1, 2, 3, 4), (1, 2, 3, 5), (3, 4, 5, 1) 和 (3, 4, 5, 2)].该列表不包含 (1, 2, 5) 因为它的长度不是 3+1，并且它不包含 (1, 2, 4, 5) 因为它不包含任何原始集合.

For example, from (1, 2, 3) and (3, 4, 5) we obtain [(1, 2, 3, 4), (1, 2, 3, 5), (3, 4, 5, 1) and (3, 4, 5, 2)]. The list does not contain (1, 2, 5) because it is not of length 3+1, and it does not contain (1, 2, 4, 5) because it contains neither of the original sets.

推荐答案

您需要创建一个自定义的用户定义函数来执行转换，创建一个与 spark 兼容的 UserDefinedFunction 来自它，然后使用 withColumn 应用.所以实际上，这里有两个问题:(1)如何进行您描述的集合转换，以及(2)如何使用用户定义的函数在 DataFrame 中创建一个新列.

You need to create a custom user-defined function to perform the transformation, create a spark-compatible UserDefinedFunction from it, then apply using withColumn. So really, there are two questions here: (1) how to do the set transformation you described, and (2) how to create a new column in a DataFrame using a user-defined function.

这是设置逻辑的第一次尝试，如果它符合您的要求，请告诉我:

Here's a first shot at the set logic, let me know if it does what you're looking for:

def combo[A](a: Set[A], b: Set[A]): Set[Set[A]] = 
    a.diff(b).map(b+_) ++ b.diff(a).map(a+_)

现在创建 UDF 包装器.请注意，在幕后，这些集合都由 WrappedArrays 表示，因此我们需要处理它.通过定义一些隐式转换，可能有一种更优雅的方法来处理这个问题，但这应该有效:

Now create the UDF wrapper. Note that under the hood these sets are all represented by WrappedArrays, so we need to handle this. There's probably a more elegant way to deal with this by defining some implicit conversions, but this should work:

import scala.collection.mutable.WrappedArray
val comboWrap: (WrappedArray[Int],WrappedArray[Int])=>Array[Array[Int]] = 
    (x,y) => combo(x.toSet,y.toSet).map(_.toArray).toArray
val comboUDF = udf(comboWrap)

最后，通过创建一个新列将其应用到 DataFrame:

Finally, apply it to the DataFrame by creating a new column:

val data = Seq((Set(1,2,3),Set(3,4,5))).toDF("elements_1","elements_2")
val result = data.withColumn("result", 
    comboUDF(col("elements_1"),col("elements_2")))
result.show

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