将新列添加到Dataframe.新列我希望它是一个UUID生成器

Question

我想在Dataframe(UUID生成器)中添加一个新列.

UUID值看起来像 21534cf7-cff9-482a-a3a8-9e7244240da7

我的研究:

我试过withColumn火花的方法.

val DF2 = DF1.withColumn("newcolname", DF1("existingcolname" + 1)

因此DF2将newcolname在所有行中添加一个额外的列,并添加1.

根据我的要求,我想要一个可以生成UUID的新列.

Answer 1

你应该尝试这样的事情:

val sc: SparkContext = ...
val sqlContext = new SQLContext(sc)

import sqlContext.implicits._

val generateUUID = udf(() => UUID.randomUUID().toString)
val df1 = Seq(("id1", 1), ("id2", 4), ("id3", 5)).toDF("id", "value")
val df2 = df1.withColumn("UUID", generateUUID())

df1.show()
df2.show()

输出将是:

+---+-----+
| id|value|
+---+-----+
|id1|    1|
|id2|    4|
|id3|    5|
+---+-----+

+---+-----+--------------------+
| id|value|                UUID|
+---+-----+--------------------+
|id1|    1|f0cfd0e2-fbbe-40f...|
|id2|    4|ec8db8b9-70db-46f...|
|id3|    5|e0e91292-1d90-45a...|
+---+-----+--------------------+

Answer 2

您可以使用内置的 Spark SQL uuid 函数：

.withColumn("uuid", expr("uuid()"))

Scala 中的完整示例：

import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.functions._
import org.apache.spark.sql.types._

object CreateDf extends App {

  val spark = SparkSession.builder
    .master("local[*]")
    .appName("spark_local")
    .getOrCreate()
  import spark.implicits._

  Seq(1, 2, 3).toDF("col1")
    .withColumn("uuid", expr("uuid()"))
    .show(false)

}

输出：

+----+------------------------------------+
|col1|uuid                                |
+----+------------------------------------+
|1   |24181c68-51b7-42ea-a9fd-f88dcfa10062|
|2   |7cd21b25-017e-4567-bdd3-f33b001ee497|
|3   |1df7cfa8-af8a-4421-834f-5359dc3ae417|
+----+------------------------------------+