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Spark chaîne de colonne remplacer lorsqu'elle est présente dans une autre colonne (ligne)

Je souhaite supprimer des chaînes de col1 présents dans col2:

val df = spark.createDataFrame(Seq(
("Hi I heard about Spark", "Spark"),
("I wish Java could use case classes", "Java"),
("Logistic regression models are neat", "models")
)).toDF("sentence", "label")

en utilisant regexp_replace ou translate ref: fonctions spark api

val res = df.withColumn("sentence_without_label", regexp_replace 
(col("sentence") , "(?????)", "" ))

de sorte que res ressemble à ci-dessous:

enter image description here

scalaapache-sparkuser-defined-functions
8
Karol Sudol

Vous pouvez simplement utiliser regexp_replace

df5.withColumn("sentence_without_label", regexp_replace($"sentence" , lit($"label"), lit("" )))

ou vous pouvez utiliser la fonction udf simple comme ci-dessous

val df5 = spark.createDataFrame(Seq(
  ("Hi I heard about Spark", "Spark"),
  ("I wish Java could use case classes", "Java"),
  ("Logistic regression models are neat", "models")
)).toDF("sentence", "label")

val replace = udf((data: String , rep : String)=>data.replaceAll(rep, ""))

val res = df5.withColumn("sentence_without_label", replace($"sentence" , $"label"))

res.show()

Production:

+-----------------------------------+------+------------------------------+
|sentence                           |label |sentence_without_label        |
+-----------------------------------+------+------------------------------+
|Hi I heard about Spark             |Spark |Hi I heard about              |
|I wish Java could use case classes |Java  |I wish  could use case classes|
|Logistic regression models are neat|models|Logistic regression  are neat |
+-----------------------------------+------+------------------------------+
8
Shankar Koirala

Si label c'est juste un littéral c'est assez simple:

import org.Apache.spark.sql.functions._

df.withColumn("sentence_without_label", 
  regexp_replace(col("sentence"), col("label"), lit(""))).show(false)

+-----------------------------------+------+------------------------------+
|sentence                           |label |sentence_without_label        |
+-----------------------------------+------+------------------------------+
|Hi I heard about Spark             |Spark |Hi I heard about              |
|I wish Java could use case classes |Java  |I wish  could use case classes|
|Logistic regression models are neat|models|Logistic regression  are neat |
+-----------------------------------+------+------------------------------+

Dans Spark 1.6, vous pouvez faire de même avec expr:

df.withColumn(
  "sentence_without_label",
  expr("regexp_replace(sentence, label, '')"))
8
hi-zir