How do I split a column by using delimiters from another column in Spark/Scala

You need to use expr with split() to make the split dynamic

df = spark.createDataFrame([("50000.0#0#0#","#"),("0@1000.0@","@")],["VALUES","Delimiter"])
df = df.withColumn("split", F.expr("""split(VALUES, Delimiter)"""))
df.show()

+------------+---------+-----------------+
|      VALUES|Delimiter|            split|
+------------+---------+-----------------+
|50000.0#0#0#|        #|[50000.0, 0, 0, ]|
|   0@1000.0@|        @|    [0, 1000.0, ]|
+------------+---------+-----------------+

EDIT: Please check the bottom of the answer for scala version.

You can use a custom user-defined function (pyspark.sql.functions.udf) to achieve this.

from typing import List

from pyspark.sql import SparkSession
from pyspark.sql.functions import udf
from pyspark.sql.types import StringType, ArrayType


def split_col(value: StringType, delimiter: StringType) -> List[str]:
    return str(value).split(str(delimiter))


udf_split = udf(lambda x, y: split_col(x, y), ArrayType(StringType()))

spark = SparkSession.builder.getOrCreate()

df = spark.createDataFrame([
    ('50000.0#0#0#', '#'), ('0@1000.0@', '@'), ('1$', '$'), ('1000.00^Test_string', '^')
], schema='VALUES String, Delimiter String')

df = df.withColumn("split_values", udf_split(df['VALUES'], df['Delimiter']))

df.show(truncate=False)

Output

+-------------------+---------+----------------------+
|VALUES             |Delimiter|split_values          |
+-------------------+---------+----------------------+
|50000.0#0#0#       |#        |[50000.0, 0, 0, ]     |
|0@1000.0@          |@        |[0, 1000.0, ]         |
|1$                 |$        |[1, ]                 |
|1000.00^Test_string|^        |[1000.00, Test_string]|
+-------------------+---------+----------------------+

Note that the split_values column contains a list of strings. You can also update split_col function to do more changes to values.

EDIT : Scala version

import org.apache.spark.sql.functions.udf

import spark.implicits._

val data = Seq(("50000.0#0#0#", "#"), ("0@1000.0@", "@"), ("1$", "$"), ("1000.00^Test_string", "^"))
var df = data.toDF("VALUES", "Delimiter")

val udf_split_col = udf {(x:String,y:String)=> x.split(y)}

df = df.withColumn("split_values", udf_split_col(df.col("VALUES"), df.col("Delimiter")))

df.show(false)

Edit 2

To avoid the issue with special characters used in regexes, you can use char instead of a String when using the split() method as follow.

val udf_split_col = udf { (x: String, y: String) => x.split(y.charAt(0)) }

This is another way of handling this,using sparksql

df.createOrReplaceTempView("test")

spark.sql("""select VALUES,delimiter,split(values,case when delimiter in ("$","^") then concat("\\",delimiter) else delimiter end) as split_value from test""").show(false)

Note that I included the case when statement to add escape characters to handle cases for '$' and '^',otherwise it doesn't split.

+-------------------+---------+----------------------+
|VALUES             |delimiter|split_value           |
+-------------------+---------+----------------------+
|50000.0#0#0#       |#        |[50000.0, 0, 0, ]     |
|0@1000.0@          |@        |[0, 1000.0, ]         |
|1$                 |$        |[1, ]                 |
|1000.00^Test_string|^        |[1000.00, Test_string]|
+-------------------+---------+----------------------+

This is my lately solution

import java.util.regex.Pattern
val split_udf = udf((value: String, delimiter: String) => value.split(Pattern.quote(delimiter), -1))
val solution = dept.withColumn("split_values", split_udf(col("VALUES"),col("Delimiter")))
solution.show(truncate = false)

it will skip special characters in Delimiter column. Other answers not work for

("50000.0\\0\\0\\", "\\")

and linusRian's answer need to add special characters manually