[FEA] Support REGEXP_REPLACE to replace null values

[viadea]

Is your feature request related to a problem? Please describe.
Need to support REGEXP_REPLACE to replace null values.
Such as REGEXP_REPLACE(col, '\000', '')

For example, here is a minimum reproduce:

val address = Seq((1,"xxxx"),
(2,"fdsfds \u0000dfsdfs"),
(3,"sdfdsf sdfdsf \u0000sdf"))

import spark.implicits._
val df = address.toDF("id","txt")
df.write.mode("overwrite").format("parquet").save("/tmp/testparquet")
val df2=spark.read.parquet("/tmp/testparquet")
df2.createOrReplaceTempView("df2")
spark.sql("select id,txt, REGEXP_REPLACE(txt, '\000', '') AS new_txt from df2").collect()

It will fallback with below driver log:

!Expression <RegExpReplace> regexp_replace(txt#336, , , 1) cannot run on GPU because Only non-null, non-empty String literals that are not regex patterns are supported by RegExpReplace on the GPU

Below is the result:

res36: Array[org.apache.spark.sql.Row] = Array([3,sdfdsf sdfdsf ?sdf,sdfdsf sdfdsf sdf], [2,fdsfds ?dfsdfs,fdsfds dfsdfs], [1,xxxx,xxxx])

[sameerz]

Related issue: rapidsai/cudf#6196

[sameerz]

Correction, 6196 is about treating a null character as end of string. This is about replacing null values.

[viadea]

@andygrove another variation query is

REGEXP_REPLACE(col, '\000', '')

[viadea]

Confirmed it works.