Back-reference in Spark DataFrame `regexp_replace`

Question

I was recently trying to answer a question, when I realised I didn't know how to use a back-reference in a regexp with Spark DataFrames.

For instance, with sed, I could do

> echo 'a1
b22
333' | sed "s/\([0-9][0-9]*\)/;\1/"                                                                                                   

a;1
b;22
;333

But with Spark DataFrames I can't:

val df = List("a1","b22","333").toDF("str")
df.show

+---+
|str|
+---+
| a1|
|b22|
|333|
+---+

val res = df  .withColumn("repBackRef",regexp_replace('str,"(\\d+)$",";\\1"))
res.show

+---+-----------+
|str|repBackRef|
+---+----------+
| a1|       a;1|
|b22|       b;1|
|333|        ;1|
+---+----------+

Just to make it clear: I don't want the result in this particular case, I would like a solution that would be as generic as back reference in, for instance, sed.

Note also that using regexp_extract is lacking since it behaves badly when no matching:

val res2 = df
  .withColumn("repExtract",regexp_extract('str,"^([A-z])+?(\\d+)$",2))
res2.show

So that you are forced to use one column per pattern to extract as I did in the said answer.

Thanks!

Answer 1

You need to use the $+numeric_ID backreference syntax:

.withColumn("repBackRef",regexp_replace('str,"(\\d+)$",";$1"))
                                                         ^^