Pyspark, find substring as whole word(s)-CodePudding

I would like to see if a string column is contained in another column as a whole word. There are few approaches like using contains as described here or using array_contains as described here.

The first approach fails in the following edge case:

 --------- ----------------------- 
|candidate| sentence              |
 --------- ----------------------- 
|  su     |We saw the survivors.  |
 --------- -----------------------

su should be found as a separate word and not as a pure substring of the sentence column.

The second approach fails when the candidate is a compound word. An example is:

 ---------------- ------------------------ 
|candidate       | sentence               |
 ---------------- ------------------------ 
|  Roman emperor | He was a Roman emperor.|
 ---------------- ------------------------

The second approach fails here because it turns the sentence column to an array of tokens: [He, was, a, Roman, emperor] and none of them is equal to Roman emperor.

Is there any way to resolve this issue?

CodePudding user response：

This probably still has edge cases but I hope you get some ideas. I would use regex_extract to match the candidate against the sentence.

First, I convert the candidate to regex (ie, convert space to \s), then use regex_extract with word boundary (\b).

df = (df.withColumn('regex', F.regexp_replace(F.col('candidate'), ' ', '\\\s'))
      .withColumn('match', F.expr(r"regexp_extract(sentence, concat('\\b', regex, '\\b'), 0)")))

Result

 ------------- ----------------------- -------------- ------------- 
|    candidate|               sentence|         regex|        match|
 ------------- ----------------------- -------------- ------------- 
|           su|  We saw the survivors.|            su|             |
|Roman emperor|He was a Roman emperor.|Roman\semperor|Roman emperor|
 ------------- ----------------------- -------------- -------------