Home > Mobile >  How does regexp_replace function in PySpark?
How does regexp_replace function in PySpark?

Time:05-09

I can't seem to find anything online about how this function works. I have the following code which I'm trying to understand:

new_df = df.withColumn('a_col', regexp_replace('b_col','\\{(.*)\\}', '\\[$1\\]'))

What is being replaced here? Also, where can I find a documentation for defining the pattern to be replaced?

CodePudding user response:

Your call to REGEXP_REPLACE will find elements in curly braces and replace with the same elements in square brackets.

Here is an {ELEMENT}.

becomes

Here is an [ELEMENT].

As a side note, you probably want to use lazy dot in your regex pattern, to avoid crossing across matches. If so, then use this version:

new_df = df.withColumn('a_col', regexp_replace('b_col','\\{(.*?)\\}', '\\[$1\\]'))

CodePudding user response:

You should read about what a regular expression is and how it works. Briefly, a regular expression can check if an input string matches what the regular expression expects. For instance, your regex might be something like this:

[0-9] 

Which means the input string must be one or any number of characters between zero and 9. When you use groups in your regex (those parenthesis), the regex engine will return the substring that matches the regex inside the group. Now in your regex, anything between those curly braces( {<ANYTHING HERE>} ) will be matched and returned as the result, as the first (note the first word here) group value. regexp_replace receives a column, a regular expression, and needs to know what to do with the return value of expression, which will be the 3rd element. Now what this \\[$1\\] means, is to take the result of first group (which would be <ANYTHING HERE> in our case), and wrap it around braces. [, ], {, } must be escaped using \\ because those are valid regular expression terms, which can define class or amount of a character.

  • Related