pyspark replace regex with regex

Question

I am trying to replaces a regex (in this case a space with a number) with

I have a Spark dataframe that contains a string column. I want to replace a regex (space plus a number) with a comma without losing the number. I have tried both of these with no luck:

df.select("A", f.regexp_replace(f.col("A"), "\s+[0-9]", ' , ').alias("replaced"))

df.select("A", f.regexp_replace(f.col("A"), "\s+[0-9]", '\s+[0-9] , ').alias("replaced"))

Any help is appreciated.

Some example inputs/outputs would be helpful. How to create good reproducible apache spark dataframe examples. — pault
– pault, Commented Aug 17, 2018 at 13:42
Can you elaborate the example with data. What is the actual value and what do u want to derive. — Manu Gupta
– Manu Gupta, Commented Aug 17, 2018 at 18:23
Possible duplicate of PySpark - String matching to create new column — Luis A.G.
– Luis A.G., Commented Mar 20, 2019 at 15:17

Luis A.G. · Accepted Answer · 2019-03-20 15:17:05Z

0

What you need is another function, regex_extract

So, you have to divide the regex and get the part you need. It could be something like this:

df.select("A", f.regexp_extract(f.col("A"), "(\s+)([0-9])", 2).alias("replaced"))

answered Mar 20, 2019 at 15:17

Luis A.G.

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