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There is a column batch in dataframe. It has values like '9%','$5', etc.

I need use regex_replace in a way that it removes the special characters from the above example and keep just the numeric part.

Examples like 9 and 5 replacing 9% and $5 respectively in the same column.

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df.withColumn("batch",regexp_replace(col("batch"), "/[^0-9]+/", ""))
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8 Comments

creating a column batch with only numbers as value... Every thing apart from numbers will be replaced with blank
Sorry for the delay in response. I tried your code and it worked fine. Thanks a lott
what if we want to do it for all columns in single command, please
@dileepvarma regex_replace can be used in select... df.select(regexp_replace(col("col1"), "/[^0-9]+/", ""),regexp_replace(col("col2"), "/[^0-9]+/", ""))
val cols = df.columns.map(x=>regexp_replace(col(x), "/[^0-9]+/", ""))... df.select(cols:_*)
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You can use this regex:

\W+

\W - matches any non-word character (equal to [^a-zA-Z0-9_])

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What have you tried so far?

select regexp_replace("'$5','9%'","[^0-9A-Za-z]","")
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