[jira] [Assigned] (SPARK-38904) Low cost DataFrame schema swap util

["Apache Spark (Jira)" <ji...@apache.org>]

     [ https://issues.apache.org/jira/browse/SPARK-38904?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel ]

Apache Spark reassigned SPARK-38904:
------------------------------------

    Assignee: Apache Spark

> Low cost DataFrame schema swap util
> -----------------------------------
>
>                 Key: SPARK-38904
>                 URL: https://issues.apache.org/jira/browse/SPARK-38904
>             Project: Spark
>          Issue Type: New Feature
>          Components: SQL
>    Affects Versions: 3.2.1
>            Reporter: Rafal Wojdyla
>            Assignee: Apache Spark
>            Priority: Major
>
> This question is related to [https://stackoverflow.com/a/37090151/1661491]. Let's assume I have a pyspark DataFrame with certain schema, and I would like to overwrite that schema with a new schema that I *know* is compatible, I could do:
> {code:python}
> df: DataFrame
> new_schema = ...
> df.rdd.toDF(schema=new_schema)
> {code}
> Unfortunately this triggers computation as described in the link above. Is there a way to do that at the metadata level (or lazy), without eagerly triggering computation or conversions?
> Edit, note:
>  * the schema can be arbitrarily complicated (nested etc)
>  * new schema includes updates to description, nullability and additional metadata (bonus points for updates to the type)
>  * I would like to avoid writing a custom query expression generator, *unless* there's one already built into Spark that can generate query based on the schema/{{{}StructType{}}}
> Copied from: [https://stackoverflow.com/questions/71610435/how-to-overwrite-pyspark-dataframe-schema-without-data-scan]
> See POC of workaround/util in [https://github.com/ravwojdyla/spark-schema-utils]
> Also posted in [https://lists.apache.org/thread/5ds0f7chzp1s3h10tvjm3r96g769rvpj]



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