Applying function to Spark Dataframe Column

2020-02-23 08:38发布

Coming from R, I am used to easily doing operations on columns. Is there any easy way to take this function that I've written in scala

def round_tenths_place( un_rounded:Double ) : Double = {
    val rounded = BigDecimal(un_rounded).setScale(1, BigDecimal.RoundingMode.HALF_UP).toDouble
    return rounded
}

And apply it to a one column of a dataframe - kind of what I hoped this would do:

 bid_results.withColumn("bid_price_bucket", round_tenths_place(bid_results("bid_price")) )

I haven't found any easy way and am struggling to figure out how to do this. There's got to be an easier way than converting the dataframe to and RDD and then selecting from rdd of rows to get the right field and mapping the function across all of the values, yeah? And also something more succinct creating a SQL table and then doing this with a sparkSQL UDF?

1条回答
Explosion°爆炸
2楼-- · 2020-02-23 09:21

You can define an UDF as follows:

val round_tenths_place_udf = udf(round_tenths_place _)
bid_results.withColumn(
  "bid_price_bucket", val round_tenths_place_udf($"bid_price"))

although built-in Round expression is using exactly the same logic as your function and should be more than enough, not to mention much more efficient:

import org.apache.spark.sql.functions.round

bid_results.withColumn("bid_price_bucket", round($"bid_price", 1))

See also:

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