Hi MoTao,
What about broadcasting the model?
Cheers,
Ndjido.
> On 08 Aug 2016, at 11:00, MoTao <mo...@sensetime.com> wrote:
>
> Hi all,
>
> I'm trying to append a column to a df.
> I understand that the new column must be created by
> 1) using literals,
> 2) transforming an existing column in df,
> or 3) generated from udf over this df
>
> In my case, the column to be appended is created by processing each row,
> like
>
> val df = spark.createDataFrame(Seq(1.0, 2.0, 3.0)).toDF("value")
> val func = udf {
> v: Double => {
> val model = initModel()
> model.process(v)
> }
> }
> val df2 = df.withColumn("valueWithBias", func(col("value")))
>
> This works fine. However, for performance reason, I want to avoid
> initModel() for each row.
> So I come with mapParitions, like
>
> val df = spark.createDataFrame(Seq(1.0, 2.0, 3.0)).toDF("value")
> val df2 = df.mapPartitions(rows => {
> val model = initModel()
> rows.map(row => model.process(row.getAs[Double](0)))
> })
> val df3 = df.withColumn("valueWithBias", df2.col("value")) // FAIL
>
> But this is wrong as a column of df2 *CANNOT* be appended to df.
>
> The only solution I got is to force mapPartitions to return a whole row
> instead of the new column,
> ( Something like "row => Row.fromSeq(row.toSeq ++
> Array(model.process(...)))" )
> which requires a lot of copy as well.
>
> I wonder how to deal with this problem with as few overhead as possible?
>
>
>
>
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