Basically you are creating a dataframe (a dataframe is a *Dataset* organized
into named columns. It is conceptually equivalent to a table in a
relational database) out of RDD here.
scala> val rdd = sc.parallelize( List(3, 2, 1, 4, 0))
rdd: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[19] at
parallelize at <console>:24
scala> // convert it to a dataframe
scala> val df = rdd.toDF
df: org.apache.spark.sql.DataFrame = [value: int]
scala> df.filter('value > 2).show
+-----+
|value|
+-----+
| 3|
| 4|
+-----+
HTH
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On Sun, 6 Feb 2022 at 11:51, <capitnfrak...@free.fr> wrote:
> for example, this work for RDD object:
>
> scala> val li = List(3,2,1,4,0)
> li: List[Int] = List(3, 2, 1, 4, 0)
>
> scala> val rdd = sc.parallelize(li)
> rdd: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[0] at
> parallelize at <console>:24
>
> scala> rdd.filter(_ > 2).collect()
> res0: Array[Int] = Array(3, 4)
>
>
> After I convert RDD to the dataframe, the filter won't work:
>
> scala> val df = rdd.toDF
> df: org.apache.spark.sql.DataFrame = [value: int]
>
> scala> df.filter(_ > 2).show()
> <console>:24: error: value > is not a member of org.apache.spark.sql.Row
> df.filter(_ > 2).show()
>
>
> But this can work:
>
> scala> df.filter($"value" > 2).show()
> +-----+
> |value|
> +-----+
> | 3|
> | 4|
> +-----+
>
>
> Where to check all the methods supported by dataframe?
>
>
> Thank you.
> Frakass
>
>
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