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Posted to user@spark.apache.org by onmstester onmstester <on...@zoho.com> on 2018/05/15 10:29:39 UTC

Scala's Seq:* equivalent in java

I could not find how to pass a list to isin() filter in java, something like this could be done with scala:

val ids = Array(1,2) df.filter(df("id").isin(ids:_*)).show
But in java everything that converts java list to scala Seq fails with unsupported literal type exception:
JavaConversions.asScalaBuffer(list).toSeq()
JavaConverters.asScalaIteratorConverter(list.iterator()).asScala().toSeq().seq()


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Re: Scala's Seq:* equivalent in java

Posted by Koert Kuipers <ko...@tresata.com>.

Isn't _* varargs? So you should be able to use Java array?

On Tue, May 15, 2018, 06:29 onmstester onmstester <on...@zoho.com>
wrote:

> I could not find how to pass a list to isin() filter in java, something
> like this could be done with scala:
>
> val ids = Array(1,2)
> df.filter(df("id").isin(ids:_*)).show
>
> But in java everything that converts java list to scala Seq fails with
> unsupported literal type exception:
> JavaConversions.asScalaBuffer(list).toSeq()
>
> JavaConverters.asScalaIteratorConverter(list.iterator()).asScala().toSeq().seq()
>
> Sent using Zoho Mail <https://www.zoho.com/mail/>
>
>
>
On May 15, 2018 06:29, "onmstester onmstester" <on...@zoho.com> wrote:

I could not find how to pass a list to isin() filter in java, something
like this could be done with scala:

val ids = Array(1,2)
df.filter(df("id").isin(ids:_*)).show

But in java everything that converts java list to scala Seq fails with
unsupported literal type exception:
JavaConversions.asScalaBuffer(list).toSeq()
JavaConverters.asScalaIteratorConverter(list.iterator()).asScala().toSeq().seq()

Sent using Zoho Mail <https://www.zoho.com/mail/>