Convert Seq[Any] to Seq[String]

Hi,

Can someone please help me how to convert Seq[Any] to Seq[String]

For line

val df = row.toSeq.toDF(newCol.toSeq: _*)

I get that error message.

I converted Map "val aMap = Map("admit" -> ("description","comments"))"

to Seq

var atrb = ListBuffer[(String,String,String)]()

for((key,value) <- aMap){
  atrb += ((key, value._1, value._2))
}

var newCol = atrb.head.productIterator.toList.toSeq

Please someone help me on this.

Thanks

[srowen]

It's not really a Spark question. .toDF() takes column names. 

atrb.head.toSeq.map(_.toString)? but it's not clear what you mean the col names to be

On Fri, Dec 18, 2020 at 8:37 AM Vikas Garg <[hidden email]> wrote:

Hi,

Can someone please help me how to convert Seq[Any] to Seq[String]

For line

val df = row.toSeq.toDF(newCol.toSeq: _*)

I get that error message.

I converted Map "val aMap = Map("admit" -> ("description","comments"))"

to Seq

var atrb = ListBuffer[(String,String,String)]()

for((key,value) <- aMap){
  atrb += ((key, value._1, value._2))
}

var newCol = atrb.head.productIterator.toList.toSeq

Please someone help me on this.

Thanks

I am getting the table schema through Map which I have converted to Seq and passing to toDF

On Fri, 18 Dec 2020 at 20:13, Sean Owen <[hidden email]> wrote:

It's not really a Spark question. .toDF() takes column names. 

atrb.head.toSeq.map(_.toString)? but it's not clear what you mean the col names to be

On Fri, Dec 18, 2020 at 8:37 AM Vikas Garg <[hidden email]> wrote:

Hi,

Can someone please help me how to convert Seq[Any] to Seq[String]

For line

val df = row.toSeq.toDF(newCol.toSeq: _*)

I get that error message.

I converted Map "val aMap = Map("admit" -> ("description","comments"))"

to Seq

var atrb = ListBuffer[(String,String,String)]()

for((key,value) <- aMap){
  atrb += ((key, value._1, value._2))
}

var newCol = atrb.head.productIterator.toList.toSeq

Please someone help me on this.

Thanks

[Roland Johann]

Your code looks overly complicated and the relevant parts are missing. If possible please post the complete snippet including the retrieval/type if rows so we get the complete picture and can try to help.

For first simplification you can just convert aMap to Seq[(String, (String, String))] and further map it to flatten the nested tuple into a Seq which you then pass to toDF via var arg expansion.

Val colNames: Seq[String] = aMap.toSeq.map(kv => Seq(kv._1, kv._2._1, kv._2._2))

Depending on the type of aMap this leads to problems as we assume it to be Map[String, (String, String)].

Best Regards

Vikas Garg <[hidden email]> schrieb am Fr. 18. Dez. 2020 um 15:46:

I am getting the table schema through Map which I have converted to Seq and passing to toDF

On Fri, 18 Dec 2020 at 20:13, Sean Owen <[hidden email]> wrote:

It's not really a Spark question. .toDF() takes column names. 

atrb.head.toSeq.map(_.toString)? but it's not clear what you mean the col names to be

On Fri, Dec 18, 2020 at 8:37 AM Vikas Garg <[hidden email]> wrote:

Hi,

Can someone please help me how to convert Seq[Any] to Seq[String]

For line

val df = row.toSeq.toDF(newCol.toSeq: _*)

I get that error message.

I converted Map "val aMap = Map("admit" -> ("description","comments"))"

to Seq

var atrb = ListBuffer[(String,String,String)]()

for((key,value) <- aMap){
  atrb += ((key, value._1, value._2))
}

var newCol = atrb.head.productIterator.toList.toSeq

Please someone help me on this.

Thanks

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