CODEWITHSUNDEEP
scala
Jake
Jake

Reputation: 4650

scala: foldLeft with zipWithIndex

The following code works:

likertRoundDfSeq:Seq[DataFrame] =    ......
likertRoundDfSeq match
  {

    case head :: tail => tail.foldLeft(head){(dforg,df1)=>
      DataFrameUtils.join(dforg,devianceFromAverageOneRound(df1),"A_RowId")
    }
  }

BUT, I need to add an index as an additional argument to devianceFromAverageOneRound

I thought of doing this with zipWithIndex Perhaps, like this:

 likertRoundDfSeq match
  {

    case head :: tail => tail.zipWithIndex.foldLeft(head){(dforg,df1)=>
      DataFrameUtils.join(dforg,devianceFromAverageOneRound(df1,*myzipindex*),"A_RowId" )
    }
  }

But I am not sure how to break out the dataframe and idx in this case. Intellij does not seem to guide me on this, so I'm a bit lost

Any advice would be appreciated

Upvotes: 1

Views: 940

Answers (1)

Leo C
Leo C

Reputation: 22439

The tail of your DF Seq is now a list of Tuple2[DataFrame, Long], hence your foldLeft should look like the following:

case head :: tail => tail.zipWithIndex.foldLeft(head){ (dforg, df1) =>
  DataFrameUtils.join(dforg, devianceFromAverageOneRound(df1._1, df1._2), "A_RowId")

This assumes your new devianceFromAverageOneRound(DataFrame, Long) still returns a DataFrame (and not Tuple2[DataFrame, Long]).

Upvotes: 2

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