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scala:用zipwithindex左折叠

scala

1

Jake · 技术社区 · 7 年前

作品

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

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

devianceFromAverageOneRound

我想用这个 zipWithIndex

 likertRoundDfSeq match
  {

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

但在这种情况下,我不确定如何分解数据帧和IDX。Intellij似乎没有指导我,所以我有点迷路了。

1 回复 | 直到 7 年前

1

2

Leo C 7 年前

df-seq的尾部现在是tuple2[dataframe,long]的列表,因此 foldLeft

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

这假设您的新 devianceFromAverageOneRound(DataFrame, Long) DataFrame (而不是 Tuple2[DataFrame, Long]