Make .csv files from arrays in Scala -

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i have 2 tuples in scala of following form:

val array1 = (bucket1, seq((datea, amount11), (dateb, amount12), (datec, amount13)))  val array2 = (bucket2, seq((datea, amount21), (dateb, amount22), (datec, amount23)))

what quickest way make .csv file in scala such that:

  1. date* pivot.
  2. bucket* column name.
  3. amount* fill table.

it needs this:

dates______________bucket1__________bucket2

datea______________amount11________amount21

dateb______________amount12________amount22

datec______________amount13________amount23

you can make shorter chaining operations, :

scala> val array1 = ("bucket1", seq(("datea", "amount11"), ("dateb", "amount12"), ("datec", "amount13")))  array1: (string, seq[(string, string)]) =           (bucket1,list((datea,amount11), (dateb,amount12), (datec,amount13)))  scala> val array2 = ("bucket2", seq(("datea", "amount21"), ("dateb", "amount22"), ("datec", "amount23"))) array2: (string, seq[(string, string)]) =            (bucket2,list((datea,amount21), (dateb,amount22), (datec,amount23)))  // single array work scala> val arrays = list(array1, array2) arrays: list[(string, seq[(string, string)])] = list(           (bucket1,list((datea,amount11), (dateb,amount12), (datec,amount13))),           (bucket2,list((datea,amount21), (dateb,amount22), (datec,amount23)))         ) 
// split between buckets , values scala> val (buckets, values) = arrays.unzip buckets: list[string] = list(bucket1, bucket2) values: list[seq[(string, string)]] = list(           list((datea,amount11), (dateb,amount12), (datec,amount13)),           list((datea,amount21), (dateb,amount22), (datec,amount23))         )  // format data // note not keep 'datex' order scala> val grouped = values.flatten                            .groupby(_._1)                            .map { case (date, list) => date::(list.map(_._2)) } grouped: scala.collection.immutable.iterable[list[string]] = list(           list(datec, amount13, amount23),            list(dateb, amount12, amount22),           list(datea, amount11, amount21)         )  // join everything, , add "dates" column in front of buckets scala> val table = ("dates"::buckets)::grouped.tolist table: list[list[string]] = list(           list(dates, bucket1, bucket2),           list(datec, amount13, amount23),           list(dateb, amount12, amount22),           list(datea, amount11, amount21)         )  // join rows ',' , lines "\n" scala> val string = table.map(_.mkstring(",")).mkstring("\n") string: string = dates,bucket1,bucket2 datec,amount13,amount23 dateb,amount12,amount22 datea,amount11,amount21

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