Question

我想使用没有Pivot功能的spark scala转换下表

我使用Spark 1.5.1并且在1.5.1中不支持Pivot功能。请建议适当的方法转换下表：

Customer Day   Sales
1        Mon    12
1        Tue    10
1        Thu    15
1        Fri     2
2        Sun    10
2        Wed     5
2        Thu     4
2        Fri     3

输出表：

Customer Sun Mon Tue Wed Thu Fri
   1      0   12  10   0  15  2
   2     10    0   0   5  4   3

以下代码无效，因为我使用的是Spark 1.5.1，并且可以从Spark 1.6获得数据透视功能：

    var Trans = Cust_Sales.groupBy("Customer").Pivot("Day").sum("Sales")

Answer 1

不确定效率如何，但您可以使用collect获取所有不同的日期，然后添加这些列，然后使用groupBy和sum：

// get distinct days from data (this assumes there are not too many of them):
val days: Array[String] = df.select("Day")
    .distinct()
    .collect()
    .map(_.getAs[String]("Day"))

// add column for each day with the Sale value if days match:
val withDayColumns = days.foldLeft(df) { 
    case (data, day) => data.selectExpr("*", s"IF(Day = '$day', Sales, 0) AS $day")
}

// wrap it up 
val result = withDayColumns
   .drop("Day")
   .drop("Sales")
   .groupBy("Customer")
   .sum(days: _*)

result.show()

打印（几乎）你想要的东西：

+--------+--------+--------+--------+--------+--------+--------+
|Customer|sum(Tue)|sum(Thu)|sum(Sun)|sum(Fri)|sum(Mon)|sum(Wed)|
+--------+--------+--------+--------+--------+--------+--------+
|       1|      10|      15|       0|       2|      12|       0|
|       2|       0|       4|      10|       3|       0|       5|
+--------+--------+--------+--------+--------+--------+--------+

如果需要，我会留给您重命名/重新排序。

Answer 2

如果你正在使用python下面的代码可能会有所帮助。假设您要转置spark DataFrame df：

pandas_df = df.toPandas().transpose().reset_index()
transposed_df = sqlContext.createDataFrame(pandas_df)
transposed_df.show()

Answer 3

考虑一个有6列的数据框，我们希望按前4列分组并在col5上进行聚合，同时在col6上聚合（比如它上面的总和）。因此，假设你不能使用spark 1.6版本，那么下面的代码可以写成（在spark 1.5中）：

val pivotedDf = df_to_pivot
    .groupBy(col1,col2,col3,col4)
    .pivot(col5)
    .agg(sum(col6))

以下是具有相同输出但不使用内置数据透视功能的代码：

import scala.collection.SortedMap

//Extracting the col5 distinct values to create the new columns
val distinctCol5Values = df_to_pivot
    .select(col(col5))
    .distinct
    .sort(col5)  // ensure that the output columns are in a consistent logical order
    .map(_.getString(0))
    .toArray
    .toSeq

//Grouping by the data frame to be pivoted on col1-col4
val pivotedAgg = df_to_pivot.rdd
      .groupBy{row=>(row.getString(col1Index),
      row.getDate(col2Index),
      row.getDate(col3Index),
      row.getString(col4Index))}

//Initializing a List of tuple of (String, double values) to be filled in the columns that will be created
val pivotColListTuple = distinctCol5Values.map(ft=> (ft,0.0))
// Using Sorted Map to ensure the order is maintained
var distinctCol5ValuesListMap = SortedMap(pivotColListTuple : _*)
//Pivoting the data on col5 by opening the grouped data
val pivotedRDD = pivotedAgg.map{groupedRow=>
    distinctCol5ValuesListMap = distinctCol5ValuesListMap.map(ft=> (ft._1,0.0))
     groupedRow._2.foreach{row=>
//Updating the distinctCol5ValuesListMap values to reflect the changes
//Change this part accordingly to what you want
       distinctCol5ValuesListMap = distinctCol5ValuesListMap.updated(row.getString(col5Index),
         distinctCol5ValuesListMap.getOrElse(row.getString(col5Index),0.0)+row.getDouble(col6Index))
     }        
    Row.fromSeq(Seq(groupedRow._1._1,groupedRow._1._2,groupedRow._1._3,groupedRow._1._4) ++ distinctCol5ValuesListMap.values.toSeq)
  }

//Consructing the structFields for new columns
val colTypesStruct = distinctCol5ValuesListMap.map(colName=>StructField(colName._1,DoubleType))
//Adding the first four column structFields with the new columns struct
val opStructType = StructType(Seq(StructField(col1Name,StringType),
                                     StructField(col2Name,DateType),
                                     StructField(col3Name,DateType),
                                     StructField(col4Name,StringType)) ++ colTypesStruct )

//Creating the final data frame
val pivotedDF = sqlContext.createDataFrame(pivotedRDD,opStructType)

如何在Spark 1.5中转置数据帧（没有可用的枢轴运算符）？

3 个答案: