Question

我有一个Dataframe，希望将它分成相同数量的行。

换句话说，我想要一个数据帧列表，其中每个数据帧都是原始数据帧的脱节子集。

让我们说输入dataframer如下：

  +------------------+-----------+-----+--------------------+
  |         eventName|original_dt|count|            features|
  +------------------+-----------+-----+--------------------+
  |15.509775004326936|          0|  100|[15.5097750043269...|
  |15.509775004326936|          0|  101|[15.5097750043269...|
  |15.509775004326936|          0|  102|[15.5097750043269...|
  |15.509775004326936|          0|  103|[15.5097750043269...|
  |15.509775004326936|          0|  104|[15.5097750043269...|
  |15.509775004326936|          0|  105|[15.5097750043269...|
  |15.509775004326936|          0|  106|[15.5097750043269...|
  |15.509775004326936|          0|  107|[15.5097750043269...|
  |15.509775004326936|          0|  108|[15.5097750043269...|
  |15.509775004326936|          0|  109|[15.5097750043269...|
  |15.509775004326936|          0|  110|[15.5097750043269...|
  |15.509775004326936|          0|  111|[15.5097750043269...|
  |15.509775004326936|          0|  112|[15.5097750043269...|
  |15.509775004326936|          0|  113|[15.5097750043269...|
  |15.509775004326936|          0|  114|[15.5097750043269...|
  |15.509775004326936|          0|  115|[15.5097750043269...|
  | 43.01955000865387|          0|  116|[43.0195500086538...|
  +------------------+-----------+-----+--------------------+

我希望将其拆分为K个相等大小的数据帧。如果k = 4，则可能的结果是：

  +------------------+-----------+-----+--------------------+
  |         eventName|original_dt|count|            features|
  +------------------+-----------+-----+--------------------+
  |15.509775004326936|          0|  106|[15.5097750043269...|
  |15.509775004326936|          0|  107|[15.5097750043269...|
  |15.509775004326936|          0|  110|[15.5097750043269...|
  |15.509775004326936|          0|  111|[15.5097750043269...|
  +------------------+-----------+-----+--------------------+

  +------------------+-----------+-----+--------------------+
  |         eventName|original_dt|count|            features|
  +------------------+-----------+-----+--------------------+
  |15.509775004326936|          0|  104|[15.5097750043269...|
  |15.509775004326936|          0|  108|[15.5097750043269...|
  |15.509775004326936|          0|  112|[15.5097750043269...|
  |15.509775004326936|          0|  114|[15.5097750043269...|
  +------------------+-----------+-----+--------------------+


  +------------------+-----------+-----+--------------------+
  |         eventName|original_dt|count|            features|
  +------------------+-----------+-----+--------------------+
  |15.509775004326936|          0|  100|[15.5097750043269...|
  |15.509775004326936|          0|  105|[15.5097750043269...|
  |15.509775004326936|          0|  109|[15.5097750043269...|
  |15.509775004326936|          0|  115|[15.5097750043269...|
  +------------------+-----------+-----+--------------------+


  +------------------+-----------+-----+--------------------+
  |         eventName|original_dt|count|            features|
  +------------------+-----------+-----+--------------------+
  |15.509775004326936|          0|  101|[15.5097750043269...|
  |15.509775004326936|          0|  102|[15.5097750043269...|
  |15.509775004326936|          0|  103|[15.5097750043269...|
  |15.509775004326936|          0|  113|[15.5097750043269...|
  | 43.01955000865387|          0|  116|[43.0195500086538...|
  +------------------+-----------+-----+--------------------+

Answer 1

根据我对您的输入和所需输出的理解，您可以row numbers grouping dataframe创建one groupId。

然后，您可以filter dataframe根据您的需要在其他地方比较row number和storing。

以下是满足您需求的临时解决方案。您可以根据需要进行更改

val k = 4

val windowSpec = Window.partitionBy("grouped").orderBy("original_dt")

val newDF = dataFrame.withColumn("grouped", lit("grouping"))

var latestDF = newDF.withColumn("row", row_number() over windowSpec)

val totalCount = latestDF.count()
var lowLimit = 0
var highLimit = lowLimit + k

while(lowLimit < totalCount){
  latestDF.where(s"row <= ${highLimit} and row > ${lowLimit}").show(false)
  lowLimit = lowLimit + k
  highLimit = highLimit + k
}

我希望这会给你一个良好的开端。

Answer 2

另一种解决方案是使用限制和除外。以下程序将返回一个数组具有相同行数的数组。除了可能包含较少行的第一个。

var numberOfNew = 4
var input = List(1,2,3,4,5,6,7,8,9).toDF
var newFrames = 0 to numberOfNew map (_ => Seq.empty[Int].toDF) toArray
var size = input.count();
val limit = (size / numberOfNew).toInt

while (size > 0) {
    newFrames(numberOfNew) = input.limit(limit)
    input = input.except(newFrames(numberOfNew))
    size = size - limit
    numberOfNew = numberOfNew - 1
}

newFrames.foreach(_.show)

+-----+
|value|
+-----+
|    7|
+-----+

+-----+
|value|
+-----+
|    4|
|    8|
+-----+

+-----+
|value|
+-----+
|    5|
|    9|
+-----+

...

Answer 3

这是对斯蒂芬·施密茨（Steffen Schmitz）的改进答案，实际上这是不正确的。我已经对其进行了改进，以使其后代化。但是，我确实对大规模的性能感到好奇。

var numberOfNew = 4  
var input = Seq((1,2),(3,4),(5,6),(7,8),(9,10),(11,12)).toDF
var newFrames = 0 to numberOfNew-1 map (_ => Seq.empty[(Int, Int)].toDF) toArray
var size = input.count();
val limit = (size / numberOfNew).toInt
val limit_fract = (size / numberOfNew.toFloat)
val residual = ((limit_fract.toDouble - limit.toDouble) * size).toInt
var limit_to_use = limit

while (numberOfNew > 0) {

    if (numberOfNew == 1 && residual != 0) limit_to_use = residual  

    newFrames(numberOfNew-1) = input.limit(limit_to_use)
    input = input.except(newFrames(numberOfNew-1))
    size = size - limit
    numberOfNew = numberOfNew - 1
}

newFrames.foreach(_.show)

val singleDF = newFrames.reduce(_ union _)
singleDF.show(false)

返回单个数据帧：

+---+---+
| _1| _2|
+---+---+
|  7|  8|
|  3|  4|
| 11| 12|
+---+---+

+---+---+
| _1| _2|
+---+---+
|  5|  6|
+---+---+

+---+---+
| _1| _2|
+---+---+
|  9| 10|
+---+---+

+---+---+
| _1| _2|
+---+---+
|  1|  2|
+---+---+

Answer 4

如果您要将一个数据集划分为n个相等的数据集

double[] arraySplit = {1,1,1...,n}; //you can also divide into ratio if you change the numbers.

List<Dataset<String>> datasetList = dataset.randomSplitAsList(arraySplit,1);

Spark Scala将数据帧拆分为相同的行数

4 个答案: