DataFrame 化的 zipWithIndex

Question

我正在尝试解决向数据集添加序列号这一由来已久的问题。我正在使用 DataFrame，但似乎没有等效于 RDD.zipWithIndex 的 DataFrame。另一方面，以下内容或多或少地按照我想要的方式工作：

val origDF = sqlContext.load(...)    

val seqDF= sqlContext.createDataFrame(
    origDF.rdd.zipWithIndex.map(ln => Row.fromSeq(Seq(ln._2) ++ ln._1.toSeq)),
    StructType(Array(StructField("seq", LongType, false)) ++ origDF.schema.fields)
)

在我的实际应用程序中，origDF 不会直接从文件中加载 -- 它将通过将 2-3 个其他 DataFrame 连接在一起来创建，并将包含超过 1 亿行。

有更好的方法吗？我可以做些什么来优化它？

Answer 1

以下内容是代表 David Griffin 发布的（有问题的编辑）。

能歌善舞的dfZipWithIndex方法。可以设置起始偏移量（默认为1），索引列名（默认为"id"），将列放在前面或后面：

import org.apache.spark.sql.DataFrame
import org.apache.spark.sql.types.{LongType, StructField, StructType}
import org.apache.spark.sql.Row


def dfZipWithIndex(
  df: DataFrame,
  offset: Int = 1,
  colName: String = "id",
  inFront: Boolean = true
) : DataFrame = {
  df.sqlContext.createDataFrame(
    df.rdd.zipWithIndex.map(ln =>
      Row.fromSeq(
        (if (inFront) Seq(ln._2 + offset) else Seq())
          ++ ln._1.toSeq ++
        (if (inFront) Seq() else Seq(ln._2 + offset))
      )
    ),
    StructType(
      (if (inFront) Array(StructField(colName,LongType,false)) else Array[StructField]()) 
        ++ df.schema.fields ++ 
      (if (inFront) Array[StructField]() else Array(StructField(colName,LongType,false)))
    )
  ) 
}

Answer 2

从 Spark 1.5 开始，Window 表达式被添加到 Spark。您现在可以使用 org.apache.spark.sql.expressions.row_number 而不是必须将 DataFrame 转换为 RDD。请注意，我发现上述 dfZipWithIndex 的性能明显快于以下算法。但我发布它是因为：

其他人会被诱惑尝试这个
也许有人可以优化下面的表达式

无论如何，这对我有用：

import org.apache.spark.sql.expressions._

df.withColumn("row_num", row_number.over(Window.partitionBy(lit(1)).orderBy(lit(1))))

请注意，我使用 lit(1) 进行分区和排序——这使得所有内容都在同一个分区中，并且似乎保留了 DataFrame 的原始顺序，但我想这就是它减慢速度的原因。

我在一个有 7,000,000 行的 4 列 DataFrame 上测试了它，这个和上面的 dfZipWithIndex 之间的速度差异很大（就像我说的，RDD 函数是快得多）。

Answer 3

PySpark 版本：

from pyspark.sql.types import LongType, StructField, StructType

def dfZipWithIndex (df, offset=1, colName="rowId"):
    '''
        Enumerates dataframe rows is native order, like rdd.ZipWithIndex(), but on a dataframe 
        and preserves a schema

        :param df: source dataframe
        :param offset: adjustment to zipWithIndex()'s index
        :param colName: name of the index column
    '''

    new_schema = StructType(
                    [StructField(colName,LongType(),True)]        # new added field in front
                    + df.schema.fields                            # previous schema
                )

    zipped_rdd = df.rdd.zipWithIndex()

    new_rdd = zipped_rdd.map(lambda (row,rowId): ([rowId +offset] + list(row)))

    return spark.createDataFrame(new_rdd, new_schema)

还创建了一个 jira 以在 Spark 中本地添加此功能：https://issues.apache.org/jira/browse/SPARK-23074

Answer 4

从 Spark 1.6 开始，有一个名为 monotonically_increasing_id()
的函数它为每一行生成一个具有唯一 64 位单调索引的新列
但这并不重要，每个分区都开始一个新的范围，所以我们必须在使用之前计算每个分区偏移量。
试图提供一个“无rdd”的解决方案，我最终得到了一些collect()，但它只收集偏移量，每个分区一个值，所以它不会导致OOM

def zipWithIndex(df: DataFrame, offset: Long = 1, indexName: String = "index") = {
    val dfWithPartitionId = df.withColumn("partition_id", spark_partition_id()).withColumn("inc_id", monotonically_increasing_id())

    val partitionOffsets = dfWithPartitionId
        .groupBy("partition_id")
        .agg(count(lit(1)) as "cnt", first("inc_id") as "inc_id")
        .orderBy("partition_id")
        .select(sum("cnt").over(Window.orderBy("partition_id")) - col("cnt") - col("inc_id") + lit(offset) as "cnt" )
        .collect()
        .map(_.getLong(0))
        .toArray
        
     dfWithPartitionId
        .withColumn("partition_offset", udf((partitionId: Int) => partitionOffsets(partitionId), LongType)(col("partition_id")))
        .withColumn(indexName, col("partition_offset") + col("inc_id"))
        .drop("partition_id", "partition_offset", "inc_id")
}

此解决方案不重新打包原始行，也不对原始巨大的数据帧进行重新分区，因此在现实世界中速度相当快： 200GB 的 CSV 数据（4300 万行，150 列）在 240 个内核上在 2 分钟内读取、索引和打包到 parquet
测试我的解决方案后，我运行慢了 20 秒
您可能想要或不想使用 dfWithPartitionId.cache()，取决于任务

Answer 5

@Evgeny，很有趣。请注意，当您有空分区时会出现错误（数组缺少这些分区索引，至少我使用 spark 1.6 时会发生这种情况），因此我将数组转换为 Map(partitionId -> offsets)。

此外，我取出了 monotonically_increasing_id 的来源，使每个分区中的 "inc_id" 从 0 开始。

这是一个更新版本：

import org.apache.spark.sql.catalyst.expressions.LeafExpression
import org.apache.spark.sql.catalyst.InternalRow
import org.apache.spark.sql.types.LongType
import org.apache.spark.sql.catalyst.expressions.Nondeterministic
import org.apache.spark.sql.catalyst.expressions.codegen.GeneratedExpressionCode
import org.apache.spark.sql.catalyst.expressions.codegen.CodeGenContext
import org.apache.spark.sql.types.DataType
import org.apache.spark.sql.DataFrame
import org.apache.spark.sql.functions._
import org.apache.spark.sql.Column
import org.apache.spark.sql.expressions.Window

case class PartitionMonotonicallyIncreasingID() extends LeafExpression with Nondeterministic {

  /**
   * From org.apache.spark.sql.catalyst.expressions.MonotonicallyIncreasingID
   *
   * Record ID within each partition. By being transient, count's value is reset to 0 every time
   * we serialize and deserialize and initialize it.
   */
  @transient private[this] var count: Long = _

  override protected def initInternal(): Unit = {
    count = 1L // notice this starts at 1, not 0 as in org.apache.spark.sql.catalyst.expressions.MonotonicallyIncreasingID
  }

  override def nullable: Boolean = false

  override def dataType: DataType = LongType

  override protected def evalInternal(input: InternalRow): Long = {
    val currentCount = count
    count += 1
    currentCount
  }

  override def genCode(ctx: CodeGenContext, ev: GeneratedExpressionCode): String = {
    val countTerm = ctx.freshName("count")
    ctx.addMutableState(ctx.JAVA_LONG, countTerm, s"$countTerm = 1L;")
    ev.isNull = "false"
    s"""
      final ${ctx.javaType(dataType)} ${ev.value} = $countTerm;
      $countTerm++;
    """
  }
}

object DataframeUtils {
  def zipWithIndex(df: DataFrame, offset: Long = 0, indexName: String = "index") = {
    // from 
    val dfWithPartitionId = df.withColumn("partition_id", spark_partition_id()).withColumn("inc_id", new Column(PartitionMonotonicallyIncreasingID()))

    // collect each partition size, create the offset pages
    val partitionOffsets: Map[Int, Long] = dfWithPartitionId
      .groupBy("partition_id")
      .agg(max("inc_id") as "cnt") // in each partition, count(inc_id) is equal to max(inc_id) (I don't know which one would be faster)
      .select(col("partition_id"), sum("cnt").over(Window.orderBy("partition_id")) - col("cnt") + lit(offset) as "cnt")
      .collect()
      .map(r => (r.getInt(0) -> r.getLong(1)))
      .toMap

    def partition_offset(partitionId: Int): Long = partitionOffsets(partitionId)
    val partition_offset_udf = udf(partition_offset _)
    // and re-number the index
    dfWithPartitionId
      .withColumn("partition_offset", partition_offset_udf(col("partition_id")))
      .withColumn(indexName, col("partition_offset") + col("inc_id"))
      .drop("partition_id")
      .drop("partition_offset")
      .drop("inc_id")
  }
}

Answer 6

Spark Java API 版本：

我已经实现了@Evgeny 的来对 Java 中的 DataFrames 执行 zipWithIndex 并想分享代码。

它还包含@fylb 在他的 . I can confirm for Spark 2.4 that the execution fails when the entries returned by spark_partition_id() do not start with 0 or do not increase sequentially. As this function is documented 中提供的改进是非确定性的，很可能会出现上述情况之一。一个示例是通过增加分区计数来触发的。

java实现如下：

public static Dataset<Row> zipWithIndex(Dataset<Row> df, Long offset, String indexName) {
        Dataset<Row> dfWithPartitionId = df
                .withColumn("partition_id", spark_partition_id())
                .withColumn("inc_id", monotonically_increasing_id());

        Object partitionOffsetsObject = dfWithPartitionId
                .groupBy("partition_id")
                .agg(count(lit(1)).alias("cnt"), first("inc_id").alias("inc_id"))
                .orderBy("partition_id")
                .select(col("partition_id"), sum("cnt").over(Window.orderBy("partition_id")).minus(col("cnt")).minus(col("inc_id")).plus(lit(offset).alias("cnt")))
                .collect();
        Row[] partitionOffsetsArray = ((Row[]) partitionOffsetsObject);
        Map<Integer, Long> partitionOffsets = new HashMap<>();
        for (int i = 0; i < partitionOffsetsArray.length; i++) {
            partitionOffsets.put(partitionOffsetsArray[i].getInt(0), partitionOffsetsArray[i].getLong(1));
        }

        UserDefinedFunction getPartitionOffset = udf(
                (partitionId) -> partitionOffsets.get((Integer) partitionId), DataTypes.LongType
        );

        return dfWithPartitionId
                .withColumn("partition_offset", getPartitionOffset.apply(col("partition_id")))
                .withColumn(indexName, col("partition_offset").plus(col("inc_id")))
                .drop("partition_id", "partition_offset", "inc_id");
    }

Answer 7

我已经在 Python 3.7 上将@Tagar 的版本修改为运行，想要分享：

def dfZipWithIndex (df, offset=1, colName="rowId"):
'''
    Enumerates dataframe rows is native order, like rdd.ZipWithIndex(), but on a dataframe
    and preserves a schema

    :param df: source dataframe
    :param offset: adjustment to zipWithIndex()'s index
    :param colName: name of the index column
'''

new_schema = StructType(
                [StructField(colName,LongType(),True)]        # new added field in front
                + df.schema.fields                            # previous schema
            )

zipped_rdd = df.rdd.zipWithIndex()

new_rdd = zipped_rdd.map(lambda args: ([args[1] + offset] + list(args[0])))      # use this for python 3+, tuple gets passed as single argument so using args and [] notation to read elements within args
return spark.createDataFrame(new_rdd, new_schema)

Answer 8

这是我的建议，优点是：

不涉及我们DataFrame的[=13]serialization/deserialization^[1] =]s.
它的逻辑是极简主义的，只依赖于RDD.zipWithIndex。

它的主要缺点是：

无法直接从非 JVM API（pySpark、SparkR）使用它。
它必须在 package org.apache.spark.sql;.

进口：

import org.apache.spark.rdd.RDD
import org.apache.spark.sql.catalyst.InternalRow
import org.apache.spark.sql.execution.LogicalRDD
import org.apache.spark.sql.functions.lit

/**
  * Optimized Spark SQL equivalent of RDD.zipWithIndex.
  *
  * @param df
  * @param indexColName
  * @return `df` with a column named `indexColName` of consecutive unique ids.
  */
def zipWithIndex(df: DataFrame, indexColName: String = "index"): DataFrame = {
  import df.sparkSession.implicits._

  val dfWithIndexCol: DataFrame = df
    .drop(indexColName)
    .select(lit(0L).as(indexColName), $"*")

  val internalRows: RDD[InternalRow] = dfWithIndexCol
    .queryExecution
    .toRdd
    .zipWithIndex()
    .map {
      case (internalRow: InternalRow, index: Long) =>
        internalRow.setLong(0, index)
        internalRow
    }

  Dataset.ofRows(
    df.sparkSession,
    LogicalRDD(dfWithIndexCol.schema.toAttributes, internalRows)(df.sparkSession)
  )

^[1]: (from/to InternalRow的底层字节数组<--> GenericRow的底层JVM对象集合Array[Any]).

DataFrame 化的 zipWithIndex

DataFrame-ified zipWithIndex

scala

apache-spark

apache-spark-sql