Apache Spark 的主键 [英] Primary keys with Apache Spark

问题描述

我与 Apache Spark 和 PostgreSQL 建立了 JDBC 连接，我想将一些数据插入到我的数据库中.当我使用 append 模式时，我需要为每个 DataFrame.Row 指定 id.Spark有没有办法创建主键?

I am having a JDBC connection with Apache Spark and PostgreSQL and I want to insert some data into my database. When I use append mode I need to specify id for each DataFrame.Row. Is there any way for Spark to create primary keys?

推荐答案

Scala:

如果您只需要唯一的数字，您可以使用 zipWithUniqueId 并重新创建 DataFrame.首先是一些导入和虚拟数据:

If all you need is unique numbers you can use zipWithUniqueId and recreate DataFrame. First some imports and dummy data:

import sqlContext.implicits._
import org.apache.spark.sql.Row
import org.apache.spark.sql.types.{StructType, StructField, LongType}

val df = sc.parallelize(Seq(
    ("a", -1.0), ("b", -2.0), ("c", -3.0))).toDF("foo", "bar")

提取架构以供进一步使用:

Extract schema for further usage:

val schema = df.schema

添加id字段:

val rows = df.rdd.zipWithUniqueId.map{
   case (r: Row, id: Long) => Row.fromSeq(id +: r.toSeq)}

创建数据帧:

val dfWithPK = sqlContext.createDataFrame(
  rows, StructType(StructField("id", LongType, false) +: schema.fields))

Python 中的相同内容:

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

row = Row("foo", "bar")
row_with_index = Row(*["id"] + df.columns)

df = sc.parallelize([row("a", -1.0), row("b", -2.0), row("c", -3.0)]).toDF()

def make_row(columns):
    def _make_row(row, uid):
        row_dict = row.asDict()
        return row_with_index(*[uid] + [row_dict.get(c) for c in columns])
    return _make_row

f = make_row(df.columns)

df_with_pk = (df.rdd
    .zipWithUniqueId()
    .map(lambda x: f(*x))
    .toDF(StructType([StructField("id", LongType(), False)] + df.schema.fields)))

如果您更喜欢连续数字，您可以将 zipWithUniqueId 替换为 zipWithIndex，但它会贵一点.

If you prefer consecutive number your can replace zipWithUniqueId with zipWithIndex but it is a little bit more expensive.

直接使用DataFrame API:

Directly with DataFrame API:

(通用 Scala、Python、Java、R，语法几乎相同)

以前我错过了 monotonicallyIncreasingId 函数，只要您不需要连续数字，它就可以正常工作:

Previously I've missed monotonicallyIncreasingId function which should work just fine as long as you don't require consecutive numbers:

import org.apache.spark.sql.functions.monotonicallyIncreasingId

df.withColumn("id", monotonicallyIncreasingId).show()
// +---+----+-----------+
// |foo| bar|         id|
// +---+----+-----------+
// |  a|-1.0|17179869184|
// |  b|-2.0|42949672960|
// |  c|-3.0|60129542144|
// +---+----+-----------+

虽然有用的 monotonicallyIncreasingId 是不确定的.不仅 ids 可能因执行而异，而且当后续操作包含过滤器时，如果没有额外的技巧，就不能用于识别行.

While useful monotonicallyIncreasingId is non-deterministic. Not only ids may be different from execution to execution but without additional tricks cannot be used to identify rows when subsequent operations contain filters.

注意:

也可以使用rowNumber窗口函数:

from pyspark.sql.window import Window
from pyspark.sql.functions import rowNumber

w = Window().orderBy()
df.withColumn("id", rowNumber().over(w)).show()

不幸的是:

警告窗口:没有为窗口操作定义分区！将所有数据移动到单个分区，这会导致严重的性能下降.

WARN Window: No Partition Defined for Window operation! Moving all data to a single partition, this can cause serious performance degradation.

因此，除非您有一种自然的方法来分区数据并确保唯一性，否则目前并不是特别有用.

So unless you have a natural way to partition your data and ensure uniqueness is not particularly useful at this moment.

这篇关于Apache Spark 的主键的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！