12

I'm having a hard time implementing something that seems like it should be very easy:

My goal is to make translations in an RDD/dataframe using a second RDD/dataframe as a lookup table or translation dictionary. I want to make these translations in multiple columns.

The easiest way to explain the problem is by example. Let's say I have as my input the following two RDDs:

Route SourceCityID DestinationCityID
A     1            2
B     1            3
C     2            1

and

CityID CityName
1      London
2      Paris
3      Tokyo

My desired output RDD is:

Route SourceCity DestinationCity
A     London     Paris
B     London     Tokyo
C     Paris      London

How should I go about it producing it?

This is an easy problem in SQL, but I don't know of obvious solutions with RDDs in Spark. The join, cogroup, etc methods seem to not be well-suited to multi-column RDDs and don't allow specifying which column to join on.

Any ideas? Is SQLContext the answer?

4

2 回答 2

7

rdd方式：

routes = sc.parallelize([("A", 1, 2),("B", 1, 3), ("C", 2, 1) ])
cities = sc.parallelize([(1, "London"),(2, "Paris"), (3, "Tokyo")])


print routes.map(lambda x: (x[1], (x[0], x[2]))).join(cities) \
.map(lambda x: (x[1][0][1], (x[1][0][0], x[1][1]))).join(cities). \
map(lambda x: (x[1][0][0], x[1][0][1], x[1][1])).collect()

哪个打印：

[('C', 'Paris', 'London'), ('A', 'London', 'Paris'), ('B', 'London', 'Tokyo')]

和 SQLContext 方式：

from pyspark.sql import HiveContext
from pyspark.sql import SQLContext

df_routes = sqlContext.createDataFrame(\
routes, ["Route", "SourceCityID", "DestinationCityID"])
df_cities = sqlContext.createDataFrame(\
cities, ["CityID", "CityName"])

temp =  df_routes.join(df_cities, df_routes.SourceCityID == df_cities.CityID) \
.select("Route", "DestinationCityID", "CityName")
.withColumnRenamed("CityName", "SourceCity")

print temp.join(df_cities, temp.DestinationCityID == df_cities.CityID) \
.select("Route", "SourceCity", "CityName")
.withColumnRenamed("CityName", "DestinationCity").collect()

哪个打印：

[Row(Route=u'C', SourceCity=u'Paris', DestinationCity=u'London'),
Row(Route=u'A', SourceCity=u'London', DestinationCity=u'Paris'),
Row(Route=u'B', SourceCity=u'London', DestinationCity=u'Tokyo')]
于 2015-10-13T08:32:44.207 回答
4

假设我们有两个包含路线和城市的 RDD：

val routes = sc.parallelize(List(("A", 1, 2),("B", 1, 3),("C", 2, 1)))
val citiesByIDRDD = sc.parallelize(List((1, "London"), (2, "Paris"), (3, "Tokyo")))

有几种方法可以实现城市查找。假设与包含许多项目的路线相比，城市查找包含的项目很少。在这种情况下，让我们从收集城市作为地图开始，该地图由驱动程序发送给每个任务。

val citiesByID = citiesByIDRDD.collectAsMap

routes.map{r => (r._1, citiesByID(r._2), citiesByID(r._3))}.collect
=> Array[(String, String, String)] = Array((A,London,Paris), (B,London,Tokyo), (C,Paris,London))

为了避免将查找表发送给每个任务，但只发送给工作人员一次，您可以扩展现有代码广播查找图。

val bCitiesByID = sc.broadcast(citiesByID)

routes.map{r => (r._1, bCitiesByID.value(r._2), bCitiesByID.value(r._3))}.collect
=> Array[(String, String, String)] = Array((A,London,Paris), (B,London,Tokyo), (C,Paris,London))

我认为这里不需要数据框，但如果你愿意，你可以：

import sqlContext.implicits._

case class Route(id: String, from: Int, to: Int)
case class City(id: Int, name: String)

val cities = List(City(1, "London"), City(2, "Paris"), City(3, "Tokyo"))
val routes = List(Route("A", 1, 2), Route("B", 1, 3), Route("C", 2, 1))

val citiesDf = cities.df
citiesDf.registerTempTable("cities")
val routesDf = routes.df
citiesDf.registerTempTable("routes")

routesDf.show
+---+----+---+
| id|from| to|
+---+----+---+
|  A|   1|  2|
|  B|   1|  3|
|  C|   2|  1|
+---+----+---+

citiesDf.show
+---+------+
| id|  name|
+---+------+
|  1|London|
|  2| Paris|
|  3| Tokyo|
+---+------+

你提到这在 SQL 中是一个简单的问题，所以我假设你可以从这里开始。执行 SQL 是这样的：

sqlContext.sql ("SELECT COUNT(*) FROM routes")
于 2015-10-13T08:20:05.353 回答