scala-spark:如何在groupby之后过滤RDD

Question

我已经开始使用具有管道分隔字符串的RDD.我已处理数据并采用以下格式:

((0001F46468,239394055),(7665710590658745,-414963169),0,1420276980302)
((0001F46468,239394055),(8016905020647641,183812619),1,1420347885727)
((0001F46468,239394055),(6633110906332136,294201185),1,1420398323110)
((0001F46468,239394055),(6633110906332136,294201185),0,1420451687525)
((0001F46468,239394055),(7722056727387069,1396896294),1,1420537469065)
((0001F46468,239394055),(7722056727387069,1396896294),1,1420623297340)
((0001F46468,239394055),(8045651092287275,-4814845),1,1420720722185)
((0001F46468,239394055),(5170029699836178,-1332814297),0,1420750531018)
((0001F46468,239394055),(7722056727387069,1396896294),0,1420807545137)
((0001F46468,239394055),(4784119468604853,1287554938),1,1421050087824) 

只是为了对数据描述给出高级别的观点.您可以将主元组中的第一个元素(第一个元组)视为用户标识,将第二个元组视为产品标识,第三个元素是用户对产品的偏好.(为了将来的参考,我将上面的数据集标记为val userData)

我的目标是,如果用户已经为产品投放了正(1)和负(0)偏好,则只记录正数.例如:

((0001F46468,239394055),(6633110906332136,294201185),1,1420398323110)
((0001F46468,239394055),(6633110906332136,294201185),0,1420451687525)

我只想保留

((0001F46468,239394055),(6633110906332136,294201185),1,1420398323110) 

所以我按用户产品元组对用户进行了分组 (0001F46468,239394055),(6633110906332136,294201185

val groupedFiltered = userData.groupBy(x => (x._1, x._2)).map(u => {
      for(k <- u._2) {
        if(k._3 > 0)
          u
      }
    })

但那返回空元组.

所以我采取了以下方法:

val groupedFiltered = userData. groupBy(x => (x._1, x._2)).flatMap(u => u._2).filter(m => m._3 > 0)

((47734739656882457,-1782798434),(7585453414177905,-461779195),1,1422013413082)
((47734739656882457,-1782798434),(7585453414177905,-461779195),1,1422533237758)
((55218449094787901,-1374432022),(6227831620534109,1195766703),1,1420410603596)
((71212122719822610,-807015489),(6769904840922490,1642054117),1,1422549467554)
((75414197560031509,1830213715),(6724015489416254,-1389654186),1,1420196951100)
((60422797294995441,734266951),(6335216393920738,1528026712),1,1421161253600)
((35091051395844216,451349158),(8135854751464083,-1751839326),1,1422083101033)
((16647193023519619,990937787),(5384884550662007,-910998857),1,1420659873572)
((43355867025936022,-945669937),(7336240855866885,518993644),1,1420880078266)
((12188366927481231,-2007889717),(5336507724485344,363519858),1,1420827788022)

这很有希望,但看起来它所有的记录都是零,如果用户对同一项目只有1和0,我只想要保持1为1.

Answer 1

您只能保留分组结果中的最大用户首选项.

userData
 // group by user and product
 .groupBy(x => (x._1, x._2))
 // only keep the maximum user preference per user/product
 .mapValues(_.maxBy(_._3))
 // only keep the values
 .values