Python 如何找到路径流并使用pig或hive对其进行排序？

下面是我的用例示例

您可以参考OP询问类似问题的地方。如果我正确理解了您的问题，那么您希望从路径中删除重复项，但仅当它们彼此相邻时。所以

1->1->2->1

将变成

1->2->1

。如果这是正确的，那么您不能仅仅分组和

区分（我相信您已经注意到），因为它将删除所有重复项。一个简单的解决方案是编写一个UDF来删除这些重复项，同时保留用户的不同路径

UDF：

package something;

import java.util.ArrayList;
import org.apache.hadoop.hive.ql.exec.UDF;
import org.apache.hadoop.io.Text;

public class RemoveSequentialDuplicatesUDF extends UDF {
    public ArrayList<Text> evaluate(ArrayList<Text> arr) {
        ArrayList<Text> newList = new ArrayList<Text>();
        newList.add(arr.get(0));
        for (int i = 1; i < arr.size(); i++) {

            String front = arr.get(i).toString();
            String back  = arr.get(i-1).toString();

            if (!back.equals(front)) {
                newList.add(arr.get(i));
            }
        }
        return newList;
    }
}

add jar /path/to/jars/brickhouse-0.7.1.jar;
add jar /path/to/jars/hive_common-SNAPSHOT.jar;
create temporary function collect as "brickhouse.udf.collect.CollectUDAF";
create temporary function remove_dups as "something.RemoveSequentialDuplicatesUDF";

select screen_flow, count
  , dense_rank() over (order by count desc) rank
from (
  select screen_flow
    , count(*) count
  from (
    select session_id
      , concat_ws("->", remove_dups(screen_array)) screen_flow
    from (
      select session_id
        , collect(screen_name) screen_array
      from (
        select *
        from database.table
        order by screen_launch_time ) a
      group by session_id ) b
    ) c
  group by screen_flow ) d

s1->s2->s3      2       1
s1->s2          1       2
s1->s2->s3->s1  1       2

输出：

package something;

import java.util.ArrayList;
import org.apache.hadoop.hive.ql.exec.UDF;
import org.apache.hadoop.io.Text;

public class RemoveSequentialDuplicatesUDF extends UDF {
    public ArrayList<Text> evaluate(ArrayList<Text> arr) {
        ArrayList<Text> newList = new ArrayList<Text>();
        newList.add(arr.get(0));
        for (int i = 1; i < arr.size(); i++) {

            String front = arr.get(i).toString();
            String back  = arr.get(i-1).toString();

            if (!back.equals(front)) {
                newList.add(arr.get(i));
            }
        }
        return newList;
    }
}

add jar /path/to/jars/brickhouse-0.7.1.jar;
add jar /path/to/jars/hive_common-SNAPSHOT.jar;
create temporary function collect as "brickhouse.udf.collect.CollectUDAF";
create temporary function remove_dups as "something.RemoveSequentialDuplicatesUDF";

select screen_flow, count
  , dense_rank() over (order by count desc) rank
from (
  select screen_flow
    , count(*) count
  from (
    select session_id
      , concat_ws("->", remove_dups(screen_array)) screen_flow
    from (
      select session_id
        , collect(screen_name) screen_array
      from (
        select *
        from database.table
        order by screen_launch_time ) a
      group by session_id ) b
    ) c
  group by screen_flow ) d

s1->s2->s3      2       1
s1->s2          1       2
s1->s2->s3->s1  1       2

希望这有帮助。
输入

990004916946605-1404157897784,S1,1404157898275
990004916946605-1404157897784,S1,1404157898286
990004916946605-1404157897784,S2,1404157898337
990004947764274-1435162269418,S1,1435162274044
990004947764274-1435162269418,S2,1435162274057
990004947764274-1435162269418,S3,1435162274081
990004947764274-1435162287965,S2,1435162690002
990004947764274-1435162287965,S1,1435162690001
990004947764274-1435162287965,S3,1435162690003
990004947764274-1435162287965,S1,1435162690004
990004947764274-1435162212345,S1,1435168768574
990004947764274-1435162212345,S2,1435168768585
990004947764274-1435162212345,S3,1435168768593


register /home/cloudera/jar/ScreenFilter.jar;

screen_records =  LOAD '/user/cloudera/inputfiles/screen.txt' USING PigStorage(',') AS(session_id:chararray,screen_name:chararray,launch_time:long);

screen_rec_order =  ORDER screen_records  by launch_time ASC;

session_grped = GROUP screen_rec_order BY session_id;

eached = FOREACH session_grped
                      {
                         ordered = ORDER screen_rec_order by launch_time;

                        GENERATE group as session_id, REPLACE(BagToString(ordered.screen_name),'_','-->') as screen_str;

                      };

screen_each  =  FOREACH eached GENERATE session_id, GetOrderedScreen(screen_str) as screen_pattern;

screen_grp   = GROUP screen_each by screen_pattern;

screen_final_each = FOREACH screen_grp GENERATE group as screen_pattern, COUNT(screen_each) as pattern_cnt;

ranker = RANK screen_final_each BY pattern_cnt DESC DENSE;

output_data = FOREACH ranker GENERATE screen_pattern, pattern_cnt, $0 as rank_value;

dump output_data;

我无法找到使用Pig内置函数删除同一会话id的相邻屏幕的方法，因此我使用JAVA UDF来删除相邻的屏幕名称

我创建了一个名为GetOrderedScreen的JAVA UDF，并将该UDF添加到jar中，将该jar命名为ScreenFilter.jar，并在这个Pig脚本中注册了该jar

下面是GetOrderedScreen Java UDF的代码

public class GetOrderedScreen extends EvalFunc<String> {


@Override
public String exec(Tuple input) throws IOException {


    String incoming_screen_str= (String)input.get(0);
    String outgoing_screen_str ="";
    String screen_array[] =incoming_screen_str.split("-->");

    String full_screen=screen_array[0];

  for (int i=0; i<screen_array.length;i++)
  {
     String prefix_screen=  screen_array[i];
     String suffix_screen="";
     int j=i+1;

     if(j< screen_array.length)
     {
         suffix_screen  = screen_array[j];
     }


 if (!prefix_screen.equalsIgnoreCase(suffix_screen))
     {
     full_screen = full_screen+ "-->" +suffix_screen;
     }

  }
  outgoing_screen_str =full_screen.substring(0, full_screen.lastIndexOf("-->")); 


  return outgoing_screen_str;

}

希望这对你有帮助！。。再等一段时间，一些看到这个问题的好头脑会有效地回答（没有JAVA UDF）
Ok。。我得到了要求。。尝试了一段时间以获得实际输出。。看起来我们需要一个java UDF来删除相邻的相同屏幕名。。让我再考虑一下。我想给出一个解决方案，而不是UDF@explorethis：是否可以添加纯文本格式的输入和预期输出？我想试试这个用例。@MuraliRao:我不能按同样的顺序格式化。它以纯文本的形式放置。如果你能分享你的电子邮件，我将为此发送一份word文档：）@Surender Raja-你能给我关于以下场景Geat的建议吗！！我将尝试一下，并让你知道：）根据我所看到的，我理解你的方法，希望能解决我的问题。太棒了！！它很好吃。感谢GoBrewerI我需要你的专业知识来解决我的新帖子中的问题。你能用pig重写上面的蜂巢逻辑吗？太棒了！！让我试试这个。一个小问题。类似于排名，我也需要级别。如何从上面的输出中找到级别？例如，（S1-->S2-->S3为级别2）（S1-->S2为级别1）（S1-->S2-->S3-->S1为级别3）…基本上计算箭头的数量（->）。。我编写了一个python脚本来计算子字符串以获得结果。谢谢你的帮助！我现在正试图在上面的pig脚本中添加两个新的列application_name和date，但却遇到了麻烦。你能帮忙吗？你的申请名称和日期是什么。。是否要对所有输出记录保持相同