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;

我找不到使用pig内置函数删除同一会话id的相邻屏幕的方法，因此我使用了javaudf来删除相邻的屏幕名称。
我创建了一个名为getorderedscreen的JavaUDF，将该udf合并到jar中，并将该jar命名为screenfilter.jar，并在这个pig脚本中注册了该jar
下面是getOrderedScreenJavaUDF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;
}

}
输出

(S1-->S2-->S3,2,1)
(S1-->S2,1,2)
(S1-->S2-->S3-->S1,1,2)

希望这对你有帮助！。。还要再等一段时间，一些看到这个问题的好头脑会有效地回答（没有javaudf）

你可以参考这个问题，一个特工问了类似的问题。如果我正确地理解了您的问题，那么您希望从路径中删除重复项，但仅当它们彼此相邻时。所以呢 1 -> 1 -> 2 -> 1 会变成 1 -> 2 -> 1 . 如果这是正确的，那么你不能只是分组和 distinct （我相信你已经注意到了）因为它会删除所有的副本。一个简单的解决方案是编写一个udf来删除那些重复项，同时保留用户的不同路径。
自定义项：

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;
    }
}

要建造这个jar，你需要一个 hive-core.jar 以及 hadoop-core.jar ，您可以在maven存储库中找到这些。确保您获得了在您的环境中使用的hive和hadoop版本。另外，如果您计划在生产环境中运行这个，我建议您向udf添加一些异常处理。构建jar之后，导入它并运行以下查询：
查询：

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

输出：

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

希望这有帮助。