问题
Below is the example for my use case.
回答1:
You can reference this question where an OP was asking something similar. If I am understanding your problem correctly, you want to remove duplicates from the path, but only when they occur next to each other. So 1 -> 1 -> 2 -> 1
would become 1 -> 2 -> 1
. If this is correct, then you can't just group and distinct
(as I'm sure you have noticed) because it will remove all duplicates. An easy solution is to write a UDF to remove those duplicates while preserving the distinct path of the user.
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;
}
}
To build this jar you will need a hive-core.jar
and hadoop-core.jar
, you can find these here in the Maven Repository. Make sure you get the version of Hive and Hadoop that you are using in your environment. Also, if you plan to run this in a production environment, I'd suggest adding some exception handling to the UDF. After the jar is built, import it and run this query:
Query:
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
Output:
s1->s2->s3 2 1
s1->s2 1 2
s1->s2->s3->s1 1 2
Hope this helps.
回答2:
Input
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;
I am not able to find a way to use Pig Builtin function to remove adjacent screens for a same session_id ,hence I have used JAVA UDF inorder to remove the adjacent screen names.
I created a JAVA UDF called GetOrderedScreen and covrerted that UDF in to jar and named that jar as ScreenFilter.jar and registered that jar in this Pig Script
Below is the Code for that GetOrderedScreen Java UDF
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;
}
}
Output
(S1-->S2-->S3,2,1)
(S1-->S2,1,2)
(S1-->S2-->S3-->S1,1,2)
Hope this helps you!.. Also Wait for some more time, Some good brains who see this Question will answer effectively(without JAVA UDF)
来源:https://stackoverflow.com/questions/32681157/how-to-find-the-pathing-flow-and-rank-them-using-pig-or-hive