Reduce elements in a java list using Java8 streams

Question

I have a java list that contains elements like below:

Input List name is itemsList:

List<Items> itemsList ;

Items class:

List<String> Identifier;
List<String> websites;

Input Elements for the list:

Identifier    websites
 id1,id2      site1
 id2,id3       site3
 id5           site5
 id1           site6
 id5          site7 
 id6           site8

Result list:

   Identifier         websites
     id1,id2,id3      site1,site3,site6   
     id5                site5,site7
     id6                site8

As you can see from the result : Identifier should be grouped together if anyone Identifier is present from the other row and all websites should be combined together as well

Here is what I tried :

    itemsList.stream().reduce((v1, v2) ->
    {
        //see if identofier overlaps
        if (!Collections.disjoint(v1.getIdentifier(), (v2.getIdentifier()))) {
            v1.getIdentifier().addAll(v2.getIdentifier());
            v1.getwebsites().addAll(v2.getwebsites());
     
        } 
        return v1;
    });

my solution doesn't help much..As it only reduces the first row. I know it is not an easy one to solve.

I am reading your input as id1 -> site1, site2, id2 -> site1, site2 and then : union id1 and id2. Is that correct? — jrook
– jrook, Commented Sep 24, 2020 at 16:58
but how do you know that id1 and id2 are to be joined because it is now : id1 -> site1, site2, id2 -> site1, site2, the ids has to be in the other row for them to be joined. if they are separated as 1 id per row , join is not possible correct — Neethu Lalitha
– Neethu Lalitha, Commented Sep 24, 2020 at 17:02
The relation between ids and sites is many to one.. sorry I updated my input data in the question. — Neethu Lalitha
– Neethu Lalitha, Commented Sep 24, 2020 at 17:08
They rather seem to be many-to-many. I haven’t thought it through, but my immediate idea is that streams are poorly suited for this. — Anonymous
– Anonymous, Commented Sep 24, 2020 at 18:23

Anonymous · Accepted Answer · 2020-09-24 18:43:57Z

2

A simple approach:

Create a result list and initialize it to empty.
Iterate over the input list. For each input element:
1. Find all elements in the result list that holds either an ID or a site (or both) from the input element.
2. Combine (reduce) them into one result list member.
3. Also add the input element.

Use a classical loop, no stream operation.

answered Sep 24, 2020 at 18:43

Anonymous

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