Java How to compare 2 Arraylist with different objects And separate the list

Question

Previously I asked on this and I realized that it was a mistake on my side. The list was actually tag3.add instead. Java how to compare 2 ArrayList of different objects

Anyway, here's the second part of the story... Now that I have 2 lists, how to I actually elegantly separate the 2 lists? Now it seems like I have the lists joined together. For the results I am using it to pace in different columns based on the id itself:

package com.java;
import java.util.*;
public class TestList {
 public static void main(String []args) {
    Tag tag1 = new Tag();
    tag1.setId(15);
    tag1.setTag("Test");

    Tag tag2 = new Tag();
    tag2.setId(15);
    tag2.setTag("Oracle");

    Tag tag3 = new Tag();
    tag3.setId(15);
    tag3.setTag("OHNO CANNOE");

    Tag tag4 = new Tag();
    tag4.setId(16);
    tag4.setTag("Test");

    Tag tag5 = new Tag();
    tag5.setId(16);
    tag5.setTag("Oracle");

    Tag tag6 = new Tag();
    tag6.setId(16);
    tag6.setTag("OHNO CANNOE");


    List<Tag> tagList = new ArrayList<Tag>();

    tagList.add(tag1);
    tagList.add(tag2);
    tagList.add(tag3);
    tagList.add(tag4);
    tagList.add(tag5);
    tagList.add(tag6);

    System.out.println(tagList.size());

    AnotherTest test1 = new AnotherTest();
    test1.setId(15);
    test1.setTestcol("Another test col");

    AnotherTest test2 = new AnotherTest();
    test2.setId(15);
    test2.setTestcol("HAHAHA");

    AnotherTest test3 = new AnotherTest();
    test3.setId(16);
    test3.setTestcol("name it");

    AnotherTest test4 = new AnotherTest();
    test4.setId(16);
    test4.setTestcol("checkmate");

    List<AnotherTest> anotherTests = new ArrayList<AnotherTest>();
    anotherTests.add(test1);
    anotherTests.add(test2);
    anotherTests.add(test3);
    anotherTests.add(test4);

    System.out.println(anotherTests.size());
    List<String> getTaglist = new ArrayList<String>(tagList.size());
    for(AnotherTest anotherTest : anotherTests) {
        for(Tag tag: tagList) {
            if(tag.getId()==anotherTest.getId()) {
                //getTaglist = new ArrayList<String>();
                getTaglist.add(tag.getTag());
            }
        }
    }

    for (String str: getTaglist) {
        System.out.println(str);
    }
    /* returns:

    Test
    Oracle
    OHNO CANNOE
    Test
    Oracle
    OHNO CANNOE
    Test
    Oracle
    OHNO CANNOE
    Test
    Oracle
    OHNO CANNOE
    But I want 2 separate lists.
    */
}
}

How do I do this?

What do you mean "separate"? You have two List(s); anotherTests and tagList - they are separate and distinct. — Elliott Frisch
– Elliott Frisch, Commented Apr 8, 2014 at 23:31
Well, the system.out.println at the last portion of the code was originally a list. — prog rice bowl
– prog rice bowl, Commented Apr 8, 2014 at 23:37
Three List(s) then when you add getTaglist; but what are you trying to do? — Elliott Frisch
– Elliott Frisch, Commented Apr 8, 2014 at 23:39
I am trying to get 2 lists from the "taglist" because of the ids. I want to separate id15 and id16 when I compare with anotherTest list. Is it possible? — prog rice bowl
– prog rice bowl, Commented Apr 8, 2014 at 23:59
If I understand you, then yes. And you're already doing it. Do you want to make getTaglist (which is a horrible variable name) into a Map<Integer, List<String>> where Integer is your id and the List<String> are your tag(s)? — Elliott Frisch
– Elliott Frisch, Commented Apr 9, 2014 at 0:25

Elliott Frisch · Accepted Answer · 2014-04-09 00:40:55Z

0

Something with a Map<Integer, List<String>> like so -

Map<Integer, List<String>> tagMap = new HashMap<Integer, List<String>>();
for (AnotherTest anotherTest : anotherTests) {
  int id = anotherTest.getId();
  for (Tag tag : tagList) {
    if (id == tag.getId()) {
      List<String> al = null;
      if (tagMap.containsKey(id)) { // is this `id` already populated?
        al = tagMap.get(id);
      } else { // No! Create a new List.
        al = new ArrayList<String>();
        tagMap.put(id, al);
      }
      // Add the tag to the List.
      al.add(tag.getTag());
    }
  }
}

answered Apr 9, 2014 at 0:40

Elliott Frisch

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