Java: how to sort and group a list of objects by their attributes

Question

For Example I have a java object called RecordGroup. class signature is give below:

public class RecordGroup {

private String owner;
private Integer startRow;
private Integer recordCount;

public RecordGroup() {
}

public RecordGroup(String owner, Integer startRow, Integer recordCount) {
    this.owner = owner;
    this.startRow = startRow;
    this.recordCount = recordCount;
}

public String getOwner() {
    return owner;
}

public void setOwner(String owner) {
    this.owner = owner;
}

public Integer getRecordCount() {
    return recordCount;
}

public void setRecordCount(Integer recordCount) {
    this.recordCount = recordCount;
}

public Integer getStartRow() {
    return startRow;
}

public void setStartRow(Integer startRow) {
    this.startRow = startRow;
}

}

And, i have a List which holds a list of above object as given below.

public class Test {

List<'RecordGroup'> mergerMap = new ArrayList<'RecordGroup'>();

    mergerMap.add(new RecordGroup("RECORD", 1, 6));
    mergerMap.add(new RecordGroup("RECORD", 7, 9));
    mergerMap.add(new RecordGroup("RECORD", 3, 4));
    mergerMap.add(new RecordGroup("ZONE", 3, 1));
    mergerMap.add(new RecordGroup("MODULE", 5, 6));
    mergerMap.add(new RecordGroup("ZONE", 14, 28));
    mergerMap.add(new RecordGroup("ZONE", 6, 30));
    mergerMap.add(new RecordGroup("MODULE", 1, 60));
    mergerMap.add(new RecordGroup("OFFICE", 2, 4));
    mergerMap.add(new RecordGroup("OFFICE", 8, 6));
    mergerMap.add(new RecordGroup("USER", 1, 6));
    mergerMap.add(new RecordGroup("USER", 9, 8));
    mergerMap.add(new RecordGroup("USER", 5, 7));
    mergerMap.add(new RecordGroup("OFFICE", 3, 1));

}

My Question is, how to sort the above List of RecordGroup objects by their 'owner' and 'startRow' so that it can group the records by owner i.e first "ZONE" group and then "OFFICE" group and then "USER" group and then "MODULE" and finally "RECORD" group should appear in the list. It should also consider the "startRow" field while sorting and grouping i.e arrange each group by value of the "startRow" field in ascending order.

Out put should be like this:

mergerMap.add(new RecordGroup("ZONE", 3, 1));
mergerMap.add(new RecordGroup("ZONE", 6, 30));
mergerMap.add(new RecordGroup("ZONE", 14, 28));
mergerMap.add(new RecordGroup("OFFICE", 2, 4));
mergerMap.add(new RecordGroup("OFFICE", 3, 1));
mergerMap.add(new RecordGroup("OFFICE", 8, 6));
mergerMap.add(new RecordGroup("USER", 1, 6));
mergerMap.add(new RecordGroup("USER", 5, 7));
mergerMap.add(new RecordGroup("USER", 9, 8));
mergerMap.add(new RecordGroup("MODULE", 1, 60));
mergerMap.add(new RecordGroup("OFFICE", 2, 4));
mergerMap.add(new RecordGroup("MODULE", 5, 6));
mergerMap.add(new RecordGroup("RECORD", 1, 6));
mergerMap.add(new RecordGroup("RECORD", 3, 4));
mergerMap.add(new RecordGroup("RECORD", 7, 9));

possible duplicate of [Sort ArrayList of custom Objects by property](http://stackoverflow.com/questions/2784514/sort-arraylist-of-custom-objects-by-property) — sanbhat, Oct 09 '13 at 09:17
This will solve your purpose. http://www.thejavageek.com/2013/06/17/sorting-user-defined-objects-part-2/ — Prasad Kharkar, Oct 09 '13 at 09:22

LMG · Answer 1 · 2013-10-09T09:30:56.813

0

you have to write your own comparator, so that it does what you want with the object of your specific and specialized collection..

try to give a look here Interface Comparator

EDIT: probably comparable is even better as the other answer suggests

edited Oct 09 '13 at 09:30

answered Oct 09 '13 at 09:18

LMG

966
1
12
28

score 0 · Answer 2 · answered Oct 09 '13 at 09:19

Simple implement the Comparable Interface of Java, override the compareTo function as required and push your RecordGroups to a SortedSet, like TreeSet or sth.

http://docs.oracle.com/javase/6/docs/api/java/lang/Comparable.html

http://docs.oracle.com/javase/6/docs/api/java/util/TreeSet.html

Update: If you need different sortings for different Locations, a Entity-bound compare function would ofc. not be what you want to implement. (It will be equal through the whole system). If this is the Case, just use a Comperator per sorting case.

score 0 · Answer 3 · answered Oct 09 '13 at 09:26

0

You should probably implement the java.lang.Comparable interface:

public class RecordGroup implements Comparable<RecordGroup> {
  //Rest of your implementation
  @Override
  public int compareTo(RecordGroup o) {
     //logic to compare two RecordGroup objects
  }
}

answered Oct 09 '13 at 09:26

Panagiotis Koutsourakis

261
1
8

1

compareTo starts with lowercase – RokL Oct 09 '13 at 09:28
You are right! I edited the answer. Thanks for the correction. – Panagiotis Koutsourakis Oct 09 '13 at 09:29

Hemant Metalia · Answer 4 · 2013-10-09T09:39:17.353

Comparator does this for you Comparator

use following code.

public int compare(RecordGroup o1, RecordGroup o2) {
if (o1.getOwner().compareTo(o2.getOwner()) == 0) {
return o1.getStartRow() - o2.getStartRow();
} else {
return o1.getOwner().compareTo(o2.getOwner());
}
}
});

to check working demo go to http://ideone.com/rsM9Un for demo Here startRow is in ascending Order

Output :

[MODULE , 1
, MODULE , 5
, OFFICE , 2
, OFFICE , 3
, OFFICE , 8
, RECORD , 1
, RECORD , 3
, RECORD , 7
, USER , 1
, USER , 5
, USER , 9
, ZONE , 3
, ZONE , 6
, ZONE , 14
]

Java: how to sort and group a list of objects by their attributes

4 Answers4