[英]Java: how to sort and group a list of objects by their attributes
例如,我有一個名為RecordGroup的java對象。 類簽名如下:
公共類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;
}
}
並且,我有一個List,其中包含上面提到的上述對象的列表。
公共課測試{
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));
}
我的問題是,如何按照'owner'和'startRow'對RecordGroup對象的上述列表進行排序,以便它可以按所有者分組記錄,即首先是“ZONE”組,然后是“OFFICE”組,然后是“USER”組和然后“MODULE”和最后“RECORD”組應出現在列表中。 它還應該在排序和分組時考慮“startRow”字段,即按“startRow”字段的值按升序排列每個組。
出局應該是這樣的:
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));
簡單地實現Java的Comparable
Interface,根據需要覆蓋compareTo
函數,並將RecordGroups推送到SortedSet,如TreeSet
或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
更新:如果您需要為不同的位置進行不同的排序,則實體綁定的比較功能將會出現。 不是你想要實現的。 (整個系統將是平等的)。 如果是這種情況,只需在每個排序情況下使用一個Comperator。
您應該實現java.lang.Comparable接口:
public class RecordGroup implements Comparable<RecordGroup> {
//Rest of your implementation
@Override
public int compareTo(RecordGroup o) {
//logic to compare two RecordGroup objects
}
}
比較器為您做比較
使用以下代碼。
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());
}
}
});
要查看工作演示,請訪問http://ideone.com/rsM9Un進行演示此處startRow按升序排列
輸出:
[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
]
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