JAVA：如何基於屬性值合並arrayList中的對象？

Question

我正在使用Spring Boot，Java和MySQL構建（或學習如何）運動REST API。 我正在構建一個方法，該方法當前從匹配項集合中獲取每個匹配項，並為完整的匹配項列表返回TeamStandings的ArrayList 。

方法如下：

public List<TeamStanding> createStandingsTable(Match[] matches){
        List<TeamStanding> teamStandings = new ArrayList<TeamStanding>();
        for(int i = 0;i < matches.length; i++) {
            TeamStanding firstTeam = new TeamStanding();
            TeamStanding secondTeam = new TeamStanding();

            //set team ids
            firstTeam.setIdTeam(matches[i].getWcmHome());
            secondTeam.setIdTeam(matches[i].getWcmAway());


            //first team stats
            firstTeam.setTeamPlayed((long) 1);
            firstTeam.setTeamGoalsFavor(matches[i].getWcmHomeGoals());
            firstTeam.setTeamGoalsAgainst(matches[i].getWcmAwayGoals());
            firstTeam.setTeamGoalDif(firstTeam.getTeamGoalsFavor() - firstTeam.getTeamGoalsAgainst());

            //second team stats
            secondTeam.setTeamPlayed((long) 1);
            secondTeam.setTeamGoalsFavor(matches[i].getWcmAwayGoals());
            secondTeam.setTeamGoalsAgainst(matches[i].getWcmHomeGoals());
            secondTeam.setTeamGoalDif(secondTeam.getTeamGoalsFavor() - secondTeam.getTeamGoalsAgainst());

            //combined team stats

            if(firstTeam.getTeamGoalsFavor() > secondTeam.getTeamGoalsFavor()) {
                firstTeam.setTeamWins((long) 1);
                firstTeam.setTeamLoses((long) 0);
                firstTeam.setTeamDraws((long) 0);
                firstTeam.setTeamPoints((long) 3);
                secondTeam.setTeamWins((long) 0);
                secondTeam.setTeamLoses((long) 1);
                secondTeam.setTeamDraws((long) 0);
                secondTeam.setTeamPoints((long) 0);
            } else if (firstTeam.getTeamGoalsFavor() == secondTeam.getTeamGoalsFavor()) {
                firstTeam.setTeamWins((long) 0);
                firstTeam.setTeamLoses((long) 0);
                firstTeam.setTeamDraws((long) 1);
                firstTeam.setTeamPoints((long) 1);
                secondTeam.setTeamWins((long) 0);
                secondTeam.setTeamLoses((long) 0);
                secondTeam.setTeamDraws((long) 1);
                secondTeam.setTeamPoints((long) 1);
            } else {
                firstTeam.setTeamWins((long) 0);
                firstTeam.setTeamLoses((long) 1);
                firstTeam.setTeamDraws((long) 0);
                firstTeam.setTeamPoints((long) 0);
                secondTeam.setTeamWins((long) 1);
                secondTeam.setTeamLoses((long) 0);
                secondTeam.setTeamDraws((long) 0);
                secondTeam.setTeamPoints((long) 3);
            }
            teamStandings.add(firstTeam);
            teamStandings.add(secondTeam);
        }
        return teamStandings;
    }

結果是這樣的：

[
    {
        "idTeam": 7,
        "teamPoints": 3,
        "teamPlayed": 1,
        "teamWins": 1,
        "teamDraws": 0,
        "teamLoses": 0,
        "teamGoalsFavor": 4,
        "teamGoalsAgainst": 1,
        "teamGoalDif": 3
    },
    {
        "idTeam": 13,
        "teamPoints": 0,
        "teamPlayed": 1,
        "teamWins": 0,
        "teamDraws": 0,
        "teamLoses": 1,
        "teamGoalsFavor": 1,
        "teamGoalsAgainst": 4,
        "teamGoalDif": -3
    },
    {
        "idTeam": 4,
        "teamPoints": 3,
        "teamPlayed": 1,
        "teamWins": 1,
        "teamDraws": 0,
        "teamLoses": 0,
        "teamGoalsFavor": 1,
        "teamGoalsAgainst": 0,
        "teamGoalDif": 1
    },
    {
        "idTeam": 7,
        "teamPoints": 0,
        "teamPlayed": 1,
        "teamWins": 0,
        "teamDraws": 0,
        "teamLoses": 1,
        "teamGoalsFavor": 0,
        "teamGoalsAgainst": 1,
        "teamGoalDif": -1
    }
]

我的問題是如何基於idTeam合並這些對象？ 我試圖實現的結果是，在idTeam保持不變的情況下，將所有其他屬性加起來。 在給定的示例中，預期的是：

[
        {
            "idTeam": 7,
            "teamPoints": 3,
            "teamPlayed": 2,
            "teamWins": 1,
            "teamDraws": 0,
            "teamLoses": 1,
            "teamGoalsFavor": 4,
            "teamGoalsAgainst": 2,
            "teamGoalDif": 2
        },
        {
            "idTeam": 13,
            "teamPoints": 0,
            "teamPlayed": 1,
            "teamWins": 0,
            "teamDraws": 0,
            "teamLoses": 1,
            "teamGoalsFavor": 1,
            "teamGoalsAgainst": 4,
            "teamGoalDif": -3
        },
        {
            "idTeam": 4,
            "teamPoints": 3,
            "teamPlayed": 1,
            "teamWins": 1,
            "teamDraws": 0,
            "teamLoses": 0,
            "teamGoalsFavor": 1,
            "teamGoalsAgainst": 0,
            "teamGoalDif": 1
        }
    ]

同樣，只是一個細節，我首先構建了TeamStandings的ArrayList ，現在嘗試合並它們，但是也許我應該將它們堆疊成Matches數組中的循環，但在上面的相同方法中，但我不確定。

Answer 1

遍歷TeamStanding的列表，注意團隊ID並執行添加。 您可能希望使用地圖將一對團隊ID保存為鍵，將團隊本身保存為值，以便於操作。 這是片段（我尚未測試過，因此您可能需要對其進行一些修改）。

List<TeamStanding> list = createStandingsTable(matches);
Map<Integer, TeamStanding> map = new HashMap<>();

for (TeamStanding team: list) {
    int id = team.getIdTeam();
    if (map.containsKey(id)) {
        TeamStanding other = map.get(id);
        other.setTeamPoints(team.getTeamPoints());
        other.setTeamPlayed(team.getTeamPlayed());
        // and so on...
    } else {
        map.put(id, team);
    }
}

List<TeamStanding> merged = new ArrayList<>(map.values());

如果要直接從Match[]創建合並的List<TeamStanding> ，則必須使用相同的想法，但是，將兩個迭代合並在一起可能會有些復雜。 然后，我建議您堅持使用這兩個單獨的迭代。 簡短性，可讀性和可維護性優於性能-此外，此處的性能並不是真正的問題。

Answer 2

您可以使用HashMap。 使用“ idTeam”作為鍵，使用對象TeamStanding作為值。 現在，您可以在結果列表上進行迭代，如果在地圖上找到對象，則只需更新其字段即可；如果找不到，則插入對象。 迭代完成后，您可以調用map.values（），它將為您提供對象的集合（TeamStanding），然后您可以使用此集合創建一個新的ArrayList。

該代碼將如下所示：

public List<TeamStanding> mergeTeamStandingList(List<TeamStanding> teamStandingList) {
    final Map<Integer, TeamStanding> idTeamVsTeamStandingMap = new HashMap<Integer, TeamStanding>();
    teamStandingList.forEach(teamStanding -> {
        if(idTeamVsTeamStandingMap.containsKey(teamStanding.getIdTeam())) {
            TeamStanding teamStanding1 = idTeamVsTeamStandingMap.get(teamStanding.getIdTeam());
            teamStanding1.setTeamDraws(teamStanding1.getTeamDraws() + teamStanding.getTeamDraws());
            //so on
        } else {
            idTeamVsTeamStandingMap.put(teamStanding.getIdTeam(), teamStanding);
        }
    });

    return new ArrayList<>(idTeamVsTeamStandingMap.values());
}

Answer 3

在您的Teamstanding對象上創建合並方法。

public TeamStanding merge(TeamStanding other) {
     this.teamPoints += other.getTeamPoints();
     this.teamPlayed += other.getTeamPlayed();
     this.teamWins += other.getTeamWins();
     this.teamDraws += other.getTeamDraws();
     this.teamGoalsFavor += other.getTeamGoalsFavor();
     this.teamLoses += other.getTeamLoses();
     this.teamGoalDif += other.getTeamGoalDif();
     return this;
}

---

然后使用Streams按teamId分組，並使用merge方法減少常見項目。

Map<Integer, Optional<TeamStanding>> mapReduced = teamStandings
.stream()
.collect(groupingBy(TeamStanding::getIdTeam, Collectors.reducing(TeamStanding::merge)));