Question

我试图创建一个Flask应用程序（Postgresql数据库），显示各个玩家的最新得分，包括团队的平均得分和目标。因此，玩家可以拥有多个分数，但在排行榜上我只想显示她最近的分数，并根据每位玩家的最新分数显示整个团队的平均目标数。

我的Models.py

class Player(db.Model):
    __tablename__ = 'player'
    id = db.Column(db.Integer, primary_key=True)
    firstname = db.Column(db.String, nullable=False)
    score = db.relationship('Score', backref='player', lazy='dynamic')

    def __init__(self, firstname):
        self.firstname = firstname

    def __repr__(self):
        return '<id {}>'.format(self.id)


class Score(db.Model):
    __tablename__ = 'score'
    id = db.Column(db.Integer, primary_key=True)
    timestamp = db.Column(db.DateTime, nullable=False)
    score = db.Column(db.Integer, nullable=True)
    goals = db.Column(db.Integer, nullable=True)
    player_id = db.Column(db.Integer, db.ForeignKey('player.id'))

    def __init__(self, score, player_id):
        self.timestamp = datetime.now()
        self.score = score
        self.player_id = player_id

    def __repr__(self):
        return '<id {}>'.format(self.id)

我的app.py：

@app.route('/', methods=['GET', 'POST'])
@login_required
def home():

    #Latest score per player
    latest_scores_per_player = db.session.query(Player, Score).\
        join(Score).\
        distinct(Player.id).\
        filter(Player.user_id == current_user.id).\
        order_by(Player.id, Score.timestamp.desc())

    #Average goals per player
    avarage_goals = db.session.query(Player, Score).\
        join(Score).\
        distinct(Player.id).\
        filter(Player.user_id == current_user.id).\
        order_by(Player.id, Score.timestamp.desc()).\
        func.avg(Score.goals).label('average').scalar()

    return render_template(
        'home.html',
        latest_scores_per_player=latest_scores_per_player,
        avarage_goals=avarage_goals)

＆＃39;每位玩家的最新得分＆＃39;正确显示每位玩家最新输入的分数。但是＆＃39;每位玩家的平均目标＆＃39;返回错误：

AttributeError：＆＃39; BaseQuery＆＃39;对象没有属性＆＃39; func＆＃39;

如何仅针对每位玩家最新输入的分数查询所有玩家的平均目标？

更新我觉得我离得更近了：

#Average goals per player
avarage_goals = db.session.query(Player, Score, func.avg(Score.goals)).\
  join(Score).\
  distinct(Player.id).\
  filter(Player.user_id == current_user.id).\
  order_by(Player.id, Score.timestamp.desc()).scalar()

但还没到。努力让func.avg与distinct和order_by一起工作。

Answer 1

如果我理解正确，那么您只需要将每位玩家的最新分数用作子查询并对结果取平均值。由于您以每个玩家的最新分数存储Query对象，因此您可以使用：

@app.route('/', methods=['GET', 'POST'])
@login_required
def home():

    # Latest score per player
    latest_scores_per_player = db.session.query(Player, Score).\
        join(Score).\
        distinct(Player.id).\
        filter(Player.user_id == current_user.id).\
        order_by(Player.id, Score.timestamp.desc())

    # Create a subquery that queries only the goals from the groups
    subq = latest_scores_per_player.\
        with_entities(Score.goals).\
        subquery()

    #Average goals *across all players*
    average_goals = db.session.query(func.avg(subq.c.goals)).scalar()

    return render_template(
        'home.html',
        latest_scores_per_player=latest_scores_per_player,
        avarage_goals=average_goals)

这将进行两次数据库访问。您也可以只查询每个玩家的最新得分结果并计算Python中的平均值。使用最新版本的Python，您可以使用statistics模块：

import statistics

@app.route('/', methods=['GET', 'POST'])
@login_required
def home():

    # Latest score per player
    # Materialize the results to a list here for calculating average
    latest_scores_per_player = db.session.query(Player, Score).\
        join(Score).\
        distinct(Player.id).\
        filter(Player.user_id == current_user.id).\
        order_by(Player.id, Score.timestamp.desc()).\
        all()

    average_goals = None
    if latest_scores_per_player:
        average_goals = statistics.mean(
            score.goals for player, score in latest_scores_per_player)

    return render_template(
        'home.html',
        latest_scores_per_player=latest_scores_per_player,
        avarage_goals=average_goals)

或者自己动手（如果使用Python 2，则添加必要的转换以浮动）：

    average_goals = None
    if latest_scores_per_player:
        average_goals = sum(
            score.goals for player, score in latest_scores_per_player)
        average_goals /= len(latest_scores_per_player)

SQLAlchemy：获得每组最大n的平均值

1 个答案: