sqlalchemy查询时间序列数据,格式化为成对的(步骤,下一步)时间相邻的样本

问题描述 投票:4回答:2

我有一些时间序列数据,其中我有一组时间序列,每个Timeseries实例与Point实例有一对多的关系。下面是数据的简化表示。

tables.朋友:

class Timeseries(Base):
    __tablename__ = "timeseries"

    id = Column("id", Integer, primary_key=True)
    points = relationship("Point", back_populates="ts")


class Point(Base):
    __tablename__ = "point"

    id = Column("id", Integer, primary_key=True)
    t = Column("t", Float)
    v = Column("v", Float)
    ts_id = Column(Integer, ForeignKey("timeseries.id"))
    ts = relationship("Timeseries", back_populates="points")

问题:我正在尝试用这些列提出查询:“timeseries_id”,“id”,“t”,“v”,“id_next”,“t_next”,“v_next”。也就是说,我希望能够按时间顺序查看每个点的数据以及时间序列中的下一个点数据,但是我一直在努力获得一个不是来自隐式连接的元素的表吗? (编辑:重要的一点是,我希望能够在sqlalchemy中使用100%查询和子查询对象获取此列表,因为我需要在进一步的连接,过滤器等中使用此查询表。)这是什么的基本开始我得到了,(注意我没有运行此代码,因为这是我的实际数据库的简化版本,但它是相同的想法):

# The point data actually in the database.
sq = (session.query(
    Timeseries.id.label("timeseries_id"),
    Point.id,
    Point.t,
    Point.v)
.select_from(
    join(Timeseries, Point, Timeseries.id==Point.ts_id))
.group_by('timeseries_id')
.subquery())

# first point manually added to each list in query
sq_first = (session.query(
    Timeseries.id.label("timeseries_id"),
    sa.literal_column("-1", Integer).label("id"), # Some unused Point.id value
    sa.literal_column(-math.inf, Float).label("t"),
    sa.literal_column(-math.inf, Float).label("v"))
.select_from(
    join(Timeseries, Point, Timeseries.id==Point.ts_id))
.subquery())

# last point manually added to each list in query.
sq_last = (session.query(
    Timeseries.id.label("timeseries_id"),
    sa.literal_column("-2", Integer).label("id"), # Another unused Point.id value
    sa.literal_column(math.inf, Float).label("t"),
    sa.literal_column(math.inf, Float).label("v"))
.select_from(
    join(Timeseries, Point, Timeseries.id==Point.ts_id))
.subquery())

# Append each timeseries in `sq` table with last point
sq_points_curr = session.query(sa.union_all(sq_first, sq)).subquery()
sq_points_next = session.query(sa.union_all(sq, sq_last)).subquery()

假设我到目前为止所做的事情很有用,这就是我遇到困难的部分:

#I guess rename the columns in `sq_points_next` to append them by "_next"....
sq_points_next = (session.query(
    sq_points_curr.c.timeseries_id
    sq_points_curr.c.id.label("id_next"),
    sq_points_curr.c.t.label("t_next"),
    sq_points_curr.c.v.label("v_next"))
.subquery())

# ... and then perform a join along "timeseries_id" somehow to get the table I originally wanted...
sq_point_pairs = (session.query(
    Timeseries.id.label("timeseries_id")
    "id",
    "t",
    "v",
    "id_next",
    "t_next",
    "v_next"
).select_from(
    sq_points, sq_points_next, sq_points.timeseries_id==sq_points_next.timeseries_id)
)

我甚至不确定这最后是否会在此时编译,因为它再次从实际代码中进行了调整/简化,但它没有产生相邻时间点的表格等。

python sql sqlalchemy time-series
2个回答
1
投票

假设您可以获得最近版本的sqlite3 python模块(例如,通过使用Anaconda),您可以使用LEAD窗口函数来实现您的目标。为了在进一步的查询中使用LEAD函数的结果,您还需要使用CTE。以下方法适用于我给出的架构:

sq = session.query(
        Timeseries.id.label("timeseries_id"),
        Point.id.label("point_id"),
        Point.t.label("point_t"),
        Point.v.label("point_v"),
        func.lead(Point.id).over().label('point_after_id'),
        func.lead(Point.v).over().label('point_after_v'),
        func.lead(Point.t).over().label('point_after_t')).select_from(
            join(Timeseries, Point, Timeseries.id == Point.ts_id)).order_by(Timeseries.id)

with_after = sq.cte()
session.execute(with_after.select().where(
        with_after.c.point_v < with_after.c.point_after_v)).fetchall()

0
投票

而不是通过箍来获取查询以产生您正在寻找的配对结果,为什么不只是检索与特定points行相关的所有Timeseries数据,然后将数据重新组合成您正在寻找的对?例如:

from operator import attrgetter

def to_dict(a, b):
    # formats a pair of points rows into a dict object
    return {
        'timeseries_id': a.ts_id,
        'id': a.id, 't': a.t, 'v': a.v,
        'id_next': b.id, 't_next': b.t, 'v_next': b.v
    }      

def timeseries_pairs(session, ts_id):
        # queries the db for particular Timeseries row, and combines points pairs
        ts = session.query(Timeseries).\
            filter(Timeseries.id == ts_id).\
            first()

        ts.points.sort(key=attrgetter('t'))
        pairs = [to_dict(a, b) for a, b in zip(ts.points, ts.points[1:])]
        last = ts.points[-1]
        pairs.append({
            'timeseries_id': last.ts_id,
            'id': last.id, 't': last.t, 'v': last.v,
            'id_next': None, 't_next': None, 'v_next': None
            })

        return pairs

# pass the session and a timeseries id to return a list of points pairs
timeseries_pairs(session, 1)
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