我在SQLAlchemy方面实现了一个具有Partially Ordered Set数学特征的结构，我需要能够一次添加和删除一个边。
在我目前最好的设计中，我使用了两个邻接列表，一个是赋值列表（大约是Hass图中的边缘），因为我需要保留哪些节点对被显式设置为有序，另一个是邻接列表是第一个的传递闭包，因此我可以有效地查询一个节点是否相对于另一个节点进行了排序。现在，每次在赋值邻接列表中添加或删除边时，我都会重新计算传递闭包。
它看起来像这样：
assignment = Table('assignment', metadata,
Column('parent', Integer, ForeignKey('node.id')),
Column('child', Integer, ForeignKey('node.id')))
closure = Table('closure', metadata,
Column('ancestor', Integer, ForeignKey('node.id')),
Column('descendent', Integer, ForeignKey('node.id')))
class Node(Base):
__tablename__ = 'node'
id = Column(Integer, primary_key=True)
parents = relationship(Node, secondary=assignment,
backref='children',
primaryjoin=id == assignment.c.parent,
secondaryjoin=id == assignment.c.child)
ancestors = relationship(Node, secondary=closure,
backref='descendents',
primaryjoin=id == closure.c.ancestor,
secondaryjoin=id == closure.c.descendent,
viewonly=True)
@classmethod
def recompute_ancestry(cls.conn):
conn.execute(closure.delete())
adjacent_values = conn.execute(assignment.select()).fetchall()
conn.execute(closure.insert(), floyd_warshall(adjacent_values))
其中floyd_warshall()是同名算法的实现。
这导致我遇到两个问题。第一个是它看起来效率不高，但我不确定我可以使用哪种算法。
第二个更多的是关于每次分配时必须显式调用Node.recompute_ancestry()的实用性，并且只有在分配后才会刷新到会话中并使用正确的连接。如果我想查看ORM中反映的更改，我必须再次刷新会话 。我认为，如果我能用orm来表达重新计算的祖先操作会更容易。
答案 0 :(得分：1)
好吧，我去解决了我自己的问题。其中最原始的部分是将Floyd-Warshall算法应用于父节点祖先后代的交集与子节点后代的祖先，但仅将输出应用于父母的祖先和孩子的后代的联盟。我花了很多时间在上面，我最后发布了流程on my blog，但这里是代码。
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
association_table = Table('edges', Base.metadata,
Column('predecessor', Integer,
ForeignKey('nodes.id'), primary_key=True),
Column('successor', Integer,
ForeignKey('nodes.id'), primary_key=True))
path_table = Table('paths', Base.metadata,
Column('predecessor', Integer,
ForeignKey('nodes.id'), primary_key=True),
Column('successor', Integer,
ForeignKey('nodes.id'), primary_key=True))
class Node(Base):
__tablename__ = 'nodes'
id = Column(Integer, primary_key=True)
# extra columns
def __repr__(self):
return '<Node #%r>' % (self.id,)
successors = relationship('Node', backref='predecessors',
secondary=association_table,
primaryjoin=id == association_table.c.predecessor,
secondaryjoin=id == association_table.c.successor)
before = relationship('Node', backref='after',
secondary=path_table,
primaryjoin=id == path_table.c.predecessor,
secondaryjoin=id == path_table.c.successor)
def __lt__(self, other):
return other in self.before
def add_successor(self, other):
if other in self.successors:
return
self.successors.append(other)
self.before.append(other)
for descendent in other.before:
if descendent not in self.before:
self.before.append(descendent)
for ancestor in self.after:
if ancestor not in other.after:
other.after.append(ancestor)
def del_successor(self, other):
if not self < other:
# nodes are not connected, do nothing!
return
if not other in self.successors:
# nodes aren't adjacent, but this *could*
# be a warning...
return
self.successors.remove(other)
# we buld up a set of nodes that will be affected by the removal
# we just did.
ancestors = set(other.after)
descendents = set(self.before)
# we also need to build up a list of nodes that will determine
# where the paths may be. basically, we're looking for every
# node that is both before some node in the descendents and
# ALSO after the ancestors. Such nodes might not be comparable
# to self or other, but may still be part of a path between
# the nodes in ancestors and the nodes in descendents.
ancestors_descendents = set()
for ancestor in ancestors:
ancestors_descendents.add(ancestor)
for descendent in ancestor.before:
ancestors_descendents.add(descendent)
descendents_ancestors = set()
for descendent in descendents:
descendents_ancestors.add(descendent)
for ancestor in descendent.after:
descendents_ancestors.add(ancestor)
search_set = ancestors_descendents & descendents_ancestors
known_good = set() # This is the 'paths' from the
# original algorithm.
# as before, we need to initialize it with the paths we
# know are good. this is just the successor edges in
# the search set.
for predecessor in search_set:
for successor in search_set:
if successor in predecessor.successors:
known_good.add((predecessor, successor))
# We now can work our way through floyd_warshall to resolve
# all adjacencies:
for ancestor in ancestors:
for descendent in descendents:
if (ancestor, descendent) in known_good:
# already got this one, so we don't need to look for an
# intermediate.
continue
for intermediate in search_set:
if (ancestor, intermediate) in known_good \
and (intermediate, descendent) in known_good:
known_good.add((ancestor, descendent))
break # don't need to look any further for an
# intermediate, we can move on to the next
# descendent.
# sift through the bad nodes and update the links
for ancestor in ancestors:
for descendent in descendents:
if descendent in ancestor.before \
and (ancestor, descendent) not in known_good:
ancestor.before.remove(descendent)
答案 1 :(得分：0)
在插入时更新闭包，并按照orm执行此操作：
def add_assignment(parent, child):
"""And parent-child relationship between two nodes"""
parent.descendants += child.descendants + [child]
child.ancestors += parent.ancestors + [parent]
parent.children += child
如果你需要删除分配，这在纯sql中更快：
def del_assignment(parent, child):
parent.children.remove(child)
head = [parent.id] + [node.id for node in parent.ancestors]
tail = [child.id] + [node.id for node in child.descendants]
session.flush()
session.execute(closure.delete(), and_(
closure.c.ancestor.in_(head),
closure.c.descendant.in_(tail)))
session.expire_all()