node.precedes(n) now works across trees and is used in mentionA < mentionB

Author: martinpopel

udapi/core/coref.py

@@ -34,8 +34,8 @@ def __lt__(self, other):
  if node1 is node2:
  node1 = self._words[-1] if self._words else self._head
  node2 = other._words[-1] if other._words else other._head
- return node1 < node2
- return node1 < node2
+ return node1.precedes(node2)
+ return node1.precedes(node2)
 
  @property
  def head(self):

udapi/core/node.py

@@ -127,6 +127,12 @@ def ord(self, new_ord):
  self._ord = new_ord
 
  def __lt__(self, other):
+ """Calling `nodeA < nodeB` is equivalent to `nodeA.ord < nodeB.ord`.
+
+ Note that this does not work as expected for nodes from different trees
+ because `ord` is the word order within each sentence.
+ For comparing the word order across trees, use `nodeA.precedes(nodeB)` instead.
+ """
  return self._ord < other._ord
 
  @property
@@ -660,8 +666,18 @@ def next_node(self):
  return None
 
  def precedes(self, node):
- """Does this node precedes another `node` in word order (`self.ord < node.ord`)?"""
- return self._ord < node._ord
+ """Does this node precedes another `node` in word order?
+
+ This method handles correctly also nodes from different trees (but the same zone).
+ If you have nodes from the same tree, it is faster and more elegant to use just `nodeA < nodeB`,
+ which is equivalent to calling `nodeA.ord < nodeB.ord`.
+ For sorting nodes from the same tree, you can use `nodes.sort()` or `sorted(nodes)`.
+ """
+ if self._root is node._root:
+ return self._ord < node._ord
+ if self._root._zone != node._root._zone:
+ raise ValueError(f"Cannot compare word order across zones: {self} {node}")
+ return self._root._bundle.number < node._root._bundle.number
 
  def is_leaf(self):
  """Is this node a leaf, ie. a node without any children?"""