本文整理汇总了Python中nltk.parse.dependencygraph.DependencyGraph.nodes[-1]方法的典型用法代码示例。如果您正苦于以下问题：Python DependencyGraph.nodes[-1]方法的具体用法？Python DependencyGraph.nodes[-1]怎么用？Python DependencyGraph.nodes[-1]使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类nltk.parse.dependencygraph.DependencyGraph的用法示例。
在下文中一共展示了DependencyGraph.nodes[-1]方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: as_dependencygraph
# 需要导入模块: from nltk.parse.dependencygraph import DependencyGraph [as 别名]
# 或者: from nltk.parse.dependencygraph.DependencyGraph import nodes[-1] [as 别名]
def as_dependencygraph( self, keep_dummy_root=False, add_morph=True ):
''' Returns this tree as NLTK's DependencyGraph object.
Note that this method constructs 'zero_based' graph,
where counting of the words starts from 0 and the
root index is -1 (not 0, as in Malt-TAB format);
Parameters
-----------
add_morph : bool
Specifies whether the morphological information
(information about word lemmas, part-of-speech, and
features) should be added to graph nodes.
Note that even if **add_morph==True**, morphological
information is only added if it is available via
estnltk's layer token['analysis'];
Default: True
keep_dummy_root : bool
Specifies whether the graph should include a dummy
TOP / ROOT node, which does not refer to any word,
and yet is the topmost node of the tree.
If the dummy root node is not used, then the root
node is the word node headed by -1;
Default: False
For more information about NLTK's DependencyGraph, see:
http://www.nltk.org/_modules/nltk/parse/dependencygraph.html
'''
from nltk.parse.dependencygraph import DependencyGraph
graph = DependencyGraph( zero_based = True )
all_tree_nodes = [self] + self.get_children()
#
# 0) Fix the root
#
if keep_dummy_root:
# Note: we have to re-construct the root node manually,
# as DependencyGraph's current interface seems to provide
# no easy/convenient means for fixing the root node;
graph.nodes[-1] = graph.nodes[0]
graph.nodes[-1].update( { 'address': -1 } )
graph.root = graph.nodes[-1]
del graph.nodes[0]
#
# 1) Update / Add nodes of the graph
#
for child in all_tree_nodes:
rel = 'xxx' if not child.labels else '|'.join(child.labels)
address = child.word_id
word = child.text
graph.nodes[address].update(
{
'address': address,
'word': child.text,
'rel': rel,
} )
if not keep_dummy_root and child == self:
# If we do not keep the dummy root node, set this tree
# as the root node
graph.root = graph.nodes[address]
if add_morph and child.morph:
# Add morphological information, if possible
lemmas = set([analysis[LEMMA] for analysis in child.morph])
postags = set([analysis[POSTAG] for analysis in child.morph])
feats = set([analysis[FORM] for analysis in child.morph])
lemma = ('|'.join( list(lemmas) )).replace(' ','_')
postag = ('|'.join( list(postags) )).replace(' ','_')
feats = ('|'.join( list(feats) )).replace(' ','_')
graph.nodes[address].update(
{
'tag ': postag,
'ctag' : postag,
'feats': feats,
'lemma': lemma
} )
#
# 2) Update / Add arcs of the graph
#
for child in all_tree_nodes:
# Connect children of given word
deps = [] if not child.children else [c.word_id for c in child.children]
head_address = child.word_id
for dep in deps:
graph.add_arc( head_address, dep )
if child.parent == None and keep_dummy_root:
graph.add_arc( -1, head_address )
# Connect the parent of given node
head = -1 if not child.parent else child.parent.word_id
graph.nodes[head_address].update(
{
'head': head,
} )
return graph