本文整理汇总了Python中nltk.parse.dependencygraph.DependencyGraph.get_by_address(0)["deps"]方法的典型用法代码示例。如果您正苦于以下问题：Python DependencyGraph.get_by_address(0)["deps"]方法的具体用法？Python DependencyGraph.get_by_address(0)["deps"]怎么用？Python DependencyGraph.get_by_address(0)["deps"]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.get_by_address(0)["deps"]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: make_dep_tree
# 需要导入模块: from nltk.parse.dependencygraph import DependencyGraph [as 别名]
# 或者: from nltk.parse.dependencygraph.DependencyGraph import get_by_address(0)["deps"] [as 别名]
def make_dep_tree(sent, deps):
adj = merge_with(cons, [], *[{x:[m]} for x,m,_ in deps])
heads = dict([(m,h) for h,m,_ in deps])
rel = dict([(m,rel) for _,m,rel in deps])
n = len(sent["x"])
pos = sent["pos"]
x = sent["x"]
nodelist = defaultdict(lambda: {"address": -1, "head": -1, "deps": [], "rel": "", "tag": "", "word": None})
for i in range(1, n):
node = nodelist[i]
node["address"] = i
node["head"] = heads[i]
node["deps"] = adj[i] if adj.has_key(i) else []
node["tag"] = pos[i]
node["word"] = x[i]
node["rel"] = rel[i]
g = DependencyGraph()
g.get_by_address(0)["deps"] = adj[0] if adj.has_key(0) else []
[g.add_node(node) for node in nodelist.values()]
g.root = nodelist[adj[0][0]]
return g