如何将 nltk.tree 制作的数组变成另一棵树？

Question

我有一个nltk.tree.Tree制作的列表

>>>question = 'When did Beyonce start becoming popular?'
>>>questionSpacy = spacy_nlp(question)
>>>print(questionSpacy)
[Tree('start_VB_ROOT', ['When_WRB_advmod', 'did_VBD_aux', 'Beyonce_NNP_nsubj', Tree('becoming_VBG_xcomp', ['popular_JJ_acomp']), '?_._punct'])]

目标是制作另一棵树。我知道这很愚蠢，但否则我不知道如何知道代表一个句子的树是否包含在另一个代表另一个句子的树中。

我尝试过，但没有成功。我想我没有考虑到每一个案例。有时父节点必须是array[0].label()，有时必须是array[0]。

from nltk import Tree

class WordTree:
    def __init__(self, array, parent = None):
        #print("son :",array[0][i])
        self.parent = []
        self.children = [] # if parenthesis then it has son after "," analyse : include all elements until the next parenthesi
        self.data = array
        #print(array[0])
        for son in array[0]:
            print(type(son),son)
            if type(son) is Tree:
                print("sub tree creation")
                self.children.append(son.label())
                print("son:",son)
                t = WordTree(son,son.label()) # should I verify if parent is empty ?
                print("end of sub tree creation")
            elif type(son) is str:
                print("son creation")
                self.children.append(son)
            else:
                print("issue?")
                break # prolbem ?

当我运行t = WordTree(treeQuestion, treeQuestion[0].label()) 时，我得到以下输出：

<class 'str'> When_WRB_advmod
son creation
<class 'str'> did_VBD_aux
son creation
<class 'str'> Beyonce_NNP_nsubj
son creation
<class 'nltk.tree.Tree'> (becoming_VBG_xcomp popular_JJ_acomp)
sub tree creation
son: (becoming_VBG_xcomp popular_JJ_acomp)
<class 'str'> p
son creation
<class 'str'> o
son creation
<class 'str'> p
son creation
<class 'str'> u
son creation
<class 'str'> l
son creation
<class 'str'> a
son creation
<class 'str'> r
son creation
<class 'str'> _
son creation
<class 'str'> J
son creation
<class 'str'> J
son creation
<class 'str'> _
son creation
<class 'str'> a
son creation
<class 'str'> c
son creation
<class 'str'> o
son creation
<class 'str'> m
son creation
<class 'str'> p
son creation
end of sub tree creation
<class 'str'> ?_._punct
son creation

如您所见，在('becoming_VBG_xcomp', ['popular_JJ_acomp']) 中，它使用儿子的字母popular_JJ_acomp 来制造几个儿子，而不是用它的名字来制造一个儿子。这当然是一个错误。因此如何将 nltk.tree 生成的数组变成另一棵树？

Answer 1

我想我找到了一些东西，可以将 nltk.tree 制作的数组转换为 Python 制作的树，但我还不能概括它。

from anytree import Node, RenderTree

class WordTree:
    '''Tree for spaCy dependency parsing array'''
    def __init__(self, array, parent = None):
        """
        Construct a new 'WordTree' object.

        :param array: The array contening the dependency
        :param parent: The parent of the array if exists
        :return: returns nothing
        """
        self.parent = []
        self.children = []
        self.data = array

        for element in array[0]:
            print(type(element),element)
            # we check if we got a subtree
            if type(element) is Tree:
                print("sub tree creation")
                self.children.append(element.label())
                print("son:",element)
                t = WordTree([element],element.label())
                print("end of sub tree creation")
            # else if we have a string we create a son
            elif type(element) is str:
                print("son creation")
                self.children.append(element)
            # in other case we have a problem
            else:
                print("issue?")
                break 

确实，它适用于以下示例：

[Tree('start_VB_ROOT', ['When_WRB_advmod', 'did_VBD_aux', 'Beyonce_NNP_nsubj', Tree('becoming_VBG_xcomp', ['popular_JJ_acomp']), '?_._punct'])]

给予：

<class 'str'> When_WRB_advmod
son creation
<class 'str'> did_VBD_aux
son creation
<class 'str'> Beyonce_NNP_nsubj
son creation
<class 'nltk.tree.Tree'> (becoming_VBG_xcomp popular_JJ_acomp)
sub tree creation
son: (becoming_VBG_xcomp popular_JJ_acomp)
<class 'str'> popular_JJ_acomp
son creation
end of sub tree creation
<class 'str'> ?_._punct
son creation

但我在尝试时没有输出：

for i,sent in enumerate(sentences):
    i = WordTree(sentences, sentences[0].label())