python类Tree()的实例源码

util.py 文件源码 项目:FancyWord 作者: EastonLee 项目源码 文件源码 阅读 17 收藏 0 点赞 0 评论 0
def tree2conlltags(t):
    """
    Return a list of 3-tuples containing ``(word, tag, IOB-tag)``.
    Convert a tree to the CoNLL IOB tag format.

    :param t: The tree to be converted.
    :type t: Tree
    :rtype: list(tuple)
    """

    tags = []
    for child in t:
        try:
            category = child.label()
            prefix = "B-"
            for contents in child:
                if isinstance(contents, Tree):
                    raise ValueError("Tree is too deeply nested to be printed in CoNLL format")
                tags.append((contents[0], contents[1], prefix+category))
                prefix = "I-"
        except AttributeError:
            tags.append((child[0], child[1], "O"))
    return tags
relextract.py 文件源码 项目:FancyWord 作者: EastonLee 项目源码 文件源码 阅读 24 收藏 0 点赞 0 评论 0
def ieer_headlines():

    from nltk.corpus import ieer
    from nltk.tree import Tree

    print("IEER: First 20 Headlines")
    print("=" * 45)  

    trees = [(doc.docno, doc.headline) for file in ieer.fileids() for doc in ieer.parsed_docs(file)]
    for tree in trees[:20]:
        print()
        print("%s:\n%s" % tree)



#############################################
## Dutch CONLL2002: take_on_role(PER, ORG
#############################################
util.py 文件源码 项目:beepboop 作者: nicolehe 项目源码 文件源码 阅读 19 收藏 0 点赞 0 评论 0
def tree2conlltags(t):
    """
    Return a list of 3-tuples containing ``(word, tag, IOB-tag)``.
    Convert a tree to the CoNLL IOB tag format.

    :param t: The tree to be converted.
    :type t: Tree
    :rtype: list(tuple)
    """

    tags = []
    for child in t:
        try:
            category = child.label()
            prefix = "B-"
            for contents in child:
                if isinstance(contents, Tree):
                    raise ValueError("Tree is too deeply nested to be printed in CoNLL format")
                tags.append((contents[0], contents[1], prefix+category))
                prefix = "I-"
        except AttributeError:
            tags.append((child[0], child[1], "O"))
    return tags
relextract.py 文件源码 项目:beepboop 作者: nicolehe 项目源码 文件源码 阅读 20 收藏 0 点赞 0 评论 0
def ieer_headlines():

    from nltk.corpus import ieer
    from nltk.tree import Tree

    print("IEER: First 20 Headlines")
    print("=" * 45)  

    trees = [(doc.docno, doc.headline) for file in ieer.fileids() for doc in ieer.parsed_docs(file)]
    for tree in trees[:20]:
        print()
        print("%s:\n%s" % tree)



#############################################
## Dutch CONLL2002: take_on_role(PER, ORG
#############################################
util.py 文件源码 项目:kind2anki 作者: prz3m 项目源码 文件源码 阅读 19 收藏 0 点赞 0 评论 0
def tree2conlltags(t):
    """
    Return a list of 3-tuples containing ``(word, tag, IOB-tag)``.
    Convert a tree to the CoNLL IOB tag format.

    :param t: The tree to be converted.
    :type t: Tree
    :rtype: list(tuple)
    """

    tags = []
    for child in t:
        try:
            category = child.label()
            prefix = "B-"
            for contents in child:
                if isinstance(contents, Tree):
                    raise ValueError("Tree is too deeply nested to be printed in CoNLL format")
                tags.append((contents[0], contents[1], prefix+category))
                prefix = "I-"
        except AttributeError:
            tags.append((child[0], child[1], "O"))
    return tags
relextract.py 文件源码 项目:kind2anki 作者: prz3m 项目源码 文件源码 阅读 20 收藏 0 点赞 0 评论 0
def ieer_headlines():

    from nltk.corpus import ieer
    from nltk.tree import Tree

    print("IEER: First 20 Headlines")
    print("=" * 45)  

    trees = [(doc.docno, doc.headline) for file in ieer.fileids() for doc in ieer.parsed_docs(file)]
    for tree in trees[:20]:
        print()
        print("%s:\n%s" % tree)



#############################################
## Dutch CONLL2002: take_on_role(PER, ORG
#############################################
util.py 文件源码 项目:but_sentiment 作者: MixedEmotions 项目源码 文件源码 阅读 16 收藏 0 点赞 0 评论 0
def tree2conlltags(t):
    """
    Return a list of 3-tuples containing ``(word, tag, IOB-tag)``.
    Convert a tree to the CoNLL IOB tag format.

    :param t: The tree to be converted.
    :type t: Tree
    :rtype: list(tuple)
    """

    tags = []
    for child in t:
        try:
            category = child.label()
            prefix = "B-"
            for contents in child:
                if isinstance(contents, Tree):
                    raise ValueError("Tree is too deeply nested to be printed in CoNLL format")
                tags.append((contents[0], contents[1], prefix+category))
                prefix = "I-"
        except AttributeError:
            tags.append((child[0], child[1], "O"))
    return tags
relextract.py 文件源码 项目:but_sentiment 作者: MixedEmotions 项目源码 文件源码 阅读 23 收藏 0 点赞 0 评论 0
def ieer_headlines():

    from nltk.corpus import ieer
    from nltk.tree import Tree

    print("IEER: First 20 Headlines")
    print("=" * 45)  

    trees = [(doc.docno, doc.headline) for file in ieer.fileids() for doc in ieer.parsed_docs(file)]
    for tree in trees[:20]:
        print()
        print("%s:\n%s" % tree)



#############################################
## Dutch CONLL2002: take_on_role(PER, ORG
#############################################
treebank.py 文件源码 项目:word2vec-keras-in-gensim 作者: niitsuma 项目源码 文件源码 阅读 17 收藏 0 点赞 0 评论 0
def trees2label_sents(trees,only_root=False,pos_neg_label=False,remove_double_count_sentence=False):
    #print 'trees2label_sents',flag_word_lower,flag_stemmer,flag_remove_double_count_sentence,only_root,pos_neg_label
    #sys.exit()
    lss=[]
    for tree in trees:
        lss_tmp=tree2label_sent(tree)
        if pos_neg_label and lss_tmp[0][0] == 2 :
            continue
        if pos_neg_label :
            lss_tmp2 = [ [1 if ls[0] > 2 else 0  ,ls[1]] for ls in lss_tmp]
        else:
            lss_tmp2 =lss_tmp
        if len(lss_tmp2) > 0 and only_root:
            lss.append(lss_tmp2[0])
        elif len(lss_tmp2) > 0:
            lss.extend(lss_tmp2)
    if remove_double_count_sentence :
        uss=label_sents2uni_sent(lss)
        lss_new =[[np.mean([lss[id][0] for id in uss[s]]),lss[uss[s][0]][1] ] for s in uss ]
        return lss_new
    else:
        return lss
dependencygraph.py 文件源码 项目:one-day-with-cling 作者: mariana-scorp 项目源码 文件源码 阅读 21 收藏 0 点赞 0 评论 0
def tree(self):
        """
        Starting with the ``root`` node, build a dependency tree using the NLTK
        ``Tree`` constructor. Dependency labels are omitted.
        """
        node = self.root

        word = node['word']
        deps = chain.from_iterable(node['deps'].values())
        return Tree(word, [self._tree(dep) for dep in deps])
util.py 文件源码 项目:Price-Comparator 作者: Thejas-1 项目源码 文件源码 阅读 20 收藏 0 点赞 0 评论 0
def conlltags2tree(sentence, chunk_types=('NP','PP','VP'),
                   root_label='S', strict=False):
    """
    Convert the CoNLL IOB format to a tree.
    """
    tree = Tree(root_label, [])
    for (word, postag, chunktag) in sentence:
        if chunktag is None:
            if strict:
                raise ValueError("Bad conll tag sequence")
            else:
                # Treat as O
                tree.append((word,postag))
        elif chunktag.startswith('B-'):
            tree.append(Tree(chunktag[2:], [(word,postag)]))
        elif chunktag.startswith('I-'):
            if (len(tree)==0 or not isinstance(tree[-1], Tree) or
                tree[-1].label() != chunktag[2:]):
                if strict:
                    raise ValueError("Bad conll tag sequence")
                else:
                    # Treat as B-*
                    tree.append(Tree(chunktag[2:], [(word,postag)]))
            else:
                tree[-1].append((word,postag))
        elif chunktag == 'O':
            tree.append((word,postag))
        else:
            raise ValueError("Bad conll tag %r" % chunktag)
    return tree
chunked.py 文件源码 项目:Price-Comparator 作者: Thejas-1 项目源码 文件源码 阅读 15 收藏 0 点赞 0 评论 0
def _untag(self, tree):
        for i, child in enumerate(tree):
            if isinstance(child, Tree):
                self._untag(child)
            elif isinstance(child, tuple):
                tree[i] = child[0]
            else:
                raise ValueError('expected child to be Tree or tuple')
        return tree
relextract.py 文件源码 项目:Price-Comparator 作者: Thejas-1 项目源码 文件源码 阅读 27 收藏 0 点赞 0 评论 0
def tree2semi_rel(tree):
    """
    Group a chunk structure into a list of 'semi-relations' of the form (list(str), ``Tree``). 

    In order to facilitate the construction of (``Tree``, string, ``Tree``) triples, this
    identifies pairs whose first member is a list (possibly empty) of terminal
    strings, and whose second member is a ``Tree`` of the form (NE_label, terminals).

    :param tree: a chunk tree
    :return: a list of pairs (list(str), ``Tree``)
    :rtype: list of tuple
    """

    from nltk.tree import Tree

    semi_rels = []
    semi_rel = [[], None]

    for dtr in tree:
        if not isinstance(dtr, Tree):
            semi_rel[0].append(dtr)
        else:
            # dtr is a Tree
            semi_rel[1] = dtr
            semi_rels.append(semi_rel)
            semi_rel = [[], None]
    return semi_rels
util.py 文件源码 项目:PyDataLondon29-EmbarrassinglyParallelDAWithAWSLambda 作者: SignalMedia 项目源码 文件源码 阅读 19 收藏 0 点赞 0 评论 0
def conlltags2tree(sentence, chunk_types=('NP','PP','VP'),
                   root_label='S', strict=False):
    """
    Convert the CoNLL IOB format to a tree.
    """
    tree = Tree(root_label, [])
    for (word, postag, chunktag) in sentence:
        if chunktag is None:
            if strict:
                raise ValueError("Bad conll tag sequence")
            else:
                # Treat as O
                tree.append((word,postag))
        elif chunktag.startswith('B-'):
            tree.append(Tree(chunktag[2:], [(word,postag)]))
        elif chunktag.startswith('I-'):
            if (len(tree)==0 or not isinstance(tree[-1], Tree) or
                tree[-1].label() != chunktag[2:]):
                if strict:
                    raise ValueError("Bad conll tag sequence")
                else:
                    # Treat as B-*
                    tree.append(Tree(chunktag[2:], [(word,postag)]))
            else:
                tree[-1].append((word,postag))
        elif chunktag == 'O':
            tree.append((word,postag))
        else:
            raise ValueError("Bad conll tag %r" % chunktag)
    return tree
chunked.py 文件源码 项目:PyDataLondon29-EmbarrassinglyParallelDAWithAWSLambda 作者: SignalMedia 项目源码 文件源码 阅读 20 收藏 0 点赞 0 评论 0
def _untag(self, tree):
        for i, child in enumerate(tree):
            if isinstance(child, Tree):
                self._untag(child)
            elif isinstance(child, tuple):
                tree[i] = child[0]
            else:
                raise ValueError('expected child to be Tree or tuple')
        return tree
relextract.py 文件源码 项目:PyDataLondon29-EmbarrassinglyParallelDAWithAWSLambda 作者: SignalMedia 项目源码 文件源码 阅读 21 收藏 0 点赞 0 评论 0
def tree2semi_rel(tree):
    """
    Group a chunk structure into a list of 'semi-relations' of the form (list(str), ``Tree``). 

    In order to facilitate the construction of (``Tree``, string, ``Tree``) triples, this
    identifies pairs whose first member is a list (possibly empty) of terminal
    strings, and whose second member is a ``Tree`` of the form (NE_label, terminals).

    :param tree: a chunk tree
    :return: a list of pairs (list(str), ``Tree``)
    :rtype: list of tuple
    """

    from nltk.tree import Tree

    semi_rels = []
    semi_rel = [[], None]

    for dtr in tree:
        if not isinstance(dtr, Tree):
            semi_rel[0].append(dtr)
        else:
            # dtr is a Tree
            semi_rel[1] = dtr
            semi_rels.append(semi_rel)
            semi_rel = [[], None]
    return semi_rels
negation_detection.py 文件源码 项目:negation-detection 作者: gkotsis 项目源码 文件源码 阅读 17 收藏 0 点赞 0 评论 0
def getLeaves(ptree):
    import nltk
    rs = []
    if isinstance(ptree, nltk.tree.ParentedTree):
        if len(ptree)>0:
            if  isinstance(ptree[0], unicode):
                rs.append(ptree)
    for node in ptree:
        if isinstance(node, nltk.tree.ParentedTree):
            if len(node)>0:
                if  isinstance(node[0], unicode):
                    rs.append(node)
                else:
                    rs.extend(getLeaves(node))
    return rs
util.py 文件源码 项目:neighborhood_mood_aws 作者: jarrellmark 项目源码 文件源码 阅读 22 收藏 0 点赞 0 评论 0
def conlltags2tree(sentence, chunk_types=('NP','PP','VP'),
                   root_label='S', strict=False):
    """
    Convert the CoNLL IOB format to a tree.
    """
    tree = Tree(root_label, [])
    for (word, postag, chunktag) in sentence:
        if chunktag is None:
            if strict:
                raise ValueError("Bad conll tag sequence")
            else:
                # Treat as O
                tree.append((word,postag))
        elif chunktag.startswith('B-'):
            tree.append(Tree(chunktag[2:], [(word,postag)]))
        elif chunktag.startswith('I-'):
            if (len(tree)==0 or not isinstance(tree[-1], Tree) or
                tree[-1].label() != chunktag[2:]):
                if strict:
                    raise ValueError("Bad conll tag sequence")
                else:
                    # Treat as B-*
                    tree.append(Tree(chunktag[2:], [(word,postag)]))
            else:
                tree[-1].append((word,postag))
        elif chunktag == 'O':
            tree.append((word,postag))
        else:
            raise ValueError("Bad conll tag %r" % chunktag)
    return tree
chunked.py 文件源码 项目:neighborhood_mood_aws 作者: jarrellmark 项目源码 文件源码 阅读 21 收藏 0 点赞 0 评论 0
def _untag(self, tree):
        for i, child in enumerate(tree):
            if isinstance(child, Tree):
                self._untag(child)
            elif isinstance(child, tuple):
                tree[i] = child[0]
            else:
                raise ValueError('expected child to be Tree or tuple')
        return tree
relextract.py 文件源码 项目:neighborhood_mood_aws 作者: jarrellmark 项目源码 文件源码 阅读 26 收藏 0 点赞 0 评论 0
def tree2semi_rel(tree):
    """
    Group a chunk structure into a list of 'semi-relations' of the form (list(str), ``Tree``). 

    In order to facilitate the construction of (``Tree``, string, ``Tree``) triples, this
    identifies pairs whose first member is a list (possibly empty) of terminal
    strings, and whose second member is a ``Tree`` of the form (NE_label, terminals).

    :param tree: a chunk tree
    :return: a list of pairs (list(str), ``Tree``)
    :rtype: list of tuple
    """

    from nltk.tree import Tree

    semi_rels = []
    semi_rel = [[], None]

    for dtr in tree:
        if not isinstance(dtr, Tree):
            semi_rel[0].append(dtr)
        else:
            # dtr is a Tree
            semi_rel[1] = dtr
            semi_rels.append(semi_rel)
            semi_rel = [[], None]
    return semi_rels
util.py 文件源码 项目:hate-to-hugs 作者: sdoran35 项目源码 文件源码 阅读 16 收藏 0 点赞 0 评论 0
def conlltags2tree(sentence, chunk_types=('NP','PP','VP'),
                   root_label='S', strict=False):
    """
    Convert the CoNLL IOB format to a tree.
    """
    tree = Tree(root_label, [])
    for (word, postag, chunktag) in sentence:
        if chunktag is None:
            if strict:
                raise ValueError("Bad conll tag sequence")
            else:
                # Treat as O
                tree.append((word,postag))
        elif chunktag.startswith('B-'):
            tree.append(Tree(chunktag[2:], [(word,postag)]))
        elif chunktag.startswith('I-'):
            if (len(tree)==0 or not isinstance(tree[-1], Tree) or
                tree[-1].label() != chunktag[2:]):
                if strict:
                    raise ValueError("Bad conll tag sequence")
                else:
                    # Treat as B-*
                    tree.append(Tree(chunktag[2:], [(word,postag)]))
            else:
                tree[-1].append((word,postag))
        elif chunktag == 'O':
            tree.append((word,postag))
        else:
            raise ValueError("Bad conll tag {0!r}".format(chunktag))
    return tree
chunked.py 文件源码 项目:hate-to-hugs 作者: sdoran35 项目源码 文件源码 阅读 22 收藏 0 点赞 0 评论 0
def _untag(self, tree):
        for i, child in enumerate(tree):
            if isinstance(child, Tree):
                self._untag(child)
            elif isinstance(child, tuple):
                tree[i] = child[0]
            else:
                raise ValueError('expected child to be Tree or tuple')
        return tree
relextract.py 文件源码 项目:hate-to-hugs 作者: sdoran35 项目源码 文件源码 阅读 20 收藏 0 点赞 0 评论 0
def tree2semi_rel(tree):
    """
    Group a chunk structure into a list of 'semi-relations' of the form (list(str), ``Tree``). 

    In order to facilitate the construction of (``Tree``, string, ``Tree``) triples, this
    identifies pairs whose first member is a list (possibly empty) of terminal
    strings, and whose second member is a ``Tree`` of the form (NE_label, terminals).

    :param tree: a chunk tree
    :return: a list of pairs (list(str), ``Tree``)
    :rtype: list of tuple
    """

    from nltk.tree import Tree

    semi_rels = []
    semi_rel = [[], None]

    for dtr in tree:
        if not isinstance(dtr, Tree):
            semi_rel[0].append(dtr)
        else:
            # dtr is a Tree
            semi_rel[1] = dtr
            semi_rels.append(semi_rel)
            semi_rel = [[], None]
    return semi_rels
AIDA_Check_and_Rewrite.py 文件源码 项目:Extracting-Core-Claims 作者: TomJansen25 项目源码 文件源码 阅读 16 收藏 0 点赞 0 评论 0
def check_if_atomic(sentence, parsed_sentence, tags):
    counter = 0
    atomic_check = re.compile("|".join(not_atomic_list))
    tree = Tree('s', parsed_sentence)
    for child in tree:
        string = str(child)
        if string.startswith("(S"):
            counter += 1
    if atomic_check.search(sentence_lower):
        return False
    elif counter > 1:
        return False
    else:
        return True
ddi_kernels.py 文件源码 项目:IBRel 作者: lasigeBioTM 项目源码 文件源码 阅读 21 收藏 0 点赞 0 评论 0
def trainSVMTK(docs, pairs, dditype, model="svm_tk_classifier.model", excludesentences=[]):
    if os.path.isfile("ddi_models/" + model):
        os.remove("ddi_models/" + model)
    if os.path.isfile("ddi_models/" + model + ".txt"):
        os.remove("ddi_models/" + model + ".txt")

    #docs = use_external_data(docs, excludesentences, dditype)
    xerrors = 0
    with open("ddi_models/" + model + ".txt", 'w') as train:
        #print pairs
        for p in pairs:
            if dditype != "all" and pairs[p][relations.PAIR_DDI] and pairs[p][relations.PAIR_TYPE] != dditype:
                continue
            sid = relations.getSentenceID(p)
            if sid not in excludesentences:
                tree = pairs[p][relations.PAIR_DEP_TREE][:]
                #print "tree1:", tree
                #if len(docs[sid][ddi.SENTENCE_ENTITIES]) > 20:
                    #print line
                #    line = "1 |BT| (ROOT (NP (NN candidatedrug) (, ,) (NN candidatedrug))) |ET|"
                #    xerrors += 1
                #else:
                line = get_svm_train_line(tree, pairs[p], sid,
                                              docs[sid][relations.SENTENCE_PAIRS][p])
                if not pairs[p][relations.PAIR_DDI]:
                    line = '-' + line
                elif pairs[p][relations.PAIR_TYPE] != dditype and dditype != "all":
                    line = '-' + line

                train.write(line)
    #print "tree errors:", xerrors
    svmlightcall = Popen(["./svm-light-TK-1.2/svm-light-TK-1.2.1/svm_learn", "-t", "5",
                          "-L", "0.4", "-T", "2", "-S", "2", "-g", "10",
                          "-D", "0", "-C", "T", basedir + model + ".txt", basedir + model],
                         stdout = PIPE, stderr = PIPE)
    res  = svmlightcall.communicate()
    if not os.path.isfile("ddi_models/" + model):
        print "failed training model " + basedir + model
        print res
        sys.exit()
util.py 文件源码 项目:FancyWord 作者: EastonLee 项目源码 文件源码 阅读 18 收藏 0 点赞 0 评论 0
def conlltags2tree(sentence, chunk_types=('NP','PP','VP'),
                   root_label='S', strict=False):
    """
    Convert the CoNLL IOB format to a tree.
    """
    tree = Tree(root_label, [])
    for (word, postag, chunktag) in sentence:
        if chunktag is None:
            if strict:
                raise ValueError("Bad conll tag sequence")
            else:
                # Treat as O
                tree.append((word,postag))
        elif chunktag.startswith('B-'):
            tree.append(Tree(chunktag[2:], [(word,postag)]))
        elif chunktag.startswith('I-'):
            if (len(tree)==0 or not isinstance(tree[-1], Tree) or
                tree[-1].label() != chunktag[2:]):
                if strict:
                    raise ValueError("Bad conll tag sequence")
                else:
                    # Treat as B-*
                    tree.append(Tree(chunktag[2:], [(word,postag)]))
            else:
                tree[-1].append((word,postag))
        elif chunktag == 'O':
            tree.append((word,postag))
        else:
            raise ValueError("Bad conll tag %r" % chunktag)
    return tree
chunked.py 文件源码 项目:FancyWord 作者: EastonLee 项目源码 文件源码 阅读 21 收藏 0 点赞 0 评论 0
def _untag(self, tree):
        for i, child in enumerate(tree):
            if isinstance(child, Tree):
                self._untag(child)
            elif isinstance(child, tuple):
                tree[i] = child[0]
            else:
                raise ValueError('expected child to be Tree or tuple')
        return tree
relextract.py 文件源码 项目:FancyWord 作者: EastonLee 项目源码 文件源码 阅读 22 收藏 0 点赞 0 评论 0
def tree2semi_rel(tree):
    """
    Group a chunk structure into a list of 'semi-relations' of the form (list(str), ``Tree``). 

    In order to facilitate the construction of (``Tree``, string, ``Tree``) triples, this
    identifies pairs whose first member is a list (possibly empty) of terminal
    strings, and whose second member is a ``Tree`` of the form (NE_label, terminals).

    :param tree: a chunk tree
    :return: a list of pairs (list(str), ``Tree``)
    :rtype: list of tuple
    """

    from nltk.tree import Tree

    semi_rels = []
    semi_rel = [[], None]

    for dtr in tree:
        if not isinstance(dtr, Tree):
            semi_rel[0].append(dtr)
        else:
            # dtr is a Tree
            semi_rel[1] = dtr
            semi_rels.append(semi_rel)
            semi_rel = [[], None]
    return semi_rels
util.py 文件源码 项目:beepboop 作者: nicolehe 项目源码 文件源码 阅读 18 收藏 0 点赞 0 评论 0
def conlltags2tree(sentence, chunk_types=('NP','PP','VP'),
                   root_label='S', strict=False):
    """
    Convert the CoNLL IOB format to a tree.
    """
    tree = Tree(root_label, [])
    for (word, postag, chunktag) in sentence:
        if chunktag is None:
            if strict:
                raise ValueError("Bad conll tag sequence")
            else:
                # Treat as O
                tree.append((word,postag))
        elif chunktag.startswith('B-'):
            tree.append(Tree(chunktag[2:], [(word,postag)]))
        elif chunktag.startswith('I-'):
            if (len(tree)==0 or not isinstance(tree[-1], Tree) or
                tree[-1].label() != chunktag[2:]):
                if strict:
                    raise ValueError("Bad conll tag sequence")
                else:
                    # Treat as B-*
                    tree.append(Tree(chunktag[2:], [(word,postag)]))
            else:
                tree[-1].append((word,postag))
        elif chunktag == 'O':
            tree.append((word,postag))
        else:
            raise ValueError("Bad conll tag %r" % chunktag)
    return tree
chunked.py 文件源码 项目:beepboop 作者: nicolehe 项目源码 文件源码 阅读 18 收藏 0 点赞 0 评论 0
def _untag(self, tree):
        for i, child in enumerate(tree):
            if isinstance(child, Tree):
                self._untag(child)
            elif isinstance(child, tuple):
                tree[i] = child[0]
            else:
                raise ValueError('expected child to be Tree or tuple')
        return tree


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