Python tree.Tree方法代码示例

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _backtrack(self, *e):
        if self._animating_lock: return
        if self._parser.backtrack():
            elt = self._parser.tree()
            for i in self._parser.frontier()[0]:
                elt = elt[i]
            self._lastoper1['text'] = 'Backtrack'
            self._lastoper2['text'] = ''
            if isinstance(elt, Tree):
                self._animate_backtrack(self._parser.frontier()[0])
            else:
                self._animate_match_backtrack(self._parser.frontier()[0])
            return True
        else:
            self._autostep = 0
            self._lastoper1['text'] = 'Finished'
            self._lastoper2['text'] = ''
            return False 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _chunk_parse(self, grammar=None, root_label='record', trace=0, **kwargs):
        """
        Returns an element tree structure corresponding to a toolbox data file
        parsed according to the chunk grammar.

        :type grammar: str
        :param grammar: Contains the chunking rules used to parse the
            database.  See ``chunk.RegExp`` for documentation.
        :type root_label: str
        :param root_label: The node value that should be used for the
            top node of the chunk structure.
        :type trace: int
        :param trace: The level of tracing that should be used when
            parsing a text.  ``0`` will generate no tracing output;
            ``1`` will generate normal tracing output; and ``2`` or
            higher will generate verbose tracing output.
        :type kwargs: dict
        :param kwargs: Keyword arguments passed to ``toolbox.StandardFormat.fields()``
        :rtype: ElementTree._ElementInterface
        """
        from nltk import chunk
        from nltk.tree import Tree

        cp = chunk.RegexpParser(grammar, root_label=root_label, trace=trace)
        db = self.parse(**kwargs)
        tb_etree = Element('toolbox_data')
        header = db.find('header')
        tb_etree.append(header)
        for record in db.findall('record'):
            parsed = cp.parse([(elem.text, elem.tag) for elem in record])
            tb_etree.append(self._tree2etree(parsed))
        return tb_etree 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def parse(self, chunk_struct, trace=None):
        """
        Apply the chunk parser to this input.

        :type chunk_struct: Tree
        :param chunk_struct: the chunk structure to be (further) chunked
            (this tree is modified, and is also returned)
        :type trace: int
        :param trace: The level of tracing that should be used when
            parsing a text.  ``0`` will generate no tracing output;
            ``1`` will generate normal tracing output; and ``2`` or
            highter will generate verbose tracing output.  This value
            overrides the trace level value that was given to the
            constructor.
        :return: the chunked output.
        :rtype: Tree
        """
        if trace is None: trace = self._trace
        for i in range(self._loop):
            for parser in self._stages:
                chunk_struct = parser.parse(chunk_struct, trace=trace)
        return chunk_struct 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
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 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _tagged_to_parse(self, tagged_tokens):
        """
        Convert a list of tagged tokens to a chunk-parse tree.
        """
        sent = Tree('S', [])

        for (tok,tag) in tagged_tokens:
            if tag == 'O':
                sent.append(tok)
            elif tag.startswith('B-'):
                sent.append(Tree(tag[2:], [tok]))
            elif tag.startswith('I-'):
                if (sent and isinstance(sent[-1], Tree) and
                    sent[-1].label() == tag[2:]):
                    sent[-1].append(tok)
                else:
                    sent.append(Tree(tag[2:], [tok]))
        return sent 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _parse_to_tagged(sent):
        """
        Convert a chunk-parse tree to a list of tagged tokens.
        """
        toks = []
        for child in sent:
            if isinstance(child, Tree):
                if len(child) == 0:
                    print("Warning -- empty chunk in sentence")
                    continue
                toks.append((child[0], 'B-%s' % child.label()))
                for tok in child[1:]:
                    toks.append((tok, 'I-%s' % child.label()))
            else:
                toks.append((child, 'O'))
        return toks 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def trees(self, edge, tree_class=Tree, complete=False):
        """
        Return an iterator of the tree structures that are associated
        with ``edge``.

        If ``edge`` is incomplete, then the unexpanded children will be
        encoded as childless subtrees, whose node value is the
        corresponding terminal or nonterminal.

        :rtype: list(Tree)
        :note: If two trees share a common subtree, then the same
            Tree may be used to encode that subtree in
            both trees.  If you need to eliminate this subtree
            sharing, then create a deep copy of each tree.
        """
        return iter(self._trees(edge, complete, memo={}, tree_class=tree_class)) 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _shift(self, stack, remaining_text):
        """
        Move a token from the beginning of ``remaining_text`` to the
        end of ``stack``.

        :type stack: list(str and Tree)
        :param stack: A list of strings and Trees, encoding
            the structure of the text that has been parsed so far.
        :type remaining_text: list(str)
        :param remaining_text: The portion of the text that is not yet
            covered by ``stack``.
        :rtype: None
        """
        stack.append(remaining_text[0])
        remaining_text.remove(remaining_text[0])
        if self._trace: self._trace_shift(stack, remaining_text) 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _trace_stack(self, stack, remaining_text, marker=' '):
        """
        Print trace output displaying the given stack and text.

        :rtype: None
        :param marker: A character that is printed to the left of the
            stack.  This is used with trace level 2 to print 'S'
            before shifted stacks and 'R' before reduced stacks.
        """
        s = '  '+marker+' [ '
        for elt in stack:
            if isinstance(elt, Tree):
                s += unicode_repr(Nonterminal(elt.label())) + ' '
            else:
                s += unicode_repr(elt) + ' '
        s += '* ' + ' '.join(remaining_text) + ']'
        print(s) 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _setprob(self, tree, prod_probs):
        if tree.prob() is not None: return

        # Get the prob of the CFG production.
        lhs = Nonterminal(tree.label())
        rhs = []
        for child in tree:
            if isinstance(child, Tree):
                rhs.append(Nonterminal(child.label()))
            else:
                rhs.append(child)
        prob = prod_probs[lhs, tuple(rhs)]

        # Get the probs of children.
        for child in tree:
            if isinstance(child, Tree):
                self._setprob(child, prod_probs)
                prob *= child.prob()

        tree.set_prob(prob) 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _production_to_tree(self, production):
        """
        :rtype: Tree
        :return: The Tree that is licensed by ``production``.
            In particular, given the production ``[lhs -> elt[1] ... elt[n]]``
            return a tree that has a node ``lhs.symbol``, and
            ``n`` children.  For each nonterminal element
            ``elt[i]`` in the production, the tree token has a
            childless subtree with node value ``elt[i].symbol``; and
            for each terminal element ``elt[j]``, the tree token has
            a leaf token with type ``elt[j]``.

        :param production: The CFG production that licenses the tree
            token that should be returned.
        :type production: Production
        """
        children = []
        for elt in production.rhs():
            if isinstance(elt, Nonterminal):
                children.append(Tree(elt.symbol(), []))
            else:
                # This will be matched.
                children.append(elt)
        return Tree(production.lhs().symbol(), children) 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _trace_fringe(self, tree, treeloc=None):
        """
        Print trace output displaying the fringe of ``tree``.  The
        fringe of ``tree`` consists of all of its leaves and all of
        its childless subtrees.

        :rtype: None
        """

        if treeloc == (): print("*", end=' ')
        if isinstance(tree, Tree):
            if len(tree) == 0:
                print(unicode_repr(Nonterminal(tree.label())), end=' ')
            for i in range(len(tree)):
                if treeloc is not None and i == treeloc[0]:
                    self._trace_fringe(tree[i], treeloc[1:])
                else:
                    self._trace_fringe(tree[i])
        else:
            print(unicode_repr(tree), end=' ') 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def initialize(self, tokens):
        """
        Start parsing a given text.  This sets the parser's tree to
        the start symbol, its frontier to the root node, and its
        remaining text to ``token['SUBTOKENS']``.
        """

        self._rtext = tokens
        start = self._grammar.start().symbol()
        self._tree = Tree(start, [])
        self._frontier = [()]
        self._tried_e = {}
        self._tried_m = {}
        self._history = []
        self._parses = []
        if self._trace:
            self._trace_start(self._tree, self._frontier, self._rtext) 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
def _tree2conll(self, tree, wordnum, words, pos, synt):
        assert isinstance(tree, Tree)
        if len(tree) == 1 and isinstance(tree[0], compat.string_types):
            pos[wordnum] = tree.label()
            assert words[wordnum] == tree[0]
            return wordnum+1
        elif len(tree) == 1 and isinstance(tree[0], tuple):
            assert len(tree[0]) == 2
            pos[wordnum], pos[wordnum] = tree[0]
            return wordnum+1
        else:
            synt[wordnum] = '(%s%s' % (tree.label(), synt[wordnum])
            for child in tree:
                wordnum = self._tree2conll(child, wordnum, words,
                                                  pos, synt)
            synt[wordnum-1] += ')'
            return wordnum 

# 需要导入模块: from nltk import tree [as 别名]
# 或者: from nltk.tree import Tree [as 别名]
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)



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## Dutch CONLL2002: take_on_role(PER, ORG
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