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Python Tree.is_null方法代码示例

本文整理汇总了Python中tree.Tree.is_null方法的典型用法代码示例。如果您正苦于以下问题:Python Tree.is_null方法的具体用法?Python Tree.is_null怎么用?Python Tree.is_null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tree.Tree的用法示例。


在下文中一共展示了Tree.is_null方法的2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,您的评价将有助于我们的系统推荐出更棒的Python代码示例。

示例1: Sentence

# 需要导入模块: from tree import Tree [as 别名]
# 或者: from tree.Tree import is_null [as 别名]
class Sentence(object):
    __slots__ = ('leaves', 'id', 'tree', 'true_connectives', 'checked_connectives', 'depTree', 'words', 'word_ids',
            'begin_offset', 'end_offset', 'clauses')

    def __init__(self, sent_id, parse_tree, dep_tree, words):
        self.leaves = []
        self.id = sent_id
        self.tree = Tree(parse_tree, sent_id)
        self.get_leaves()
        self.words = words
        self.begin_offset = words[0][1]['CharacterOffsetBegin']
        self.end_offset = words[-1][1]['CharacterOffsetEnd']
        self.word_ids = []
        self.true_connectives = []
        self.checked_connectives = []
        self.stem_leaf()
        self.depTree = DepTree(self, dep_tree)
        self.clauses = []
        self.break_clauses()

    def get_leaves(self):
        if not self.tree.is_null():
            self.leaves = self.tree.root.get_leaves()

    def stem_leaf(self):
        for leaf in self.leaves:
            leaf.stem = stem(leaf.value)

    def check_connectives(self):
        length = len(self.leaves)
        for a in Conn_intra:
            conns = a.split('..')
            checked = [None, None]
            i1 = i2 = length
            for i in range(length):
                if self.leaves[i].lowercased == conns[0]:
                    checked[0] = self.leaves[i]
                    i1 = i+1
                    break
            for i in range(i1, length):
                if self.leaves[i].lowercased == conns[1]:
                    checked[1] = self.leaves[i]
                    i2 = i
                    break
            if checked[0] is not None and checked[1] is not None:
                for ch in checked:
                    ch.is_conn = True
                self.checked_connectives.append(checked)
                if a == 'if..then':
                    for i in range(i1, i2):
                        if self.leaves[i].lowercased == 'if':
                            self.leaves[i].is_conn = True
                            self.checked_connectives.append([self.leaves[i], checked[1]])

        for a in Conn_group:
            conns = a.split()
            len_conns = len(conns)
            for i in range(length-len_conns+1):
                ok = True
                checked = []
                for j in range(len_conns):
                    if conns[j] != self.leaves[i+j].lowercased or self.leaves[i+j].is_conn:
                        ok = False
                        break
                    else:
                        checked.append(self.leaves[i+j])

                if ok:
                    for ch in checked:
                        ch.is_conn = True
                    self.checked_connectives.append(checked)
        return self.checked_connectives

    def get_syntactic_features(self, self_node, parent_node, left_sib_node, right_sib_node):
        curr = self_node
        self_to_root = curr.value
        self_to_root2 = curr.value
        while curr != self.tree.root:
            prev = curr
            curr = curr.parent_node
            self_to_root += '_>_'+curr.value
            if prev.value != curr.value:
                self_to_root2 += '_>_'+curr.value

        return [self_to_root, self_to_root2]

    def get_production_rules(self, conn_leaves, rules={}):
        if len(conn_leaves) == 1:
            curr = conn_leaves[0].parent_node
        else:
            tree = conn_leaves[0].goto_tree()
            curr = tree.find_least_common_ancestor(conn_leaves, with_leaves = True)
        curr.get_production_rules(rules, -1)

    def get_connective_categories(self, conn_leaves):
        self_cat = parent_cat = left_cat = right_cat = 'NONE'
        tree = conn_leaves[0].goto_tree()
        curr = tree.find_highest_common_ancestor(conn_leaves)
        parent = curr.parent_node

#.........这里部分代码省略.........
开发者ID:qcl6355,项目名称:conll2016,代码行数:103,代码来源:sentence.py

示例2: Sentence

# 需要导入模块: from tree import Tree [as 别名]
# 或者: from tree.Tree import is_null [as 别名]
class Sentence(object):
    __slots__ = ('leaves', 'id', 'tree', 'true_connectives', 'checked_connectives', 'depTree', 'words', 'word_ids',
            'begin_offset', 'end_offset', 'clauses')

    def __init__(self, sent_id, parse_tree, dep_tree, words):
        self.leaves = []
        self.id = sent_id
        self.tree = Tree(parse_tree, sent_id)
        self.get_leaves()
        self.words = words
        self.begin_offset = words[0][1]['CharacterOffsetBegin']
        self.end_offset = words[-1][1]['CharacterOffsetEnd']
        self.word_ids = []
        self.true_connectives = []
        self.checked_connectives = []
        self.depTree = DepTree(self, dep_tree)
        self.clauses = []
        self.break_clauses()

    def get_leaves(self):
        if not self.tree.is_null():
            self.leaves = self.tree.root.get_leaves()

    def _find_intra_conn(self, conn_parts, n, k, offset, checked, alls):
        if n == k:
            if len(checked) == n:
                alls.append(checked)
        else:
            length = len(self.leaves)
            conns = conn_parts[k].split()
            len_conns = len(conns)
            prev_checked = copy.copy(checked)

            for i in range(offset, length - len_conns + 1):
                ok = True
                curr = []
                for j in range(len_conns):
                    if conns[j] != self.leaves[i+j].value:
                        ok = False
                        break
                    else:
                        curr.append(self.leaves[i+j])
                if ok and len(curr) != 0:
                    checked.append(curr)
                    self._find_intra_conn(conn_parts, n, k+1, i+len_conns, checked, alls)
                    checked = prev_checked

    def check_intra_connectives(self):
        for a in CONN_INTRA:
            conn_parts = a.split('..')
            alls = []
            check = []
            self._find_intra_conn(conn_parts, len(conn_parts), 0, 0, check, alls)
            if len(alls) > 0:
                for can in alls:
                    print ' '.join(n.value for n in can), '||', a

    def check_connectives(self):
        length = len(self.leaves)
        for a in CONN_INTRA:
            conn_parts = a.split('..')
            alls = []
            check = []
            self._find_intra_conn(conn_parts, len(conn_parts), 0, 0, check, alls)
            if len(alls) > 0:
                for can in alls:
                    t = []
                    for c in can:
                        t += c
                    self.checked_connectives.append(t)

        for a in CONN_GROUP:
            conns = a.split()
            len_conns = len(conns)
            for i in range(length-len_conns+1):
                ok = True
                checked = []
                for j in range(len_conns):
                    if conns[j] != self.leaves[i+j].value:
                        ok = False
                        break
                    else:
                        checked.append(self.leaves[i+j])

                if ok:
                    self.checked_connectives.append(checked)
        return self.checked_connectives

    def get_syntactic_features(self, self_node, parent_node, left_sib_node, right_sib_node):
        curr = self_node
        self_to_root = curr.value
        self_to_root2 = curr.value
        while curr != self.tree.root:
            prev = curr
            curr = curr.parent_node
            self_to_root += '_>_'+curr.value
            if prev.value != curr.value:
                self_to_root2 += '_>_'+curr.value

        return [self_to_root, self_to_root2]
#.........这里部分代码省略.........
开发者ID:qcl6355,项目名称:conll2016,代码行数:103,代码来源:sentence.py


注:本文中的tree.Tree.is_null方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。