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Python pyparsing.NoMatch方法代码示例

本文整理汇总了Python中pyparsing.NoMatch方法的典型用法代码示例。如果您正苦于以下问题:Python pyparsing.NoMatch方法的具体用法?Python pyparsing.NoMatch怎么用?Python pyparsing.NoMatch使用的例子?那么, 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pyparsing的用法示例。


在下文中一共展示了pyparsing.NoMatch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,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: _make_matcher_element

# 需要导入模块: import pyparsing [as 别名]
# 或者: from pyparsing import NoMatch [as 别名]
def _make_matcher_element(self):
        return self._set_matcher_element_attributes(pyparsing.NoMatch()) 
开发者ID:Danesprite,项目名称:pyjsgf,代码行数:4,代码来源:expansions.py

示例2: _make_matcher_element

# 需要导入模块: import pyparsing [as 别名]
# 或者: from pyparsing import NoMatch [as 别名]
def _make_matcher_element(self):
        # Handle the case where use_current_match is True.
        if self.use_current_match is True:
            current_match = self.current_match
            if current_match is None:
                result = pyparsing.NoMatch()
            elif current_match == "":
                result = pyparsing.Empty()
            else:
                result = pyparsing.Literal(self.current_match)

            # Set the parse action and return the element.
            return result.setParseAction(self._parse_action)

        # Otherwise build a list of next possible literals. Make the required stack
        # of child-parent pairs.
        stack = []
        p1, p2 = self, self.parent
        while p1 and p2:
            stack.append((p1, p2))

            # Move both pivots further up the tree.
            p1 = p1.parent
            p2 = p2.parent

        # Build a list of next literals using the stack.
        next_literals, _ = _collect_next_literals(stack, 0, True, False)

        # De-duplicate the list.
        next_literals = set(next_literals)

        word = pyparsing.Regex(_word_regex_str, re.UNICODE)
        if next_literals:
            # Check if there is a next dictation literal. If there is, only match
            # one word for this expansion.
            if _word_regex_str in next_literals:
                result = word

            # Otherwise build an element to match one or more words stopping on
            # any of the next literals so that they aren't matched as dictation.
            else:
                next_literals = list(map(pyparsing.Literal, next_literals))
                result = pyparsing.OneOrMore(
                    word, stopOn=pyparsing.Or(next_literals)
                )
        else:
            # Handle the case of no literals ahead by allowing one or more Unicode
            # words without restrictions.
            result = pyparsing.OneOrMore(word)

        return self._set_matcher_element_attributes(result) 
开发者ID:Danesprite,项目名称:pyjsgf,代码行数:53,代码来源:expansions.py


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