python - 解析文本以替换引号和嵌套引号

使用python，我想“教育”纯文本输入的引号并将其转换为Context语法。这是一个（递归）示例：

原文：

Using python, I would like "educate" quotes of
a plain text input and turn them into the Context syntax.
Here is a (recursive) example:

输出：

Using python, I would like \quotation{educate} quotes of
a plain text input and turn them into the Context syntax.
Here is a (recursive) example:

我也希望它能处理嵌套的报价：

原文：

Original text: "Using python, I would like 'educate' quotes of
a plain text input and turn them into the Context syntax.
Here is a (recursive) example:"

输出：

Original text: \quotation {Using python, I would like \quotation{educate} quotes of
a plain text input and turn them into the Context syntax.
Here is a (recursive) example:}

当然，我应该注意一些极端情况，例如：

She said "It looks like we are back in the '90s"

上下文引号的规范在这里：

http://wiki.contextgarden.net/Nested_quotations#Nested_quotations_in_MkIV

在这种情况下最敏感的方法是什么？非常感谢你！

最佳答案

尽管它不能处理您的极端情况，但它适用于嵌套引号

def quote(string):
    text = ''
    stack = []
    for token in iter_tokes(string):
        if is_quote(token):
            if stack and stack[-1] == token: # closing
                text += '}'
                stack.pop()
            else: # opening
                text += '\\quotation{'
                stack.append(token)
        else:
            text += token
    return text

def iter_tokes(string):
    i = find_quote(string)
    if i is None:
        yield string
    else:
        if i > 0:
            yield string[:i]
        yield string[i]
        for q in iter_tokes(string[i+1:]):
            yield q

def find_quote(string):
    for i, char in enumerate(string):
        if is_quote(char):
            return i
    return None

def is_quote(char):
    return char in '\'\"'

def main():
    quoted = None
    with open('input.txt') as fh:
        quoted = quote(fh.read())
    print quoted

main()

关于python - 解析文本以替换引号和嵌套引号，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16262004/