通过索引替换多个字符串的算法

Question

我在想出一个好的算法来替换文本中的某些实体时遇到了一些问题。以下是详细信息：我有一段文本需要格式化为 html，有关格式化的信息位于包含实体字典的 python 列表中。比方说原文是这样的（请注意格式）：

Lorem Ipsum 只是 dummy printing 和排版行业的文本。

我将得到的文本是这样的（没有格式化）：

Lorem Ipsum 只是印刷和排版行业的虚拟文本。

和这样的实体列表：

entities = [{"entity_text":"Lorem Ipsum", "type": "bold", "offset": 0, "length":"11"}, {"entity_text":"dummy", "type": "italic", "offset": 22, "length":"5"},{"entity_text":"printing", "type": "text_link", "offset": 41, "length":"8", "url": "google.com"}]

我的算法应该将给定的未格式化文本和实体翻译成这个 html:

<b>Lorem Ipsum</b> is simply <i>dummy</i> text of the <a href="google.com">printing</a> and typesetting industry

以便可以将其编译成原始消息。我尝试过字符串替换，但它弄乱了偏移量（实体从文本开始的位置）。请记住，文本中可能有许多带格式的单词没有格式化，所以我必须准确找到应该格式化的单词。有任何人的帮助吗？我正在用 python 编写代码，但您可以用任何语言指定算法

编辑抱歉，我忘记了 post 我试过的代码。就是这样：

format_html(text, entities):
    for entity in entities:
        try:
            entity_text = entity['entity_text']
            position = text.find(entity_text, entity['offset'])
            if position == entity['offset']:
                before = text[:position]
                after = text[min(position+entity['length'], len(text)-1):]
                if entity['type'] == 'text_link':
                    text_link = '<a href="{}">{}</a>'.format(entity['url'], entity_text)
                    text = before + text_link + after
                elif entity['type'] == 'code':
                    code = '<code>{}</code>'.format(entity_text)
                    text = before + code + after
                elif entity['type'] == 'bold':
                    bold_text = '<b>{}</b>'.format(entity_text)
                    text = before + bold_text + after
                elif entity['type'] == 'italic':
                    italic_text = '<i>{}</i>'.format(entity_text)
                    text = before + italic_text + after
                elif entity['type'] == 'pre':
                    pre_code = '<pre>{}</pre>'.format(entity_text)
                    text = before + pre_code + after
        except:
            pass

Answer 1

你的意思可能是这样的？

text = ""
for entry in entries:
    line = ""
    for key, value in entry.iteritems():
        if key == 'entity_text':
            line += value
        elif key == 'type' and value == 'bold':
            line = "<b> {} </b>".format(line)
        elif key == 'type' and value == 'italic':
            line = "<i> {} </i>".format(line)
        elif key == 'type' and value == 'text_link':
            line = '<a href="google.com">{}</a>'.format(line)
    text += line
text

转换为

'<b> Lorem Ipsum </b><i> dummy </i><a href="google.com">printing</a>'

Answer 2

好吧，我就是这样解决的。每次修改文本时，我都会根据添加到文本中的额外字符串的长度（由于标签）来调整偏移量。这在计算时间方面成本很高，但这是我见过的唯一选择

def format_html(text, entities):
    for entity in entities:
        try:
            modified = None
            entity_text = entity['entity_text']
            position = text.find(entity_text, entity['offset'])
            if position == entity['offset']:
                before = text[:position]
                after = text[min(position+entity['length'], len(text)-1):]
                if entity['type'] == 'text_link':
                    text_link = '<a href="{}">{}</a>'.format(entity['url'], entity_text)
                    text = before + text_link + after
                    modified = 15 + len(entity['url'])
                elif entity['type'] == 'code':
                    code = '<code>{}</code>'.format(entity_text)
                    text = before + code + after
                    modified = 13
                elif entity['type'] == 'bold':
                    bold_text = '<b>{}</b>'.format(entity_text)
                    text = before + bold_text + after
                    modified = 7
                elif entity['type'] == 'italic':
                    italic_text = '<i>{}</i>'.format(entity_text)
                    text = before + italic_text + after
                    modified = 7
                elif entity['type'] == 'pre':
                    pre_code = '<pre>{}</pre>'.format(entity_text)
                    text = before + pre_code + after
                    modified = 11
               if modified:
                   for other in entites:
                       if other['offset'] > entity.offset:   
                           other.offset += modified
        except:
            pass

通过索引替换多个字符串的算法

Algorithm for multiple string replacement by index

python

regex

algorithm

replace

text-formatting