通过索引替换多个字符串的算法

Question

我有一些问题想到一个好的算法来替换文本中的某些实体。这里是详细信息： 我有一个文本，我需要格式化为HTML格式的信息是在一个包含实体的字典的Python列表。比方说，例如，原来的文字是这样的（请注意格式）：

Lorem存有简直是printing和排版行业的虚拟文本。

我会得到的文本将是（不带格式）：

Lorem存有简直是印刷排版行业的虚拟文本。

像这样实体的名单：

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"}] 

我的算法应把这个给定的无格式文本和实体到这个网站：使之能

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

被编译成原始信息。 我试过字符串替换，但它弥补了偏移量（实体从文本开始的位置）。请记住，在文本中可能会有很多格式化的单词，这些单词没有格式化，所以我必须准确找到应格式化的单词。任何人的帮助？我正在用Python编写代码，但你可以指定任何语言的算法

编辑 对不起，我忘了发布我试过的代码。这是它：

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

好了，这是我如何解决它。每次修改文本时，我都会调整偏移量，并在文本中添加额外字符串（由于标记）的长度。这是在计算时间方面是昂贵的，但是这是我见过的

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 

Answer 2

你可能是这样的意思？

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 

转化为

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