Question

标题本身并不是一个明确的问题，所以我将提供一个例子：

我有一个示例字符串：

Created and managed websites for clients to communicate securely

并且有很多“版本”。这意味着字符串“版本”中的单词或短语将包含在HTML div标签中，即<div style="font-size: 0.1000000">foo bar</div>。（这些标签是任意的，给予font-size属性的数字对应于以后将被用作其他现在无关的CSS特征的分数。）以下是该字符串的4个版本：

Created and <div style="font-size: 1">managed</div> websites for clients to communicate securely
Created and <div style="font-size: 2">managed websites</div> for clients to communicate securely
Created and managed websites for clients to <div style="font-size: 3">communicate</div> securely
<div style="font-size: 4">Created</div> and managed websites for clients to communicate securely

我想将所有这些版本合并到此：

<div style="font-size: 4">Created</div> and <div style="font-size: 2"><div style="font-size: 1">managed</div> websites</div> for clients to <div style="font-size: 3">communicate</div> securely

正如我们在此处看到的，有重叠的标签（标签中包含font-size: 2和font-size: 1）。字符串的版本数量可以在1到50之间，因此可能存在多个重叠。

以下是我到目前为止使用的正则表达式：

import re

div_str = "<div style=.*</div>"  # the div tags
div_text_str = "(?<=(>)).*(?=(</div>))"  # the content inside the div tags

# compile the regexes
div_regex = re.compile(div_str)
div_text_regex = re.compile(div_text_str)

def merge_strings(str1, str2):

    # grab the div tag off the first version
    div = div_regex.search(str1).group()
    # grab the contents of that div tag
    div_text = div_text_regex.search(div).group()

    # find the div content in the second version, then substitute
    # with the div tag
    return re.sub(div_text, div, str2)

我在循环中运行此函数并尝试一次合并2个字符串，直到我得到最终输出。我面临的问题是重叠标签不能与此功能一起使用，因为正则表达式模式与它不匹配。此外，一次替换多个div标签失败。

对此有任何帮助将不胜感激！

Answer 1

这是不正确答案。

我要提到用正则表达式解析HTML通常会让生活变得不必要。最好使用一个解析器，如BeautifulSoup，lxml，scrapy等。

很容易从您提供的每一行中恢复文本作为示例。我假设每个都是更大建筑的一部分;因此，我将每个都包含在div内。

在这里，我使用BeautifulSoup从你的每一行获取文本。

>>> for line in open('temp.htm').readlines():
...     line = line.strip()
...     print(line)
...     soup = bs4.BeautifulSoup(line, 'lxml')
...     soup.find('div').text
...     
<div>Created and <div style="font-size: 1">managed</div> websites for clients to communicate securely</div>
'Created and managed websites for clients to communicate securely'
<div>Created and <div style="font-size: 2">managed websites</div> for clients to communicate securely</div>
'Created and managed websites for clients to communicate securely'
<div>Created and managed websites for clients to <div style="font-size: 3">communicate</div> securely</div>
'Created and managed websites for clients to communicate securely'
<div><div style="font-size: 4">Created</div> and managed websites for clients to communicate securely</div>
'Created and managed websites for clients to communicate securely'

不幸的是，我不明白你是如何将输入行映射到输出HTML的。

Answer 2

我明白了。用BeautifulSoup替换正则表达式使解析变得更容易，我根据div标签之间的文本长度对这些版本进行了排序，以免遇到查找子字符串的任何问题。

使用相同的样本：

Created and <div style="font-size: 1">managed</div> websites for clients to communicate securely
Created and <div style="font-size: 2">managed websites</div> for clients to communicate securely
Created and managed websites for clients to <div style="font-size: 3">communicate</div> securely
<div style="font-size: 4">Created</div> and managed websites for clients to communicate securely

这些行以列表形式表示，然后使用BeautifulSoup按其对应的div标签之间的文本长度进行排序。这是代码：

def __merge_strings(final_str, version):

    soup = BeautifulSoup(final_str, "html.parser")

    for fixed_div in soup.find_all("div"):
        if not fixed_div.text == version.text:
            return final_str.replace(
                version.text, unicode(version)
            )

    return final_str

found_terms = (
    (i, BeautifulSoup(i, "html.parser").find("div"))
    for i in found_terms
)  # list of pairs of the version and its div text
found_terms = sorted(
    found_terms, key=lambda x: len(x[-1].text), reverse=True
)  # sort on the length of the div text to avoid issues with substrings

current_div = found_terms[0][0]  # version with the largest div text
for i in xrange(1, len(found_terms)):
    current_div = __merge_strings(current_div, found_terms[i][-1])

合并具有相同内容但在Python

2 个答案: