如何遍历 Python 中的字符串列表并连接属于标签的字符串？

Question

在 Python 3 中遍历元素列表时，如何“隔离”感兴趣的元素之间的内容？

我有一个清单：

list = ["<h1> question 1", "question 1 content", "question 1 more content", "<h1> answer 1", "answer 1 content", "answer 1 more content", "<h1> question 2", "question 2 content", "<h> answer 2", "answer 2 content"]

在这个列表中，有带有标签的元素和其他没有标签的元素。这个想法是具有此标签的元素是“标题”，直到下一个标签的以下元素是它的内容。

如何将列表中属于 header 的元素连接成两个大小相等的列表：

headers = ["<h1> question 1", "<h1> answer 1", "<h1> question 2", "<h> answer 2"]
content = ["question 1 content question 1 more content", "answer 1 content answer 1 more content", "question 2 content", "answer 2 content"]

如果这两个列表的长度相同，在这种情况下，每个列表有 4 个元素。

我能够将这些部分分开，但您可以使用一些帮助来完成：

list = ["<h1> question 1", "question 1 content", "question 1 more content", "<h1> answer 1", "answer 1 content", "answer 1 more content", "<h1> question 2", "question 2 content", "<h> answer 2", "answer 2 content"]

headers = []
content = []

for i in list:
    if "<h1>" in i:
        headers.append(i)

    if "<h1>" not in i:
        tempContent = []
        tempContent.append(i)
        content.append(tempContent)

对如何组合这些文本以使它们一一对应有什么想法吗？

谢谢！

Answer 1

假设每个标题之后的所有元素都是该标题的内容，并且第一个元素始终是标题 - 您可以使用itertools.groupby。

key可以是元素是否有header标签，这样header的内容就会在它之后被分组：

from itertools import groupby

lst = ["<h1> question 1", "question 1 content", "question 1 more content", "<h1> answer 1", "answer 1 content", "answer 1 more content", "<h1> question 2", "question 2 content", "<h> answer 2", "answer 2 content"]

headers = []
content = []

for key, values in groupby(lst, key=lambda x: "<h" in x):
    if key:
        headers.append(*values)
    else:
        content.append(" ".join(values))

print(headers)
print(content)

给予：

['<h1> question 1', '<h1> answer 1', '<h1> question 2', '<h> answer 2']
['question 1 content question 1 more content', 'answer 1 content answer 1 more content', 'question 2 content', 'answer 2 content']

---

您当前方法的问题是您总是只在内容中添加一项。您要做的是累积temp_content 列表，直到遇到下一个标头，然后才添加并重置：

headers = []
content = []
temp_content = None

for i in list:
    if "<h" in i:
        if temp_content is not None:
            content.append(" ".join(temp_content))
            temp_content = []
        headers.append(i)

    else:
        temp_content.append(i)

Answer 2

您可以在迭代列表时将标题和内容收集到 collections.defaultdict 中。然后将键和值拆分为最后的headers 和content 列表。我们可以通过简单地检查字符串 str.startswith "<h" 来检测标题。

我还使用continue 语句在找到标头后立即进入下一个迭代。也可以在这里使用else 语句。

from collections import defaultdict

lst = [
    "<h1> question 1",
    "question 1 content",
    "question 1 more content",
    "<h1> answer 1",
    "answer 1 content",
    "answer 1 more content",
    "<h1> question 2",
    "question 2 content",
    "<h> answer 2",
    "answer 2 content",
]

header_map = defaultdict(list)

header = None
for item in lst:
    if item.startswith("<h"):
        header = item
        continue
    header_map[header].append(item)

headers = list(header_map)
print(headers)

content = [" ".join(v) for v in header_map.values()]
print(content)

输出：

['<h1> question 1', '<h1> answer 1', '<h1> question 2', '<h> answer 2']
['question 1 content question 1 more content', 'answer 1 content answer 1 more content', 'question 2 content', 'answer 2 content'