parsing 相关问题

我想在没有外部库的情况下有效地解析 HTML 代码。我已经尝试过使用 for 循环来检查它是哪个符号：列表=[] html =“”“... 我想在没有外部库的情况下有效地解析 HTML 代码。我已经尝试过使用 for 循环来检查它是哪个符号： list = [] html = """<html>Hello</html>""" m = 0 for a in html: if a == "<": m = 1 list.append([]) elif a == ">": m = 0 list.append([]) else: list[-1] = a print(list) 但是代码在 50KB 文件上非常慢。我可以建议从一个简单的 HTML 解析器开始，如下所示？它使用Python自带的标准库，没有外部依赖。您可能需要根据需要更改和扩展它，但它为您提供了一个基本的 DOM API，这应该是一个很好的工作起点。该代码适用于它要解决的简单情况；但根据您的需求，您可能需要添加更多功能来实现您的最终目标。 #! /usr/bin/env python3 import html.parser import pprint import xml.dom.minidom def main(): # noinspection PyPep8 document = ''' <html><head><title>The Dormouse's story</title></head> <body> The Dormouse's story Once upon a time there were three little sisters; and their names were <a href="http://example.com/elsie" class="sister" id="link1">Elsie</a>, <a href="http://example.com/lacie" class="sister" id="link2">Lacie</a> and <a href="http://example.com/tillie" class="sister" id="link3">Tillie</a>; and they lived at the bottom of a well. ... ''' parser = DocumentParser() parser.feed(document) parser.close() model = parser.document.documentElement model.normalize() print(model.toprettyxml()) first_title = model.getElementsByTagName('title')[0] print(first_title.toxml()) print(first_title.tagName) print(first_title.firstChild.data) print(first_title.parentNode.tagName) first_p = model.getElementsByTagName('p')[0] print(first_p.toxml()) print(first_p.getAttribute('class')) all_a = model.getElementsByTagName('a') print(all_a[0].toxml()) pprint.pprint([element.toxml() for element in all_a]) pprint.pprint([element.toxml() for element in find(model, id='link3')]) for element in all_a: print(element.getAttribute('href')) print(*get_text(model), sep='\n') class DocumentParser(html.parser.HTMLParser): # noinspection SpellCheckingInspection def __init__(self, *, convert_charrefs=True): super().__init__(convert_charrefs=convert_charrefs) self.document = self.focus = xml.dom.minidom.DOMImplementation() \ .createDocument(None, None, None) @property def document_has_focus(self): return self.document is self.focus def handle_starttag(self, tag, attrs): element = self.document.createElement(tag) for name, value in attrs: element.setAttribute(name, value) self.focus.appendChild(element) self.focus = element def handle_endtag(self, tag): while self.focus.tagName != tag: self.focus = self.focus.parentNode self.focus = self.focus.parentNode def handle_data(self, data): if not self.document_has_focus and not data.isspace(): self.focus.appendChild(self.document.createTextNode(data.strip())) def error(self, message): raise RuntimeError(message) def close(self): super().close() while not self.document_has_focus: self.focus = self.focus.parentNode def find(element, **kwargs): get_attribute = getattr(element, 'getAttribute', None) if get_attribute and \ all(get_attribute(key) == value for key, value in kwargs.items()): yield element for child in element.childNodes: yield from find(child, **kwargs) def get_nodes_by_type(node, node_type): if node.nodeType == node_type: yield node for child in node.childNodes: yield from get_nodes_by_type(child, node_type) def get_text(node): return (node.data for node in get_nodes_by_type(node, node.TEXT_NODE)) if __name__ == '__main__': main()

python parsing

回答 1 投票 0

如何在Python中高效解析Html？

我将在没有外部库的情况下解析Html代码效率。我已经准备好尝试并检查它是哪个符号。原来是这样的：列表=[] html =“”“<...

python-3.x parsing

回答 1 投票 0

Bison 平衡括号语法

使用当前的解析器，我能够做到这两点：让 (x, y, z) = (10, true, 5); 让 (x) = (10); 但我不能这样做：让 (((x)), y, z) = (10, true, 5); 这是我的语法：标识符列表：标识符

parsing compiler-construction bison yacc bisonc++

回答 1 投票 0

从响应中获取数组元素

我想从 HTTP 响应中获取数组中的第二个元素，但我不明白如何做。 // 发出 GET 请求响应，错误 := http.Get("https://api.binance.com/api/v3/klines?