docs: def parse_resources(html):
      """Return a list of resource names found in the html metadata section."""
      xpath = '//*[@data-type="resources"]//xhtml:li/xhtml:a'
      for resource in html.xpath(xpath, namespaces=HTML_DOCUMENT_NAMESPACES):
          yield {
              'id': resource.get('href'),
              'filename': resource.text.strip(),
              }