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(),
            }
Return a list of resource names found in the html metadata section.