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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 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(),
            }