python lxml etree.iterparse。检查当前元素是否符合XPath

Question

我想阅读相当大的XML作为流。但是找不到任何使用我的旧XPathes找到元素的方法。 以前文件的大小适中，因此足以：

all_elements = []
for xpath in list_of_xpathes:
    all_elements.append(etree.parse(file).getroot().findall(xpath))

现在我在iterparse上挣扎。理想情况下，解决方案是将当前元素的路径与所需的XPath进行比较：

import lxml.etree as et

xml_file = r"my.xml" # quite big xml, that i should read
xml_paths = ['/some/arbitrary/xpath', '/another/xpath']

all_elements = []
iter = et.iterparse(xml_file, events = ('end',))
for event, element in iter:
    for xpath in xml_paths:
        if element_complies_with_xpath(element, xpath):
            all_elements.append(element)
            break

如何使用lxml实现element_complies_with_xpath函数？

Answer 1

如果可以提取XPATH的第一部分，则其余部分可以如下测试。可以使用＆lt;第一个元素名称＆gt;：＆lt; 的dist，而不是字符串列表。父元素也可以用作dict键。
完整的XPath：/some/nutary/xpath
dict：{'some'：'./arbitrary/xpath'}

import lxml.etree as et

def element_complies_with_xpath(element, xpath):
    children = element.xpath(xpath)
    print([ "child:" + x.tag for x in children])
    return len(children) > 0

xml_file = r"/home/lmc/tmp/test.xml" # quite big xml, that i should read
xml_paths = [{'membership': './users/user'}, {'entry':'author/name'}]

all_elements = []
iter1 = et.iterparse(xml_file, events = ('end',))

for event, element in iter1:
    for d in xml_paths:
        if element.tag in d and element_complies_with_xpath(element, d[element.tag]):
            all_elements.append(element)
            break

print([x.tag for x in all_elements])

count（） xpath函数也可以使用

def element_complies_with_xpath(element, xpath):
    children = element.xpath(xpath)
    print( f"child exist: {children}")
    return children

xml_file = r"/home/luis/tmp/test.xml" # quite big xml, that i should read
xml_paths = [{'membership': 'count(./users/user) > 0'}, {'entry':'count(author/name) > 0'}]