[英]Python LXML etree.iterparse. Check if current element complies with XPath
I would like to read quite big XML as a stream.我想以流的形式阅读相当大的 XML。 But could not find any way to use my old XPathes to find elements.
但是找不到任何方法来使用我的旧 XPathes 来查找元素。 Previously files were of moderate size, so in was enough to:
以前的文件大小适中,因此 in 足以:
all_elements = []
for xpath in list_of_xpathes:
all_elements.append(etree.parse(file).getroot().findall(xpath))
Now I am struggling with iterparse.现在我正在为 iterparse 苦苦挣扎。 Ideally the solution would be to compare path of current element with desired xpath:
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
How is it possible to implement element_complies_with_xpath function using lxml?如何使用 lxml 实现 element_complies_with_xpath 函数?
If first part of the xpath can be extracted then the rest could be tested as follows.如果可以提取 xpath 的第一部分,则可以如下测试其余部分。 Instead of a list of strings, a dict of
<first element name>: <rest of the xpath>
could be used.可以使用
<first element name>: <rest of the xpath>
的字典来代替字符串列表。 Parent element could be used as dict key also.父元素也可以用作字典键。
Full xpath: /some/arbitrary/xpath
完整的 xpath:
/some/arbitrary/xpath
dict : {'some': './arbitrary/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 function could be used also 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'}]
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