无法从python中的xml中提取文本

Question

I have an xml file that comes from a doc (MS Word 2003, so I can't use docx library). I'm using lxml to parse it. I can get most of the text (everything is in <txt> nodes) but there are some nodes with the following structure:

<txt ptr="0x7f6354043000" id="3" symbol="8SwTxtFrm" next="4" upper="2" txtNodeIndex="9">
 <infos>
  <bounds left="1521" top="851" width="10517" height="322"/>
 </infos>
 The text I want to extract    <Special nLength="0" nType="POR_MARGIN" rText="" nWidth="2396"/>
 <Text nLength="1" nType="POR_TXT" nHeight="322" nWidth="78"/>
 <Text nLength="42" nType="POR_TXT" nHeight="322" nWidth="5647"/>
 <Special nLength="0" nType="POR_MARGIN" rText="" nWidth="2397"/>
 <LineBreak nWidth="10518"/>
 <Finish/>
</txt>

When I iter over the <txt> to extract the text part with:

for txt in tree.iter('txt'):
    print(txt.text)

I realized that it's the <infos> node that causes the problem. I tried to remove it:

for elt in tree.iter('txt'):
for info in elt.findall('infos'):
    elt.remove(info)

But this remove the targeted text along with the <infos> node, even though it is outside.

Can someone help me understand why?

Answer 1

根据我对原始帖子的评论，OP通过如下更改xpath解决了该问题

tree.xpath('//text()')

Answer 2

You can extract text this way:

In [31]: txt = """<txt ptr="0x7f6354043000" id="3" symbol="8SwTxtFrm" next="4" upper="2" txtNodeIndex="9">
   ....:  <infos>
   ....:   <bounds left="1521" top="851" width="10517" height="322"/>
   ....:  </infos>
   ....:  The text I want to extract    <Special nLength="0" nType="POR_MARGIN" rText="" nWidth="2396"/>
   ....:  <Text nLength="1" nType="POR_TXT" nHeight="322" nWidth="78"/>
   ....:  <Text nLength="42" nType="POR_TXT" nHeight="322" nWidth="5647"/>
   ....:  <Special nLength="0" nType="POR_MARGIN" rText="" nWidth="2397"/>
   ....:  <LineBreak nWidth="10518"/>
   ....:  <Finish/>
   ....: </txt>"""

In [32]: node = etree.fromstring(txt)

In [33]: ''.join(node.itertext())
Out[33]: '\n \n  \n \n The text I want to extract    \n \n \n \n \n \n'

UPD:

Answer suggested by Murali actually returns list , so you still need to join strings. And my solution is a little bit faster:

In [13]: %timeit ''.join(node.itertext())
100000 loops, best of 3: 11.7 µs per loop

In [14]: %timeit ''.join(node.xpath('//text()'))
10000 loops, best of 3: 26.3 µs per loop