如何在Python 3.3中使用ElementTree选择特定元素的所有后代？

Question

这是一个示例数据。

输入文件是 input.xml

<root>
    <entry id="1">
    <headword>go</headword>
    <example>I <hw>go</hw> to school.</example>
</entry>
</root>

我想把某个节点和它的所有子节点放到一个新的地方，也就是，

输出文件是 output.xml

<root>
    <entry id="1">
    <headword>go</headword>
            <examplegrp>
                <example>I <hw>go</hw> to school.</example>
            </examplegrp>
</entry>
</root>

我写的这个不太完整的脚本是：

import codecs
import xml.etree.ElementTree as ET

fin = codecs.open(r'input.xml', 'rb', encoding='utf-8')

data = ET.parse(fin)
root = data.getroot()

example = root.find('.//example')
for elem in example.iter():
    ---and then I don't know what to do---

Answer 1

http://docs.python.org/3/library/xml.dom.html?highlight=xml#node-objects http://docs.python.org/3/library/xml.dom.html?highlight=xml#document-objects

你可能想要遵循一种创建文档元素的方式，然后把每个结果都加到这个元素里。

group = Document.createElement(tagName)
for found in founds:
    group.appendNode(found)

或者类似这样做

Answer 2

这里有一个例子，展示了怎么做：

text = """
<root>
    <entry id="1">
        <headword>go</headword>
        <example>I <hw>go</hw> to school.</example>
    </entry>
</root>
"""

import lxml.etree
import StringIO

data = lxml.etree.parse(StringIO.StringIO(text))
root = data.getroot()

for entry in root.xpath('//example/ancestor::entry[1]'):
    examplegrp = lxml.etree.SubElement(entry,"examplegrp")
    nodes = [node for node in entry.xpath('./example')]
    for node in nodes:
        entry.remove(node)
        examplegrp.append(node)

print lxml.etree.tostring(root,pretty_print=True)

这个代码会输出：

<root>
    <entry id="1">
        <headword>go</headword>
        <examplegrp><example>I <hw>go</hw> to school.</example>
    </examplegrp></entry>
</root>