我想根据某些标准将子元素从一棵树替换到另一棵树。我可以使用理解来做到这一点吗?但是我们如何替换ElementTree中的元素呢?
您无法替换
ElementTree
中的元素,只能使用 Element
。
即使您拨打
ElementTree.find()
,它也只是 getroot().find()
的快捷方式。
所以你真的需要:
如果您的目标是根子元素,则提取父元素会很容易(只需调用
getroot()
),否则您必须找到它。
我是Python新手,但我找到了一种狡猾的方法来做到这一点:
输入文件
input1.xml
:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<import ref="input2.xml" />
<name awesome="true">Chuck</name>
</root>
输入文件
input2.xml
:
<?xml version="1.0" encoding="UTF-8"?>
<foo>
<bar>blah blah</bar>
</foo>
Python代码:(注意,凌乱且hacky)
import os
import xml.etree.ElementTree as ElementTree
def getElementTree(xmlFile):
print "-- Processing file: '%s' in: '%s'" %(xmlFile, os.getcwd())
xmlFH = open(xmlFile, 'r')
xmlStr = xmlFH.read()
et = ElementTree.fromstring(xmlStr)
parent_map = dict((c, p) for p in et.getiterator() for c in p)
# ref: https://stackoverflow.com/questions/2170610/access-elementtree-node-parent-node/2170994
importList = et.findall('.//import[@ref]')
for importPlaceholder in importList:
old_dir = os.getcwd()
new_dir = os.path.dirname(importPlaceholder.attrib['ref'])
shallPushd = os.path.exists(new_dir)
if shallPushd:
print " pushd: %s" %(new_dir)
os.chdir(new_dir) # pushd (for relative linking)
# Recursing to import element from file reference
importedElement = getElementTree(os.path.basename(importPlaceholder.attrib['ref']))
# element replacement
parent = parent_map[importPlaceholder]
index = parent._children.index(importPlaceholder)
parent._children[index] = importedElement
if shallPushd:
print " popd: %s" %(old_dir)
os.chdir(old_dir) # popd
return et
xmlET = getElementTree("input1.xml")
print ElementTree.tostring(xmlET)
给出输出:
-- Processing file: 'input1.xml' in: 'C:\temp\testing'
-- Processing file: 'input2.xml' in: 'C:\temp\testing'
<root>
<foo>
<bar>blah blah</bar>
</foo><name awesome="true">Chuck</name>
</root>
这是根据以下信息得出的结论:
是的!使用切片语法,您可以通过理解来完成它。例如:
import xml.etree.ElementTree as etree
path = r'C:\Users\XXX\myfile.xml'
tree = etree.parse(path)
root = tree.getroot()
root[:] = [child for child in root if child.get('some-attribute') == 'something']
tree.write(f'filtered_{path}')