我正在尝试合并两个xml文件。这些文件包含相同的整体结构,但细节不同。
file1.xml:
<book>
<chapter id="113">
<sentence id="1">
<word id="128160">
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPV"/>
<Number type="S"/>
</word>
<word id="128161">
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPF"/>
</word>
</sentence>
<sentence id="2">
<word id="128162">
<POS Tag="P"/>
<grammar type="PREFIX"/>
<Tag Tag="bi+"/>
</word>
</sentence>
</chapter>
</book>
file2.xml:
<book>
<chapter id="113">
<sentence id="1">
<word id="128160">
<concept English="joke"/>
</word>
<word id="128161">
<concept English="romance"/>
</word>
</sentence>
<sentence id="2">
<word id="128162">
<concept English="happiness"/>
</word>
</sentence>
</chapter>
</book>
所需的输出是:
<book>
<chapter id="113">
<sentence id="1">
<word id="128160">
<concept English="joke"/>
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPV"/>
<Number type="S"/>
</word>
<word id="128161">
<concept English="romance"/>
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPF"/>
</word>
</sentence>
<sentence id="2">
<word id="128162">
<concept English="happiness"/>
<POS Tag="P"/>
<grammar type="PREFIX"/>
<Tag Tag="bi+"/>
</word>
</sentence>
</chapter>
</book>
好吧,我尝试在路径中这样做,但我没有得到所需的输出:
import os, os.path, sys
import glob
from xml.etree import ElementTree
output = open('merge.xml','w')
files="sample"
xml_files = glob.glob(files +"/*.xml")
xml_element_tree = None
for xml_file in xml_files:
data = ElementTree.parse(xml_file).getroot()
# print ElementTree.tostring(data)
for word in data.iter('word'):
if xml_element_tree is None:
xml_element_tree = data
insertion_point = xml_element_tree.findall("book/chapter/sentence/word/*")
else:
insertion_point.extend(word)
if xml_element_tree is not None:
print>>output, ElementTree.tostring(xml_element_tree)
请,任何帮助
答案 0 :(得分:1)
这是一个解决方案。从空的合并文档开始,然后在枚举文件时,将无法找到的元素添加到合并文档中。你可以概括一下,但这是第一次削减:
import lxml.etree
merged = lxml.etree.Element('book')
for xml_file in xml_files:
for merge_chapter in lxml.etree.parse(xml_file):
try:
chapter = merged.xpath('chapter[@id=%s]' % merge_chapter.get('id'))[0]
for merge_sentence in merge_chapter:
try:
sentence = chapter.xpath('sentence[@id=%s]' % merge_sentence.get('id'))[0]
for merge_word in merge_sentence:
try:
word = sentence.xpath('word[@id=%s]' % merge_word.get('id'))[0]
for data in merge_word:
try:
word.xpath(data.tag)[0]
except IndexError:
# add newly discovered word data
word.append(data)
except IndexError:
# add newly discovered word
sentence.append(merge_word)
except IndexError:
# add newly discovered sentence
chapter.append(merge_sentence)
except IndexError:
# add newly discovered chapter
merged.append(merge_chapter)
答案 1 :(得分:0)
我过去做过类似事情的方法是创建一个xml文档,然后追加你想要的值。我不相信有一种方法可以“合并”它们
xml = ET.fromstring("<book></book>")
document = ET.parse(tempFile)
childNodeList = document.findall(xpathQuery)
for node in childNodeList:
xml.append(node)
答案 2 :(得分:0)
鉴于您要将File2合并到File1中,您可以遍历File2中的所有元素,然后将File2元素中的属性复制到File1的元素中。
我必须在我正在进行的项目上做类似的事情。这是我目前的解决方案,应该在Python 2.7下工作。
请注意,我进一步添加了在公共节点之间复制属性的要求。您将看到我将以下属性添加到A:
然后到B我补充说:
最终合并的文件包含了三个权力三人组成的所有成员。
我还添加了<sentance id='3'/>
来证明元素的顺序不再重要。
#!/usr/bin/python
from lxml import etree
from copy import deepcopy
import lxml
xmlA='''
<book>
<chapter id="113">
<sentence id="1" drums='Neil'>
<word id="128160" bass='Geddy'>
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPV"/>
<Number type="S"/>
</word>
<word id="128161">
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPF"/>
</word>
</sentence>
<sentence id="2">
<word id="128162">
<POS Tag="P"/>
<grammar type="PREFIX"/>
<Tag Tag="bi+"/>
</word>
</sentence>
</chapter>
</book>
'''
xmlB='''
<book>
<chapter id="113">
<sentence id="3">
<word id="128168">
<concept English="sadness"/>
</word>
</sentence>
<sentence id="1">
<word id="128160">
<concept English="joke"/>
</word>
<word id="128161">
<concept English="romance"/>
</word>
</sentence>
<sentence id="2" guitar='Alex'>
<word id="128162">
<concept English="happiness"/>
</word>
</sentence>
</chapter>
</book>
'''
import re
from copy import deepcopy
##
# @brief Translates the relational xpath to an explicit xpath.
# In the XML examples above, getpath will return the following for
# <sentance id='1'/>:
# - xmlA = /book/chapter/sentance[1]
# - xmlb = /book/chapter/sentance[2]
#
# A path that is explicit in both document would be:
# - xmlA = /book/chapter/sentance[@id='1']
# - xmlb = /book/chapter/sentance[@id='1']
#
def convertXpath(element):
newPath = ''
tree = element.getroottree()
path = tree.getpath(element).split('/')
root = tree.getroot()
for p in path:
if p == '':
continue
if re.search('\[[0-9]*\]', p):
# Get the element at this path
#
node = root.xpath(newPath+'/'+p)[0]
id=node.get('id')
p=re.sub('\[[0-9]*\]','', p)
newPath += '/'+p+"[@id='"+id+"']"
else:
newPath+='/'+p
return newPath
def mergeXml(a,b):
for node in a.nodes():
path = convertXpath(node)
# find the element in the other document
#
elements = b.root.xpath(path)
for e in elements:
for name, value in node.items():
if name == 'id':
continue
e.set(name,value)
if len(elements) == 0:
# Add the node to other document
#
newElement = deepcopy(node)
# Find the path to the parent
#
parent = node.getparent()
path = convertXpath(parent)
bParent = b.root.xpath(path)[0]
bParent.append(newElement)
class XmlDoc:
def __init__(self, xml):
self.root = etree.fromstring(xml)
self.tree = self.root.getroottree()
def __str__(self):
return etree.tostring(self.root, pretty_print=True)
def nodes(self):
return self.root.iter('*')
if __name__ == '__main__':
a = XmlDoc(xmlA)
b = XmlDoc(xmlB)
mergeXml(a,b)
print b
这会产生以下输出:
<book>
<chapter id="113">
<sentence id="3">
<word id="128168">
<concept English="sadness"/>
</word>
</sentence>
<sentence id="1" drums="Neil">
<word id="128160" bass="Geddy">
<concept English="joke"/>
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPV"/>
<Number type="S"/>
</word>
<word id="128161">
<concept English="romance"/>
<POS Tag="V"/>
<grammar type="STEM"/>
<Aspect type="IMPF"/>
</word>
</sentence>
<sentence id="2" guitar="Alex">
<word id="128162">
<concept English="happiness"/>
<POS Tag="P"/>
<grammar type="PREFIX"/>
<Tag Tag="bi+"/>
</word>
</sentence>
</chapter>
</book>