【发布时间】:2018-05-27 02:20:51
【问题描述】:
我有 2 个 xml 文件, 词和主题。
我需要根据主题文件解析word文件。 文件如下
文件 1 主题
<?xml version="1.0" encoding="ISO-8859-1" standalone="yes"?>
<nite:root nite:id="ES2002a.topic"
xmlns:nite="http://nite.sourceforge.net/">
<topic nite:id="ES2002a.topic.vkaraisk.1" other_description="introduction of participants and their roles">
<nite:pointer role="scenario_topic_type" href="default-topics.xml#id(top.4)"/>
<nite:child href="ES2002a.B.words.xml#id(ES2002a.B.words0)..id(ES2002a.B.words5)"/>
<nite:child href="ES2002a.D.words.xml#id(ES2002a.D.words0)..id(ES2002a.D.words3)"/>
<nite:child
文件 2 字 (ES2002a.B.words.xml)
<?xml version="1.0" encoding="ISO-8859-1" standalone="yes"?>
<nite:root nite:id="ES2002a.B.words" xmlns:nite="http://nite.sourceforge.net/">
<w nite:id="ES2002a.B.words0" starttime="50.42" endtime="50.99">Okay</w>
<w nite:id="ES2002a.B.words1" starttime="50.99" endtime="50.99" punc="true">.</w>
<w nite:id="ES2002a.B.words2" starttime="53.56" endtime="53.96">Right</w>
<w nite:id="ES2002a.B.words3" starttime="53.96" endtime="53.96" punc="true">.</w>
<vocalsound nite:id="ES2002a.B.words4" starttime="55.415" endtime="55.415" type="other"/>
<w nite:id="ES2002a.B.words5" starttime="55.98" endtime="56.53">Um</w>
文件 2 字 (ES2002a.D.words.xml)
<?xml version="1.0" encoding="ISO-8859-1" standalone="yes"?>
<nite:root nite:id="ES2002a.D.words" xmlns:nite="http://nite.sourceforge.net/">
<w nite:id="ES2002a.D.words0" starttime="67.21" endtime="67.45">Mm-hmm</w>
<w nite:id="ES2002a.D.words1" starttime="67.45" endtime="67.45" punc="true">.</w>
<w nite:id="ES2002a.D.words2" starttime="74.89" endtime="75.24">Great</w>
<w nite:id="ES2002a.D.words3" starttime="75.24" endtime="75.24" punc="true">.</w>
<w nite:id="ES2002a.D.words4" starttime="82.08" endtime="82.25">And</w>
<w nite:id="ES2002a.D.words5" starttime="82.25" endtime="82.43">I'm</w>
有多个word文件需要根据主题文件进行解析。
<nite:child href="ES2002a.B.words.xml#id(ES2002a.B.words0)..id(ES2002a.B.words5)"/>
我们看到主题文件说从文件 ES2002a.B.words 中获取单词 1-5
想要的输出是 好的 。正确的 。嗯 嗯嗯。很棒
我已经在主题文件中解析过了,虽然代码很笨重
from lxml import etree
tree = etree.parse("./ES2013a.topic.xml")
root = tree.getroot()
childA = []
elementT = []
ElementA = []
for child in root:
elementT.append(str(child.tag))
ElementA.append(str(child.attrib))
childA.append(str(child.attrib))
for element in child:
elementT.append(str(element.tag))
#childA.append(child.attrib)
ElementA.append(str(element.attrib))
childA.append(str(child.attrib))
for sub in element:
#print('***', child.attrib , ':' , element.tag, ':' , element.attrib, '***')
#childA.append(child.attrib)
elementT.append(str(sub.tag))
ElementA.append(str(sub.attrib))
childA.append(str(child.attrib))
df = pd.DataFrame()
df['c'] = np.array (childA)
df['t'] = np.array(ElementA)
df['a'] = np.array(elementT)
file = df['t'].str.extract(r'([A-Z][A-Z].*[words.xml])#')
start = df['t'].str.extract(r'words([0-9]+)')
stop = df['t'].str.extract(r'.*words([0-9]+)')
tags = df['a'].str.extract(r'.*([topic]|[pointer]|[child])')
rootTopic = df['c'].str.extract(r'ES2013a.topic.rdhillon.(\d+)')
df['f'] = file
df['start'] = start
df['stop'] = stop
df['tags'] = tags
# c= topic
# r = pointerr
# d= child
df['topicID'] = rootTopic
df = df.iloc[:,3:]
我正在考虑获取一些使用的 word 文件,然后根据开始和停止条件迭代 word 文件
【问题讨论】:
-
问题不清楚。你已经用“beautifulsoup”、“lxml”和“elementree”标记了这个问题,但你没有向我们展示任何代码。你试过什么?
-
@mzjn 我会将我尝试过的内容添加到主帖中
标签: python xml beautifulsoup lxml elementtree