堆栈内存溢出
  简体   繁体   English

Python lxml如何使用XPath重新发布处理XPath结果

[英]Python lxml how to post process XPath result again with XPath

 原文  2015-07-29 08:03:31  4   2    python/ python-2.7/ xpath/ lxml

问题描述

I have the following XML file: 我有以下XML文件： 

<?xml version="1.0" encoding="utf-8"?>
<!-- Created with Unified Streaming Platform(version=1.7.8) -->
<MPD
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
        xmlns="urn:mpeg:dash:schema:mpd:2011"
        xsi:schemaLocation="urn:mpeg:dash:schema:mpd:2011 http://standards.iso.org/ittf/PubliclyAvailableStandards/MPEG-DASH_schema_files/DASH-MPD.xsd"
        type="static"
        mediaPresentationDuration="PT1H43M36.832S"
        maxSegmentDuration="PT3S"
        minBufferTime="PT10S"
        profiles="urn:mpeg:dash:profile:isoff-live:2011,urn:com:dashif:dash264">
    <Period>
        <BaseURL>dash/</BaseURL>
        <AdaptationSet group="1" contentType="audio" lang="tr" minBandwidth="157405" maxBandwidth="157405"
                       segmentAlignment="true" audioSamplingRate="48000" mimeType="audio/mp4" codecs="mp4a.40.2">
            <AudioChannelConfiguration schemeIdUri="urn:mpeg:dash:23003:3:audio_channel_configuration:2011" value="2">
            </AudioChannelConfiguration>
            <Representation id="audio_tur=157405" bandwidth="157405">
            </Representation>
        </AdaptationSet>
        <AdaptationSet group="2" contentType="video" lang="en" par="16:9" minBandwidth="501000" maxBandwidth="9001000"
                       minWidth="512" maxWidth="1920" minHeight="288" maxHeight="1080" segmentAlignment="true"
                       frameRate="25" mimeType="video/mp4" startWithSAP="1">
            <Representation id="video_eng=501000" bandwidth="501000" width="512" height="288" codecs="avc1.4D401E"
                            scanType="progressive">
            </Representation>
            <Representation id="video_eng=851000" bandwidth="851000" width="640" height="360" codecs="avc1.4D401E"
                            scanType="progressive">
            </Representation>
            <Representation id="video_eng=1302000" bandwidth="1302000" width="640" height="480" sar="4:3"
                            codecs="avc1.4D401F"
                            scanType="progressive">
            </Representation>
            <Representation id="video_eng=2601000" bandwidth="2601000" width="1024" height="576" codecs="avc1.4D401F"
                            scanType="progressive">
            </Representation>
            <Representation id="video_eng=2701000" bandwidth="2701000" width="1280" height="720" codecs="avc1.4D401F"
                            scanType="progressive">
            </Representation>
            <Representation id="video_eng=3501000" bandwidth="3501000" width="1280" height="720" codecs="avc1.4D401F"
                            scanType="progressive">
            </Representation>
            <Representation id="video_eng=6001000" bandwidth="6001000" width="1440" height="1080" sar="4:3"
                            codecs="avc1.4D4028" scanType="progressive">
            </Representation>
            <Representation id="video_eng=9001000" bandwidth="9001000" width="1920" height="1080" codecs="avc1.4D4028"
                            scanType="progressive">
            </Representation>
        </AdaptationSet>
        <AdaptationSet
                group="2" contentType="video" lang="en" par="20:11" minBandwidth="1901000" maxBandwidth="1901000"
                minWidth="872" maxWidth="872" segmentAlignment="true" width="720" height="480" sar="40:33"
                frameRate="25" mimeType="video/mp4" codecs="avc1.4D401F" startWithSAP="1">
            <Representation id="video_eng=1901000" bandwidth="1901000" scanType="progressive">
            </Representation>
        </AdaptationSet>
    </Period>
</MPD>

And run the following Python code on it: 并在其上运行以下Python代码： 

from lxml import etree

file = "Data.xml"
namespaces = {'ns':'urn:mpeg:dash:schema:mpd:2011'}

tree = etree.parse(file)
root = tree.getroot()

for r in root.xpath('//ns:AdaptationSet[@contentType="video"]',namespaces=namespaces):
    print etree.tostring(r)
    for bandwidth in r.xpath('//ns:Representation/@bandwidth',namespaces=namespaces):
        print bandwidth

My problem is now that the second loop is not using the result from the xpath before but again the complete tree! 我的问题是，第二个循环没有使用xpath之前的结果，而是再次使用了完整的树！ That's why the results include representations for audio as well. 这就是为什么结果还包括音频表示的原因。 In detail it looks like the following: 详细地，它看起来如下所示： 

<AdaptationSet xmlns="urn:mpeg:dash:schema:mpd:2011" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" group="2" contentType="video" lang="en" par="16:9" minBandwidth="501000" maxBandwidth="9001000" minWidth="512" maxWidth="1920" minHeight="288" maxHeight="1080" segmentAlignment="true" frameRate="25" mimeType="video/mp4" startWithSAP="1">
            <Representation id="video_eng=501000" bandwidth="501000" width="512" height="288" codecs="avc1.4D401E" scanType="progressive">
            </Representation>
            <Representation id="video_eng=851000" bandwidth="851000" width="640" height="360" codecs="avc1.4D401E" scanType="progressive">
            </Representation>
            <Representation id="video_eng=1302000" bandwidth="1302000" width="640" height="480" sar="4:3" codecs="avc1.4D401F" scanType="progressive">
            </Representation>
            <Representation id="video_eng=2601000" bandwidth="2601000" width="1024" height="576" codecs="avc1.4D401F" scanType="progressive">
            </Representation>
            <Representation id="video_eng=2701000" bandwidth="2701000" width="1280" height="720" codecs="avc1.4D401F" scanType="progressive">
            </Representation>
            <Representation id="video_eng=3501000" bandwidth="3501000" width="1280" height="720" codecs="avc1.4D401F" scanType="progressive">
            </Representation>
            <Representation id="video_eng=6001000" bandwidth="6001000" width="1440" height="1080" sar="4:3" codecs="avc1.4D4028" scanType="progressive">
            </Representation>
            <Representation id="video_eng=9001000" bandwidth="9001000" width="1920" height="1080" codecs="avc1.4D4028" scanType="progressive">
            </Representation>
        </AdaptationSet>

157405
501000
851000
1302000
2601000
2701000
3501000
6001000
9001000
1901000
<AdaptationSet xmlns="urn:mpeg:dash:schema:mpd:2011" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" group="2" contentType="video" lang="en" par="20:11" minBandwidth="1901000" maxBandwidth="1901000" minWidth="872" maxWidth="872" segmentAlignment="true" width="720" height="480" sar="40:33" frameRate="25" mimeType="video/mp4" codecs="avc1.4D401F" startWithSAP="1">
            <Representation id="video_eng=1901000" bandwidth="1901000" scanType="progressive">
            </Representation>
        </AdaptationSet>

157405
501000
851000
1302000
2601000
2701000
3501000
6001000
9001000
1901000

So even the right AdaptionSet's are found, for both iterations the complete tree is processed. 因此，即使找到了正确的AdaptionSet，对于两次迭代而言，完整的树也会被处理。 I know I could build a XPath just getting the bandwith, but I need the AdaptionSet before and would love to use only the result from the first loop for the second. 我知道我可以建立一个XPath来获取带宽，但是我之前需要AdaptionSet，并且很乐意仅将第一个循环的结果用于第二个循环。 How can I do this? 我怎样才能做到这一点？

2 个解决方案

解决方案1
 1 已采纳  2015-07-29 08:08:30

You got to add . 您必须添加. at the beginning of your XPath to make it relative to current context node, which in this case referenced by the variable r : 在XPath的开头，使其相对于当前上下文节点，在这种情况下，该节点由变量r引用： 

r.xpath('.//ns:Representation/@bandwidth',namespaces=namespaces)

This behavior is mentioned in the XPath 1.0 documentation as follow : XPath 1.0文档中提到了此行为，如下所示：

  • //para selects all the para descendants of the document root and thus selects all para elements in the same document as the context node //para选择文档根目录的所有para后代，并因此选择同一文档中的所有para元素作为上下文节点

  • .//para selects the para element descendants of the context node .//para选择上下文节点的para元素后代

解决方案2
 1  2015-07-29 08:08:40

Try using relative xpath - 尝试使用相对xpath- 

for bandwidth in r.xpath('.//ns:Representation/@bandwidth',namespaces=namespaces):

. would make the xpath start at the current element. 将使xpath从当前元素开始。 If the . 如果. is not specified, as you observed, the xpath would query from the root node. 如您所观察到的那样，未指定xpath将从根节点查询。

暂无
暂无

声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

相关问题 Python，如何使用lxml XPath？ - Python, How to use lxml XPath? 在 Python 中使用 XPath 和 LXML - Using XPath in Python with LXML Python lxml XPath问题 - Python lxml XPath problem python lxml xpath：如何使此谓词正常工作 - python lxml xpath: how to get this predicate working python中的lxml xpath，如何处理缺少的标签？ - lxml xpath in python, how to handle missing tags? 如何使用 Python 和 lxml 在 xpath 中输入特定节点 - How to enter a specific node in xpath with Python and lxml 在 Python 中有没有更好的方法来分配 lxml xpath() 的第一个结果？ - In Python is there a better way to assign the first result from lxml xpath()? Python lxml XPath：前面的关键字未给出预期的结果 - Python lxml XPath : preceding keyword does not give expected result 使用 LXML 在 Python 中查找字符串的 XPath - Find the XPath of a string in Python with LXML Python＆lxml / xpath：解析XML - Python & lxml / xpath: Parsing XML
相关标签
 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM