使用 xpath、lxml、python 的父属性条件的元素路径

Question

我正在使用 lxml 进行项目。这是一个示例 xml

<PatientsTree>
  <Patient PatientID="SKU065427">    
    <Study StudyInstanceUID="25.2.9.2.1107.5.1.4.49339.30000006050107501192100000001">
      <Series SeriesInstanceUID="2.16.840.1.113669.1919.1176798690"/>
      <Series SeriesInstanceUID="2.16.840.1.113669.1919.1177084041"/>
      <Series SeriesInstanceUID="25.2.9.2.1107.5.1.4.49339.30000006050108064034300000000"/>
    </Study>    
  </Patient>
  <Patient PatientID="SKU55527">
    <Study StudyInstanceUID="25.2.9.2.1107.5.1.4.49339.30000006120407393721800000007">
      <Series SeriesInstanceUID="2.16.840.1.113669.1919.1198835144"/>
    </Study>
    <Study StudyInstanceUID="25.2.9.2.1107.5.1.4.49339.30000007010207164403100000013">
      <Series SeriesInstanceUID="2.16.840.1.113669.1919.1198835358"/>    
  </Patient>
</PatientsTree>

假设我想使用条件

1. PatientID="SKU55527"
2. StudyInstanceUID="25.2.9.2.1107.5.1.4.49339.30000007010207164403100000013"; 访问系列元素；

我的结果将是：

<Series SeriesInstanceUID="2.16.840.1.113669.1919.1198835358"/>  

如果我能理解这个解决方案，那么我将在学习 xml 方面更近一步。 PS 我正在使用 python、lxml 和 xpath

Answer 1

import lxml.etree as le
with open('data.xml') as f:
    doc=le.parse( f )
patientID="SKU55527"
studyInstanceUID="25.2.9.2.1107.5.1.4.49339.30000007010207164403100000013"
xpath='''\
    /PatientsTree
        /Patient[@PatientID="{p}"]
            /Study[@StudyInstanceUID="{s}"]
                /Series'''.format(p=patientID,s=studyInstanceUID)
seriesInstanceUID=doc.xpath(xpath)
for node in seriesInstanceUID:
    print(node.attrib)
    # {'SeriesInstanceUID': '2.16.840.1.113669.1919.1198835358'}

Answer 2

此 XPath 表达式：

/PatientsTree 
  /Patient[@PatientID='SKU55527']     
    /Study[@StudyInstanceUID =
           '25.2.9.2.1107.5.1.4.49339.30000007010207164403100000013'] 
      /Series 

选择此节点的结果：

<Series SeriesInstanceUID="2.16.840.1.113669.1919.1198835358"/>  

Answer 3

如果你想原生使用 lxml 而不是 xpath：（否则，unutbu 的解决方案是完美的）

from lxml import etree as ET
tree = ET.parse('some_file.xml')
patientID="SKU55527"
studyInstanceUID="25.2.9.2.1107.5.1.4.49339.30000007010207164403100000013"
patient_node = tree.find(patientID)
if not patient_node is None:
    study_node = patient_node.find(studyInstanceUID)
    if not study_node is None:
        for child in study_node.getchildren():
            print child.attrib
            #or do whatever useful thing you want
    else:
        #didn't find the study
else:
    #didn't find the node