使用xpath、lxml和Python根据父属性条件查找元素路径
我正在做一个使用lxml的项目。这是一个示例的xml文件:
<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>
假设我想根据以下条件找到series元素:
- PatientID="SKU55527"
- StudyInstanceUID="25.2.9.2.1107.5.1.4.49339.30000007010207164403100000013";
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,那我就离学习xml更近一步了。顺便说一下,我正在使用python、lxml和xpath。
3 个回答
1
如果你想直接使用 lxml 而不是用 xpath 的话:(否则,unutbu 的解决方案非常好)
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
2
这个XPath表达式:
/PatientsTree
/Patient[@PatientID='SKU55527']
/Study[@StudyInstanceUID =
'25.2.9.2.1107.5.1.4.49339.30000007010207164403100000013']
/Series
会选中这个节点:
<Series SeriesInstanceUID="2.16.840.1.113669.1919.1198835358"/>
3
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'}
当然可以!请把你想要翻译的内容发给我,我会帮你把它变得简单易懂。