如何在 python 中将 kml 文件解析为树结构？

Question

我有一个试图在 python 中解析的 kml 文件，我想要的是从 SimpleData 传递 val3 作为参数，并且将仅检索该地标标记的坐标，我之前在 xpath 上工作过： xpath 表达式中的一个典型示例是：

value = '..'
for val in (//Placement/ExtendedData/SimpleData[contains(text(), "+value+")]):
    print val.find_element_by_xpath(//coordinates)

但是我似乎无法通过在 Python 中使用元素树来获得相同的结果：

这是 kml 文件
:

<?xml version="1.0" encoding="utf-8" ?>
<kml xmlns="http://www.opengis.net/kml/2.2">
<Document id="root_doc">
<Schema name="" id="">
    <SimpleField name="NAME_0" type="string"></SimpleField>
    <SimpleField name="NAME_1" type="string"></SimpleField>
</Schema>
<Folder><name></name>
  <Placemark>
    <Style><LineStyle><color>ff0000ff</color></LineStyle><PolyStyle><fill>0</fill></PolyStyle></Style>
    <ExtendedData><SchemaData schemaUrl="#gadm36_IND_3">
        <SimpleData name="NAME_0">val1</SimpleData>
        <SimpleData name="NAME_1">val2</SimpleData>
        <SimpleData name="NAME_2">val3</SimpleData>
    </SchemaData></ExtendedData>
      <MultiGeometry><Polygon><outerBoundaryIs><LinearRing><coordinates>92.7877807617189,9.24416637420654 92.7888870239258,9.24305438995361 92.7897186279296,9.24306106567383 92.7902832031251,9.24250030517589 92.7905578613282,9.24250030517589 92.7911148071289,9.24194431304943 92.7913894653321,9.24194431304943 92.7922210693359,9.24110984802257 92.7922210693359,9.24083423614508 92.7930526733399,9.23999977111822
      </coordinates></LinearRing>...
      <Placemark>
    <Style><LineStyle><color>ff0000ff</color></LineStyle><PolyStyle><fill>0</fill></PolyStyle></Style>
    <ExtendedData><SchemaData schemaUrl="#gadm36_IND_3">
        <SimpleData name="NAME_0">val1</SimpleData>
        <SimpleData name="NAME_1">val2</SimpleData>
        <SimpleData name="NAME_2">val3</SimpleData>
    </SchemaData></ExtendedData>
      <MultiGeometry><Polygon><outerBoundaryIs><LinearRing><coordinates>92.7877807617189,9.24416637420654 92.7888870239258,9.24305438995361 92.7897186279296,9.24306106567383 92.7902832031251,9.24250030517589 92.7905578613282,9.24250030517589 92.7911148071289,9.24194431304943 92.7913894653321,9.24194431304943 92.7922210693359,9.24110984802257 92.7922210693359,9.24083423614508 92.7930526733399,9.23999977111822
      </coordinates></LinearRing>...

这就是我所坚持的：

import xml.etree.ElementTree as ET
tree = ET.parse('')
root = tree.getroot()
for val in root.findall('.//{http://www.opengis.net/kml/2.2}SimpleData[@text=""]//coordinates'):
    print val.text

Answer 1

使用lxml with XPath and namespaces。在 XPath 选择器中，您可以从带有文本 val3 的 SimpleData 导航回 Placemark 祖先，然后从那里导航至 coordinates.

from lxml import etree

tree = etree.parse("so.xml")
nsmap = {"kml": "http://www.opengis.net/kml/2.2"}

listOfCoordinates = tree.xpath("//kml:SimpleData[text()=\"val3\"]/ancestor::kml:Placemark//kml:coordinates", namespaces=nsmap)
print(listOfCoordinates[0].text)

输出：

92.7877807617189,9.24416637420654 92.7888870239258,9.24305438995361 92.7897186279296,9.24306106567383 92.7902832031251,9.24250030517589 92.7905578613282,9.24250030517589 92.7911148071289,9.24194431304943 92.79138
94653321,9.24194431304943 92.7922210693359,9.24110984802257 92.7922210693359,9.24083423614508 92.7930526733399,9.23999977111822

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