python - 遍历 lxml etree 中的文本和元素

Question

假设我有以下 XML 文档:

<species>
    Mammals: <dog/> <cat/>
    Reptiles: <snake/> <turtle/>
    Birds: <seagull/> <owl/>
</species>

然后我得到这样的 species 元素:

import lxml.etree
doc = lxml.etree.fromstring(xml)
species = doc.xpath('/species')[0]

现在我想打印一份按物种分组的动物列表。我如何使用 ElementTree API 来做到这一点？

Answer 1

如果您枚举所有节点，您将看到一个带有类的文本节点，后跟带有物种的元素节点:

>>> for node in species.xpath("child::node()"):
...     print type(node), node
... 
<class 'lxml.etree._ElementStringResult'> 
    Mammals: 
<type 'lxml.etree._Element'> <Element dog at 0xe0b3c0>
<class 'lxml.etree._ElementStringResult'>  
<type 'lxml.etree._Element'> <Element cat at 0xe0b410>
<class 'lxml.etree._ElementStringResult'> 
    Reptiles: 
<type 'lxml.etree._Element'> <Element snake at 0xe0b460>
<class 'lxml.etree._ElementStringResult'>  
<type 'lxml.etree._Element'> <Element turtle at 0xe0b4b0>
<class 'lxml.etree._ElementStringResult'> 
    Birds: 
<type 'lxml.etree._Element'> <Element seagull at 0xe0b500>
<class 'lxml.etree._ElementStringResult'>  
<type 'lxml.etree._Element'> <Element owl at 0xe0b550>
<class 'lxml.etree._ElementStringResult'> 

所以你可以从那里构建它:

my_species = {}
current_class = None
for node in species.xpath("child::node()"):
    if isinstance(node, lxml.etree._ElementStringResult):
        text = node.strip(' \n\t:')
        if text:
            current_class = my_species.setdefault(text, [])
    elif isinstance(node, lxml.etree._Element):
        if current_class is not None:
            current_class.append(node.tag)
print my_species

结果

{'Mammals': ['dog', 'cat'], 'Reptiles': ['snake', 'turtle'], 'Birds': ['seagull', 'owl']}

这一切都很脆弱......文本节点排列方式的微小变化可能会扰乱解析。