Question

这个xml

<Departments orgID="123" name="xmllist">
    <Department>
        <orgID>124</orgID>
        <name>A</name>
        <type>type a</type>
        <status>Active</status>
            <Department>
                <orgID>125</orgID>
                <name>B</name>
                <type>type b</type>
                <status>Active</status>
                <Department>
                    <orgID>126</orgID>
                    <name>C</name>
                    <type>type c</type>
                    <status>Active</status>
                </Department>
            </Department>
    </Department>
    <Department>
        <orgID>109449</orgID>
        <name>D</name>
        <type>type d</type>
        <status>Active</status>
    </Department>
</Departments>

如何在python中使用lxml etree获取节点的所有父节点。

预期输出：输入orgid = 126，它将返回所有父母，

{'A':124,'B':125,'C':126}

Answer 1

使用lxml和XPath：

>>> s = '''
... <Departments orgID="123" name="xmllist">
...     <Department>
...         <orgID>124</orgID>
...         <name>A</name>
...         <type>type a</type>
...         <status>Active</status>
...             <Department>
...                 <orgID>125</orgID>
...                 <name>B</name>
...                 <type>type b</type>
...                 <status>Active</status>
...                 <Department>
...                     <orgID>126</orgID>
...                     <name>C</name>
...                     <type>type c</type>
...                     <status>Active</status>
...                 </Department>
...             </Department>
...     </Department>
...     <Department>
...         <orgID>109449</orgID>
...         <name>D</name>
...         <type>type d</type>
...         <status>Active</status>
...     </Department>
... </Departments>
... '''

使用ancestor-or-self轴，您可以找到节点本身，父级，祖父母，......

>>> import lxml.etree as ET
>>> root = ET.fromstring(s)
>>> for target in root.xpath('.//Department/orgID[text()="126"]'):
...     d = {
...         dept.find('name').text: int(dept.find('orgID').text)
...         for dept in target.xpath('ancestor-or-self::Department')
...     }
...     print(d)
...
{'A': 124, 'C': 126, 'B': 125}

Answer 2

使用lxml的dim_Programacao$Classificacao_Programa[gsub("^\\s+|\\s+$", "", dim_Programacao$Classificacao) == "P"] <- "Programa"方法。

iterancestors()

输出：

from lxml import etree

doc = etree.fromstring(xml)
rval = {}
for org in doc.xpath('//orgID[text()="126"]'):
    for ancestor in org.iterancestors('Department'):
        id=ancestor.find('./orgID').text
        name=ancestor.find('./name').text
        rval[name]=id

print rval

如果你实际上试图保留元素的顺序，那么就不能使用dict，因为你无法控制dict中的键顺序。您将不得不使用OrderedDict或仅使用元组数组：

{'A': '124', 'C': '126', 'B': '125'}

输出：

doc = etree.fromstring(xml)
a = []
for org in doc.xpath('//orgID[text()="126"]'):
    for ancestor in org.iterancestors():
        if ancestor.find('./orgID') is not None:
            id=ancestor.find('./orgID').text
            name=ancestor.find('./name').text
        elif ancestor.get('orgID'):
            id=ancestor.get('orgID')
            name=ancestor.get('name')
        else:
            continue

        print id,name
        a.append((name,id))

print "In order of discovery:\n    ", a 
print "From root to child\n    ", [x for x in reversed(a)]
print "dict keys are not sorted\n    ", dict(a)

使用python获取xml节点的所有父节点

2 个答案: