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}
答案 0 :(得分:5)
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}
答案 1 :(得分:1)
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)