在 Python 中遍历 XML

分类：编程问答

如何解决在 Python 中遍历 XML

我的数据集如下：

<?xml version="1.0" encoding="UTF-8"?>
<depts xmlns="http://SOMELINK" 
        xmlns:xsd="http://www.w3.org/2001/XMLSchema" 
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
        date="2021-01-15">
 <dept dept_id="00001" 
            col_two="00001value" 
            col_three="00001false"
            name = "some_name">     
    <owners>
      <currentowner col_four="00001value" 
                    col_five="00001value" 
                    col_six="00001false"
                    name = "some_name">
        <addr col_seven="00001value" 
                col_eight="00001value" 
                col_nine="00001false"/>
      </currentowner>
      <currentowner col_four="00001bvalue" 
                    col_five="00001bvalue" 
                    col_six="00001bfalse"
                    name = "some_name">
        <addr col_seven="00001bvalue" 
                col_eight="00001bvalue" 
                col_nine="00001bfalse"/>
      </currentowner>
    </owners>
  </dept>
  <dept dept_id="00002" 
            col_two="00002value" 
            col_three="00002value"
            name = "some_name">
    <owners>
      <currentowner col_four="00002value" 
                    col_five="00002value" 
                    col_six="00002false"
                    name = "some_name">
        <addr col_seven="00002value" 
                col_eight="00002value" 
                col_nine="00002false"/>
      </currentowner>
    </owners>
  </dept> 
</depts>

目前我有两个循环，一个遍历 child 数据，另一个遍历 granchild

import pandas
import xml.etree.ElementTree as element_tree
from xml.etree.ElementTree import parse

tree = element_tree.parse('<HERE_GOES_XML>')
root = tree.getroot()
name_space = {'ns0': 'http://SOMELINK'}

#root
date_from = root.attrib['date']
print(date_from)

#child
for pharma in root.findall('.//ns0:dept',name_space):
    for key,value in pharma.items():
        print(key +': ' + value)
    
#granchild,this must be merged to above so entire script will iterate through entire dept node to move to the next
for owner in root.findall('.//ns0:dept/ns0:owners/ns0:currentowner',name_space):
    owner_dict = {}
    
    for key,value in owner.items():
        print(key +': ' + value)

当前结果是：

2021-01-15
dept_id: 00001
col_two: 00001value
col_three: 00001false
dept_id: 00002
col_two: 00002value
col_three: 00002value
col_four: 00001value
col_five: 00001value
col_six: 00001false
col_four: 00002value
col_five: 00002value
col_six: 00002false

我的目标是嵌套外观，首先迭代整个 dept 子级及其孙子，然后才移动到下一个。预期结果将低于设置，稍后将转换为 pandas' 数据帧（接下来我将尝试处理此问题）。某些列在子/孙子之间具有相同的名称，因此需要前缀或仅循环遍历特定的 children。

dept.dept_id: 00001
dept.col_two: 00001value
dept.col_three: 00001false
dept.name: some_name
currentowner.col_four: 00001value
currentowner.col_five: 00001value
currentowner.col_six: 00001false
currentowner.name: some_name

currentowner.col_four: 00001bvalue
currentowner.col_five: 00001bvalue
currentowner.col_six: 00001bfalse
currentowner.name: some_name

addr.col_seven: 00001value
addr.col_eight: 00001value
addr.col_nine: 00001false

dept.dept_id: 00002
dept.col_two: 00002value
dept.col_three: 00002value
dept.name: some_name
currentowner.col_four: 00002value
currentowner.col_five: 00002value
currentowner.col_six: 00002false
currentowner.name: some_name
addr.col_seven: 00002value
addr.col_eight: 00002value
addr.col_nine: 00002false

[UPDATE] - 我遇到了 zip 这应该可以解决问题。

dept_list = []
for item in root.iterfind('.//ns0:dept',name_space):
    #print(item.attrib)
    dept_list.append(item.attrib)
#print(dept_list)


owner_list = []
for item in root.iterfind('.//ns0:dept/ns0:owners/ns0:currentowner',name_space):
    #print(item.attrib)
    owner_list.append(item.attrib)
#print(owner_list)

zipped = zip(dept_list,owner_list)

解决方法

您可以执行深度优先搜索：

root = ElementTree.parse('data.xml').getroot()
ns = {'ns0': 'http://SOMELINK'}

date_from = root.get('date')
print(f'{date_from=}')

for dept in root.findall(f'./ns0:dept',ns):
    for key,value in dept.items():
        print(f'{key}: {value}')
    
    for node in dept.findall('.//*'):
        for key,value in node.items():
            print(f'{key}: {value}')
            
    print()
,

循环可以在列表理解中完成，然后通过导航 DOM 构建 dict。以下代码直接进入数据框。

xml = """<depts xmlns="http://SOMELINK" 
        xmlns:xsd="http://www.w3.org/2001/XMLSchema" 
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
        date="2021-01-15">
  <dept dept_id="00001" 
            col_two="00001value" 
            col_three="00001false">
    <owners>
      <currentowner col_four="00001value" 
                    col_five="00001value" 
                    col_six="00001false">
        <addr col_seven="00001value" 
                col_eight="00001value" 
                col_nine="00001false"/>
      </currentowner>
    </owners>
  </dept>
  <dept dept_id="00002" 
            col_two="00002value" 
            col_three="00002value">
    <owners>
      <currentowner col_four="00002value" 
                    col_five="00002value" 
                    col_six="00002false">
        <addr col_seven="00002value" 
                col_eight="00002value" 
                col_nine="00002false"/>
      </currentowner>
    </owners>
  </dept> 
</depts>"""

import xml.etree.ElementTree as ET
import pandas as pd

root = ET.fromstring(xml)

root.attrib
ns = {'ns0': 'http://SOMELINK'}
pd.DataFrame([{**d.attrib,**d.find("ns0:owners/ns0:currentowner",ns).attrib,**d.find("ns0:owners/ns0:currentowner/ns0:addr",ns).attrib}
 for d in root.findall("ns0:dept",ns)
])

更安全的版本

如果任何部门没有currentowner或currentowner/addr，使用.attrib都会失败。考虑这些元素是可选的，遍历 DOM。 dict 键结构更改为基于元素标签和属性名称的名称。根据您的数据设计构建理解的方式。需要考虑1对1，1对可选，1对多。真的可以追溯到 Codd 在 1970 年写的论文

import xml.etree.ElementTree as ET
import pandas as pd

root = ET.fromstring(xml)
ns = {'ns0': 'http://SOMELINK'}
pd.DataFrame([{**{f"{d.tag.split('}')[1]}.{k}":v for k,v in d.items()},**{f"{co.tag.split('}')[1]}.{k}":v  for k,v in co.items()},**{f"{addr.tag.split('}')[1]}.{k}":v for addr in co.findall("ns0:addr",ns) for k,v in addr.items()} }
 for d in root.findall("ns0:dept",ns)
 for co in d.findall("ns0:owners/ns0:currentowner",ns)
])

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