提取大型LEI XML指定字段生成CSV时报AttributeError该如何解决

我需要用Excel打开一份大型XML数据集，数据量未超出Excel工作簿上限。我尝试从XML中提取四列数据导出为CSV，代码如下：

cols = ["StartNodeID", "EndNodeID", "RelationshipType", "RelationshipDate_EndDate"]
rows = [] 

xmlparse = Xet.parse('D:\Descargas\concatenated_lei2file_StartNodeID.xml')
root = xmlparse.getroot()

Relationships = tree.findall('./RelationshipRecord/Relationship')

for elem in Relationships:
   StartNodeID = elem.find("./RelationshipRecord/Relationship/EndNode/EndNodeID").text
   EndNodeID = elem.find("EndNodeID").text
   RelationshipType = elem.find("RelationshipType").text
   RelationshipDate_EndDate = elem.find("RelationshipDate_EndDate").text

rows.append({"StartNodeID": StartNodeID,
             "EndNodeID": EndNodeID,
             "RelationshipType": RelationshipType,
             "RelationshipDate_EndDate": RelationshipDate_EndDate})

df = pd.DataFrame(rows, columns=cols)

df.to_csv('D:\Descargas\concatenated_lei2file_output.csv')

运行后报错如下：

----> 4     StartNodeID = elem.find("./RelationshipRecord/Relationship/EndNode/EndNodeID").text
  5     EndNodeID = elem.find("EndNodeID").text
  6     RelationshipType = elem.find("RelationshipType").text
AttributeError: 'NoneType' object has no attribute 'text'

该数据集体量较大，我已通过firstobjectXML Editor查看过XML树结构。我已尝试以下操作：

• 用Sublime Text简化树结构，从"Relationship"层级开始解析
• 删除其余字段，仅保留StartNode.NodeID与EndNode.NodeID
  但仍出现相同报错，请问该如何解决该问题？

你出现报错是代码存在几个基础逻辑错误，逐一修正即可：

1. 变量名未定义：你前面解析XML返回的对象是xmlparse，但后面调用findall用的是未定义的tree变量，直接会导致异常。
2. XML查找路径重复：findall('./RelationshipRecord/Relationship')已经拿到了所有<Relationship>节点作为遍历对象elem，你在elem下再查找带RelationshipRecord/Relationship前缀的路径属于路径重复，必然返回空。
3. 缩进错误：rows.append写在for循环外面，只会追加最后一次遍历的结果，无法存储全量数据。
4. 空值未处理：部分节点可能不存在目标子节点，直接调用.text会触发AttributeError。

修正后可运行的代码如下：

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

cols = ["StartNodeID", "EndNodeID", "RelationshipType", "RelationshipDate_EndDate"]
rows = [] 

# 路径加r前缀避免转义字符出错
xmlparse = Xet.parse(r'D:\Descargas\concatenated_lei2file_StartNodeID.xml')
root = xmlparse.getroot()

# 修正变量名，从根节点查找目标节点
Relationships = root.findall('./RelationshipRecord/Relationship')

for elem in Relationships:
    # 直接在当前Relationship节点下查找子节点，加判空逻辑
    StartNodeID = elem.find("./StartNode/StartNodeID").text if elem.find("./StartNode/StartNodeID") is not None else ""
    EndNodeID = elem.find("./EndNode/EndNodeID").text if elem.find("./EndNode/EndNodeID") is not None else ""
    RelationshipType = elem.find("RelationshipType").text if elem.find("RelationshipType") is not None else ""
    RelationshipDate_EndDate = elem.find("RelationshipEndDate").text if elem.find("RelationshipEndDate") is not None else ""

    # 缩进放到循环内，每条数据都追加
    rows.append({
        "StartNodeID": StartNodeID,
        "EndNodeID": EndNodeID,
        "RelationshipType": RelationshipType,
        "RelationshipDate_EndDate": RelationshipDate_EndDate
    })

df = pd.DataFrame(rows, columns=cols)
# 输出加utf-8-sig编码，保证Excel打开不会乱码
df.to_csv(r'D:\Descargas\concatenated_lei2file_output.csv', index=False, encoding='utf-8-sig')

如果运行后还是找不到节点，可以先打印前几条节点的子标签确认实际结构：

for elem in Relationships[:5]:
    print([child.tag for child in elem.iter()])
    break

内容的提问来源于stack exchange，提问作者mara875