Python XML解析-获取XML文件名

我有一个工作的Python2.7脚本，它解析在同一目录中找到的所有XML文件，并将它们保存到CSV文件中

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问题: 如何修改下面的代码以显示（在CSV输出文件中）解析的记录来自哪个XML文件

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示例XML文件 Fruit1.xml

电流输出 期望输出

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我曾尝试使用元组将XML文件名与解析的XML元素配对，但未能实现此功能。

实现此功能：

from itertools import izip
import xml.etree.ElementTree as et
import os
import csv

fruit = []
color = []
filenames = []

directory = os.path.dirname(os.path.realpath(__file__))

for filename in os.listdir(directory):
    if filename.endswith('.xml'):
        e = et.parse(filename).getroot()
        for x in e.findall('CASE'): fruit.append(x.get('Fruit'))
        for x in e.findall('CASE'): color.append(x.get('Color'))
        for x in e.findall('CASE'): filenames.append(filename)

with open('parsed.csv', 'wb') as myfile:
    fieldnames = ["fruit","color","filenames"]
    writer = csv.writer(myfile)
    writer.writerow(fieldnames)
    writer.writerows(izip(fruit,color,filenames))

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为什么代码的XML处理和文件名处理方面彼此有任何关系？整个“扎根第一元素并扩展它”的设计是。。。我想这是一个可行的选项，但是你可以让你的代码返回一个

{filename:element\u root}

格式的映射。更好的是，

{filename:（水果，颜色）}

或者诸如此类的格式——这也减少了你的内存需求，因为你一次只需要存储一个解析过的DOM……见鬼，您可以实现一个迭代器，返回

（文件名、水果、颜色）

元组。@CharlesDuffy感谢您的输入！我对Python还是相当陌生的，有一段时间我很困惑，但我找到了答案，并将发布答案

<CASES>
  <CASE
      Fruit="Banana"
      Color="Yellow">
  </CASE>
  <CASE
      Fruit="Orange"
      Color="Orange">
  </CASE>
</CASES>

from itertools import izip
import xml.etree.ElementTree as et
import os
import csv

def run(files):
    first = None
    for filename in os.listdir(files):
        if filename.endswith('.xml'):
            e = et.parse(filename).getroot()
            if first is None: first = e
            else: first.extend(e)
    return first

fruit = []
color = []

e = run(os.path.dirname(os.path.realpath(__file__)))

for x in e.findall('CASE'): fruit.append(x.get('Fruit'))
for x in e.findall('CASE'): color.append(x.get('Color'))

with open('parsed.csv', 'wb') as myfile:
    fieldnames = ["fruit","color"]
    writer = csv.writer(myfile)
    writer.writerow(fieldnames)
    writer.writerows(izip(fruit,color))

+--------+--------+
| fruit  | color  |
+--------+--------+
| Apple  | Red    |
| Pear   | Yellow |
| Banana | Yellow |
| Orange | Orange |
+--------+--------+

+--------+--------+------------+
| fruit  | color  |  filename  |
+--------+--------+------------+
| Apple  | Red    | fruit1.xml |
| Pear   | Yellow | fruit1.xml |
| Banana | Yellow | fruit2.xml |
| Orange | Orange | fruit2.xml |
+--------+--------+------------+

from itertools import izip
import xml.etree.ElementTree as et
import os
import csv

fruit = []
color = []
filenames = []

directory = os.path.dirname(os.path.realpath(__file__))

for filename in os.listdir(directory):
    if filename.endswith('.xml'):
        e = et.parse(filename).getroot()
        for x in e.findall('CASE'): fruit.append(x.get('Fruit'))
        for x in e.findall('CASE'): color.append(x.get('Color'))
        for x in e.findall('CASE'): filenames.append(filename)

with open('parsed.csv', 'wb') as myfile:
    fieldnames = ["fruit","color","filenames"]
    writer = csv.writer(myfile)
    writer.writerow(fieldnames)
    writer.writerows(izip(fruit,color,filenames))