从文件夹路径中读取特定文件夹的名称 Python

Question

我正在尝试从文件路径中读取特定位置的文件夹名称。 我当前的代码：

import os

# search for and input multiple files 
def get_files(source):
    matches = []
    for root, dirnames, filenames in os.walk(source):
        for filename in filenames:
            matches.append(os.path.join(root, filename))
    return matches

def parse(files):
    for file in files:
        xml_information = {}
        metadata = []
        # Get the file path
        filepath = os.path.dirname(file)
        xml_information['file_path'] = '%s' % filepath
        
        # Get customer name
        customer = filepath.split("\\")[5]
        xml_information['customer_name'] = '%s' % customer
        metadata.append(xml_information)
        print(metadata)

path = 'C:\\Users\\quan.nguyen\\SAGE\\Lania Thompson - Searching Project Files'

parse(get_files(path))

我的程序搜索文件夹并找到文件并报告它们的文件夹路径。 但是，我想读取文件夹路径以及第六位的文件夹名称，即客户名称。 当我运行customer = filepath.split("\\")[5]它报告错误：

Traceback (most recent call last):
  File "*hidden*", line 33, in <module>
    parse(get_files(path))
  File "*hidden*", line 26, in parse
    customer = filepath.split("\\")[5]
               ~~~~~~~~~~~~~~~~~~~~^^^
IndexError: list index out of range

但是，当我使用customer = filepath.split("\\")[4]运行时，程序会运行并读取path中指定的最后一个文件夹，即 Lania Thompson - Searching Project Files。 结果如下：

[{'file_path': 'C:\\Users\\quan.nguyen\\SAGE\\Lania Thompson - Searching Project Files\\Hazor Ltd\\PCS Mah\\Machine', 'customer_name': 'Lania Thompson - Searching Project Files'}]

我的预期结果是 Hazor Ltd：

[{'file_path': 'C:\\Users\\quan.nguyen\\SAGE\\Lania Thompson - Searching Project Files\\Hazor Ltd\\PCS Mah\\Machine', 'customer_name': 'Hazor Ltd'}]

除了我的名字以外，名字都是虚构的

Answer 1

所以我已经使用 pathlib 库编写了代码。 代码是：

import os
from pathlib import Path

# search for and input multiple files 
def get_files(source):
    matches = []
    for root, dirnames, filenames in os.walk(source):
        for filename in filenames:
            matches.append(os.path.join(root, filename))
    return matches

def parse(files):
    for file in files:
        xml_information = {}
        metadata = []
        # Get the file path
        filepath = os.path.dirname(file)
        
        # Get customer name
        p = Path(filepath)
        files = [f for f in p.rglob('*') if f.is_file()]

        for f in files:
            xml_information['Customer'] = f.parts[5]
        metadata.append(xml_information)
        print(metadata)

path = 'C:\\Users\\quan.nguyen\\SAGE\\Lania Thompson - Searching Project Files'

parse(get_files(path))

将xml_information['Customer'] = f.parts[5]中的数字[5]更改为您要从中获取文件夹名称的位置。