返回所有parts.txt文件的路径列表 [英] Return a list of the paths of all the parts.txt files

问题描述

编写一个函数list_files_walk，该函数使用os模块的walk生成器返回所有parts.txt文件的路径的列表.该函数不使用任何输入参数.

Write a function list_files_walk that returns a list of the paths of all the parts.txt files, using the os module's walk generator. The function takes no input parameters.

def list_filess_walk():
    for dirpath, dirnames, filenames in os.walk("CarItems"):
        if 'parts.txt' in dirpath:
            list_files.append(filenames)
            print(list_files)
    return list_files

当前，list_files仍然为空.输出应该看起来像这样:

Currently, list_files is still empty. The output is supposed to look similar to this:

CarItems/Chevrolet/Chevelle/2011/parts.txt
CarItems/Chevrolet/Chevelle/1982/parts.txt

如何产生此输出?

推荐答案

您几乎在这里拥有了-我唯一要做的调整是:

You pretty much have it here--the only adjustments I'd make are:

• 确保list_files在本地作用域内，以避免副作用.
• 使用参数，以便函数可以在任意路径上工作.
• 使用yield关键字返回生成器，该生成器允许延迟获取下一个文件.
如果文件名恰巧是路径中其他位置的子字符串，则
• 'parts.txt' in dirpath可能容易出错.我会使用endswith或遍历元组中的第二个项目os.walk，这是当前目录中所有项目的列表，例如'parts.txt' in dirnames.
• 沿着与上述相同的思路，您可能要确保目标是

• Make sure list_files is scoped locally to the function to avoid side effects.
• Use parameters so that the function can work on any arbitrary path.
• Return a generator with the yield keyword which allows for the next file to be fetched lazily.
• 'parts.txt' in dirpath could be error-prone if the filename happens to be a substring elsewhere in a path. I'd use endswith or iterate over the second item in the tuple that os.walk which is a list of all the items in the current directory, e.g. 'parts.txt' in dirnames.
• Along the same line of thought as above, you might want to make sure that your target is a file with os.path.isfile.

这是一个例子:

import os

def find_files_rec(path, fname):
    for dirpath, dirnames, files in os.walk(path):
        if fname in files:
            yield f"{dirpath}/{fname}"

if __name__ == "__main__":
    print(list(find_files_rec(".", "parts.txt")))

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