返回所有parts.txt文件的路径列表 [英] Return 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part.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_files_walk():
    for dirpath, dirnames, filenames in os.walk("CarItems"):
        if 'parts.txt' in dirpath:
        list_files.append(filenames)
        print(list_files)
    return list_files

输出(list_files)应该看起来像这样:

The output (list_files) 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

您已经关闭.出于某种原因，您在filenames中搜索parts.txt时应该在filenames中进行搜索:

You were close. You're searching for parts.txt in dirpath for some reason when you should be searching it in filenames:

def list_files_walk():
    results = []
    for dirpath, dirnames, filenames in os.walk("CarItems"):
        for f in filenames:
            if f.endswith('parts.txt'):
                results.append(os.path.join(dirpath, f))
    return results

我使用endswith是因为它比问"parts.txt"是否在文件名in处更准确，但是在大多数情况下也可以:

I use endswith because it is more accurate than just asking if "parts.txt" is somewhere in the filename, but that would also work in most cases:

def list_files_walk():
    results = []
    for dirpath, dirnames, filenames in os.walk("CarItems"):
        for f in filenames:
            if 'parts.txt' in f:
                results.append(os.path.join(dirpath, f))
    return results

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