如果你想要一个给定目录下的所有路径的平面列表(比如find。在壳中):

   files = [ 
       os.path.join(parent, name)
       for (parent, subdirs, files) in os.walk(YOUR_DIRECTORY)
       for name in files + subdirs
   ]

若要只包含基本目录下文件的完整路径，请省略+ subdirs。

确保你理解os.walk的三个返回值:

for root, subdirs, files in os.walk(rootdir):

具有以下含义:

root:被“遍历”的当前路径 subdirs:目录类型根目录下的文件 files:根目录下(不是subdirs目录下)的非directory类型的文件

请使用os.path.join而不是用斜杠连接!您的问题是filePath = rootdir + '/' + file -您必须连接当前“行走”的文件夹，而不是最上面的文件夹。filePath = os。path。加入(根、文件)。顺便说一句，“文件”是内置的，所以你通常不使用它作为变量名。

另一个问题是你的循环，应该是这样的，例如:

import os
import sys

walk_dir = sys.argv[1]

print('walk_dir = ' + walk_dir)

# If your current working directory may change during script execution, it's recommended to
# immediately convert program arguments to an absolute path. Then the variable root below will
# be an absolute path as well. Example:
# walk_dir = os.path.abspath(walk_dir)
print('walk_dir (absolute) = ' + os.path.abspath(walk_dir))

for root, subdirs, files in os.walk(walk_dir):
    print('--\nroot = ' + root)
    list_file_path = os.path.join(root, 'my-directory-list.txt')
    print('list_file_path = ' + list_file_path)

    with open(list_file_path, 'wb') as list_file:
        for subdir in subdirs:
            print('\t- subdirectory ' + subdir)

        for filename in files:
            file_path = os.path.join(root, filename)

            print('\t- file %s (full path: %s)' % (filename, file_path))

            with open(file_path, 'rb') as f:
                f_content = f.read()
                list_file.write(('The file %s contains:\n' % filename).encode('utf-8'))
                list_file.write(f_content)
                list_file.write(b'\n')

如果你不知道，文件的with语句是一种简写:

with open('filename', 'rb') as f:
    dosomething()

# is effectively the same as

f = open('filename', 'rb')
try:
    dosomething()
finally:
    f.close()

在我看来，os.walk()有点太复杂和啰嗦了。你可以做接受的答案清洁:

all_files = [str(f) for f in pathlib.Path(dir_path).glob("**/*") if f.is_file()]

with open(outfile, 'wb') as fout:
    for f in all_files:
        with open(f, 'rb') as fin:
            fout.write(fin.read())
            fout.write(b'\n')

同意Dave Webb的观点。Walk将为树中的每个目录生成一个项。事实上，您不需要关心子文件夹。

这样的代码应该可以工作:

import os
import sys

rootdir = sys.argv[1]

for folder, subs, files in os.walk(rootdir):
    with open(os.path.join(folder, 'python-outfile.txt'), 'w') as dest:
        for filename in files:
            with open(os.path.join(folder, filename), 'r') as src:
                dest.write(src.read())

我发现下面的方法是最简单的

from glob import glob
import os

files = [f for f in glob('rootdir/**', recursive=True) if os.path.isfile(f)]

使用glob('some/path/**'， recursive=True)获取所有文件，但也包括目录名。添加if os.path.isfile(f)条件只过滤现有文件

试试这个:

import os
import sys

for root, subdirs, files in os.walk(path):

    for file in os.listdir(root):

        filePath = os.path.join(root, file)

        if os.path.isdir(filePath):
            pass

        else:
            f = open (filePath, 'r')
            # Do Stuff