从目录及其所有子目录获取完整文件路径

import os

def get_filepaths(directory):
    """
    This function will generate the file names in a directory 
    tree by walking the tree either top-down or bottom-up. For each 
    directory in the tree rooted at directory top (including top itself), 
    it yields a 3-tuple (dirpath, dirnames, filenames).
    """
    file_paths = []  # List which will store all of the full filepaths.

    # Walk the tree.
    for root, directories, files in os.walk(directory):
        for filename in files:
            # Join the two strings in order to form the full filepath.
            filepath = os.path.join(root, filename)
            file_paths.append(filepath)  # Add it to the list.

    return file_paths  # Self-explanatory.

# Run the above function and store its results in a variable.   
full_file_paths = get_filepaths("/Users/johnny/Desktop/TEST")

我在上述函数中提供的路径包含3个文件，其中两个位于根目录中，另一个位于名为“subfolder”的子文件夹中打印将打印列表的完整文件路径：['/Users/johnny/Desktop/TEST/file1.txt'，'/Users/johnny/Desctop/TEST-file2.txt'，'/Users/johnny/Desktop/STEST/SUBFOLDER/file3.dat']

如果愿意，您可以打开并阅读内容，或者只关注扩展名为“.dat”的文件，如下面的代码所示：

for f in full_file_paths:
  if f.endswith(".dat"):
    print f

/用户/johnny/Desktop/TEST/SUBFOLDER/file3.dat

def list_files(path):
    # returns a list of names (with extension, without full path) of all files 
    # in folder path
    files = []
    for name in os.listdir(path):
        if os.path.isfile(os.path.join(path, name)):
            files.append(name)
    return files 

如果您正在寻找find的Python实现，这是我经常使用的方法：

from findtools.find_files import (find_files, Match)

# Recursively find all *.sh files in **/usr/bin**
sh_files_pattern = Match(filetype='f', name='*.sh')
found_files = find_files(path='/usr/bin', match=sh_files_pattern)

for found_file in found_files:
    print found_file

所以我用它制作了一个PyPI包，还有一个GitHub存储库。我希望有人发现它对这段代码可能有用。

一位聪明的老师曾经告诉我：

当有几种既定的方法来做某事时，没有一种方法对所有情况都有好处。

因此，我将为问题的一个子集添加一个解决方案：通常，我们只想检查文件是否匹配开始字符串和结束字符串，而不需要进入子目录。因此，我们需要一个返回文件名列表的函数，例如：

filenames = dir_filter('foo/baz', radical='radical', extension='.txt')

如果您想首先声明两个函数，可以这样做：

def file_filter(filename, radical='', extension=''):
    "Check if a filename matches a radical and extension"
    if not filename:
        return False
    filename = filename.strip()
    return(filename.startswith(radical) and filename.endswith(extension))

def dir_filter(dirname='', radical='', extension=''):
    "Filter filenames in directory according to radical and extension"
    if not dirname:
        dirname = '.'
    return [filename for filename in os.listdir(dirname)
                if file_filter(filename, radical, extension)]

这个解决方案可以很容易地用正则表达式来概括（如果您不希望模式总是停留在文件名的开头或结尾，您可能需要添加一个模式参数）。

这是我的通用函数。它返回文件路径列表而不是文件名，因为我发现这更有用。它有几个可选的参数，使其具有通用性。例如，我经常将其与pattern=“*.txt”或subfolders=True等参数一起使用。

import os
import fnmatch

def list_paths(folder='.', pattern='*', case_sensitive=False, subfolders=False):
    """Return a list of the file paths matching the pattern in the specified 
    folder, optionally including files inside subfolders.
    """
    match = fnmatch.fnmatchcase if case_sensitive else fnmatch.fnmatch
    walked = os.walk(folder) if subfolders else [next(os.walk(folder))]
    return [os.path.join(root, f)
            for root, dirnames, filenames in walked
            for f in filenames if match(f, pattern)]

对于Python 2：

pip install rglob

那就做吧

import rglob
file_list = rglob.rglob("/home/base/dir/", "*")
print file_list