这里有一种方法:

import os
import shutil

def copy_over(path, from_name, to_name):
  for path, dirname, fnames in os.walk(path):
    for fname in fnames:
      if fname == from_name:
        shutil.copy(os.path.join(path, from_name), os.path.join(path, to_name))


copy_over('.', 'index.tpl', 'index.html')

为什么没有人提到glob?glob允许您使用unix风格的路径名展开，它是我的go to函数，几乎适用于需要查找多个路径名的所有内容。这很简单:

from glob import glob
paths = glob('*/')

请注意，glob将返回带有最后斜杠的目录(就像unix一样)，而大多数基于路径的解决方案将省略最后的斜杠。

def get_folders_in_directories_recursively(directory, index=0):
    folder_list = list()
    parent_directory = directory

    for path, subdirs, _ in os.walk(directory):
        if not index:
            for sdirs in subdirs:
                folder_path = "{}/{}".format(path, sdirs)
                folder_list.append(folder_path)
        elif path[len(parent_directory):].count('/') + 1 == index:
            for sdirs in subdirs:
                folder_path = "{}/{}".format(path, sdirs)
                folder_list.append(folder_path)

    return folder_list

下面的函数可以被调用为:

get_folders_in_directores_recurrecursive (directory, index=1) ->给出了第一层的文件夹列表

get_folders_in_directores_recurrecursive (directory) ->给出所有子文件夹

我对各种函数做了一些速度测试，以返回当前所有子目录的完整路径。

tl;博士: 总是使用scandir:

List_subfolders_with_paths = [f.]f在os.scandir(Path)中的路径if f.is_dir()]

额外的好处:使用scandir，你也可以通过使用f.name而不是f.path来获取文件夹名称。

这个函数(以及下面所有其他函数)不会使用自然排序。这意味着结果将像这样排序:1,10,2。要获得自然排序(1,2,10)，请查看https://stackoverflow.com/a/48030307/2441026

结果: scandir比walk快3倍，比listdir(带过滤器)快32倍，比Pathlib快35倍，比listdir快36倍，比glob快37倍(!)

Scandir:           0.977
Walk:              3.011
Listdir (filter): 31.288
Pathlib:          34.075
Listdir:          35.501
Glob:             36.277

用W7x64测试，Python 3.8.1。文件夹，包含440个子文件夹。 如果你想知道listdir是否可以通过不执行两次os.path.join()来加速，是的，但基本上不存在区别。

代码:

import os
import pathlib
import timeit
import glob

path = r"<example_path>"



def a():
    list_subfolders_with_paths = [f.path for f in os.scandir(path) if f.is_dir()]
    # print(len(list_subfolders_with_paths))


def b():
    list_subfolders_with_paths = [os.path.join(path, f) for f in os.listdir(path) if os.path.isdir(os.path.join(path, f))]
    # print(len(list_subfolders_with_paths))


def c():
    list_subfolders_with_paths = []
    for root, dirs, files in os.walk(path):
        for dir in dirs:
            list_subfolders_with_paths.append( os.path.join(root, dir) )
        break
    # print(len(list_subfolders_with_paths))


def d():
    list_subfolders_with_paths = glob.glob(path + '/*/')
    # print(len(list_subfolders_with_paths))


def e():
    list_subfolders_with_paths = list(filter(os.path.isdir, [os.path.join(path, f) for f in os.listdir(path)]))
    # print(len(list(list_subfolders_with_paths)))


def f():
    p = pathlib.Path(path)
    list_subfolders_with_paths = [x for x in p.iterdir() if x.is_dir()]
    # print(len(list_subfolders_with_paths))



print(f"Scandir:          {timeit.timeit(a, number=1000):.3f}")
print(f"Listdir:          {timeit.timeit(b, number=1000):.3f}")
print(f"Walk:             {timeit.timeit(c, number=1000):.3f}")
print(f"Glob:             {timeit.timeit(d, number=1000):.3f}")
print(f"Listdir (filter): {timeit.timeit(e, number=1000):.3f}")
print(f"Pathlib:          {timeit.timeit(f, number=1000):.3f}")

使用Twisted的FilePath模块:

from twisted.python.filepath import FilePath

def subdirs(pathObj):
    for subpath in pathObj.walk():
        if subpath.isdir():
            yield subpath

if __name__ == '__main__':
    for subdir in subdirs(FilePath(".")):
        print "Subdirectory:", subdir

由于一些评论者问使用Twisted的库有什么好处，我将在这里稍微超出最初的问题。

分支中有一些改进的文档解释了FilePath的优点;你可能会想读一下。

More specifically in this example: unlike the standard library version, this function can be implemented with no imports. The "subdirs" function is totally generic, in that it operates on nothing but its argument. In order to copy and move the files using the standard library, you need to depend on the "open" builtin, "listdir", perhaps "isdir" or "os.walk" or "shutil.copy". Maybe "os.path.join" too. Not to mention the fact that you need a string passed an argument to identify the actual file. Let's take a look at the full implementation which will copy each directory's "index.tpl" to "index.html":

def copyTemplates(topdir):
    for subdir in subdirs(topdir):
        tpl = subdir.child("index.tpl")
        if tpl.exists():
            tpl.copyTo(subdir.child("index.html"))

上面的“subdirs”函数可以作用于任何类filepath对象。这意味着，其中包括ZipPath对象。不幸的是，ZipPath现在是只读的，但是可以扩展到支持写入。

您还可以为测试目的传递自己的对象。为了测试操作系统。在这里建议使用path-using api时，您必须处理导入的名称和隐式依赖项，并通常执行黑魔法以使您的测试正常工作。使用FilePath，你可以这样做:

class MyFakePath:
    def child(self, name):
        "Return an appropriate child object"

    def walk(self):
        "Return an iterable of MyFakePath objects"

    def exists(self):
        "Return true or false, as appropriate to the test"

    def isdir(self):
        "Return true or false, as appropriate to the test"
...
subdirs(MyFakePath(...))

import os
def get_immediate_subdirectories(a_dir):
    return [name for name in os.listdir(a_dir)
            if os.path.isdir(os.path.join(a_dir, name))]