Python 如何从文件创建嵌套字典？

假设我的fs中有这样的结构：

Directory
--name.txt
--content.html
--Subdirectory
----name.txt
----content.html
------SubSubDirectory
--------name.txt
--------content.html

在输出中，我希望：

content(Directory/name.txt):path(Directory/content.html)
--content(Directory/SubDirectory/name.txt):path(Directory/SubDirectory/content.html)
----content(Directory/SubDirectory/SubSubDirectory/name.txt):path(Directory/SubDirectory/SubSubDirectory/content.html)

我有一个功能：

def getffdirs(rootdir):
    dir = {}
    rootdir = rootdir.rstrip(os.sep)
    console.log(rootdir)
    start = rootdir.rfind(os.sep) + 1
    console.log(start)
    for path, dirs, files in os.walk(rootdir):
        folders = path[start:].split(os.sep)
        shpfiles = {}
        for file in files:

            if file.endswith(".txt"):
                if readFile(os.path.join(path, file)) != '':
                    full_path = os.path.join(path, file).replace("name.txt", "index.html")
                    # shpfiles[readFile(os.path.join(path, file))] = os.path.join(path, file)
                    shpfiles[full_path] = readFile(os.path.join(path, file))
            if file.endswith(".html"):
               shpfiles['content'] = readFile(os.path.join(path, file))

        subdir = dict.fromkeys(shpfiles)
        parent = reduce(dict.get, folders[:-1], dir)
        parent[folders[-1]] = subdir
    return dir

它生成目录、子目录和文件的列表。我尝试为我的任务重新建模，但我只在文件节点上取得了进展

如何为所有节点（包括目录和子目录）重建它

__author__ = 'pranav'

import sys
import os
from os.path import isfile
import json

try:
    myname = sys.argv[1]
except Exception, e:
    print "usage : python [listfile.py] [foldername]"
    exit(1)

def getName(myname):
    file_folder_dict = {myname: []}
    if not isfile(myname):
        try:
            for f in os.listdir(myname):
                subfolder = os.path.join(myname, f)
                file_folder_dict[myname].append(getName(subfolder))
        except:
            print "No DIR"
    else:
        return myname
    return file_folder_dict


file_folder_dict = getName(myname)

print file_folder_dict

我确实是在很久以前作为一种爱好写这个剧本的。它实际上以dict格式提供所有文件和文件夹。这应该对你有帮助

结果:

{
    '/home/likewise-open/PUNESEZ/pranav.ambhore/stack_o_list_files/': [
        {
            '/home/likewise-open/PUNESEZ/pranav.ambhore/stack_o_list_files/some1': [
                '/home/likewise-open/PUNESEZ/pranav.ambhore/stack_o_list_files/some1/some1.txt',
                '/home/likewise-open/PUNESEZ/pranav.ambhore/stack_o_list_files/some1/some1.html',
                {
                    '/home/likewise-open/PUNESEZ/pranav.ambhore/stack_o_list_files/some1/some2': [
                        '/home/likewise-open/PUNESEZ/pranav.ambhore/stack_o_list_files/some1/some2/some2.html',
                        '/home/likewise-open/PUNESEZ/pranav.ambhore/stack_o_list_files/some1/some2/some2.txt'
                    ]
                }
            ]
        }
    ]
}

---

您试图构建的字典中的键和值是什么？键可以是当前文件夹的列表[content（name.txt），path（content.html）]，值-子文件夹的嵌套字典。也可以是元组，因为不推荐将该列表用作字典中的键。不幸的是，这是一个线性结构。我需要字典或元组相互嵌套，以便以后在浏览器窗口中将其显示为节点。这里的逻辑更复杂（我很难处理）：您需要进入目录并查看-是否有扩展名为.txt的文件，如果有：提取此文件的内容并用此内容替换当前目录的名称。之后，向前移动到子目录并执行相同的操作。