Python 如何使用文件内容中的节点名称从fs创建嵌套字典？

我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

例如：

输入时：

directory
--name.txt
--subdir
----name.txt
--subdir2
----name.txt
----subdir3
-------name.txt

论产出

{'name': 'directory/name.txt', 
'content': 'Name 1', 'descendent': 
       [{'name': 'directory/subdir/name.txt', 
       'content': 'Name 2', 
       'descendent': None}, 
        {'name': 'directory/subdir2/name.txt', 
         'content': 'Name 3', 
         'descendent': 
                [{'name': 'directory/subdir2/subdir3/name.txt', 
                'content': 'Name 4', 
                'descendent': None}]}]}

我写了一个简单的实现，但有点不对劲：

__author__ = 'ivanov'

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

def readFile(filename):
    f = open(filename, 'r')
    res = f.read()
    f.close()
    return res

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

def getName(myname):
    if isfile(myname):
        if myname.split("/")[-1].endswith(".txt"):
            f = open(myname, 'r')
            res = f.read()
            f.close()
            return myname
    else:
        for i in os.listdir(myname):
            if isfile(os.path.join(myname, i)):
                return {"name": os.path.join(myname, i), "content": getName(os.path.join(myname, i))}
            else:
                return getName(os.path.join(myname, i))

file_folder_dict = getName(myname)

print file_folder_dict

---

在我的例子中，如何改进它？

根据您的示例输出，似乎您假设某个目录中只有一个txt文件，并且可能有多个子目录。下面是在此假设下使用递归的代码。getname（）的参数应该是目录的路径

import sys
import os
import json

def readfile(filepath):
    f = open(filepath)
    res = f.read()
    f.close()
    return res.strip()

def getname(dirpath):
    onlyfile = [entry for entry in os.listdir(dirpath) \
            if os.path.isfile(os.path.join(dirpath, entry)) \
            and entry.endswith(".txt")][0]
    onlyfilename = os.path.join(dirpath, onlyfile)

    decendent = []
    for entry in os.listdir(dirpath):
        entrypath = os.path.join(dirpath, entry)
        if os.path.isfile(entrypath):
            # we have handled this one, i.e. 'onlyfile'
            continue
        decendent.append(getname(entrypath))

    if len(decendent) == 0:
        decendent = None

    return {'name': onlyfilename,
            'content': readfile(onlyfilename),
            'decendent': decendent}


if __name__ == "__main__":
    dirpath = sys.argv[1]
    result = getname(dirpath)
    print json.dumps(result, indent=4)

下面是示例目录结构的输出：

$ python xx.py directory
{
    "content": "Name 1", 
    "name": "directory/name.txt", 
    "decendent": [
         {
            "content": "Name 2", 
            "name": "directory/subdir/name.txt", 
            "decendent": null
        }, 
        {
            "content": "Name 3", 
            "name": "directory/subdir2/name.txt", 
            "decendent": [
                {
                    "content": "Name 4", 
                    "name": "directory/subdir2/subdir3/name.txt", 
                    "decendent": null
                }
            ]
        }
    ]
}

---

那么os.walk（）呢？请参阅os.walk:What'sgrough with it'？