Python目录通过dict搜索和组织 [英] Python directory searching and organizing by dict

问题描述

嗯，这是我第一次最近试图进入Python的文件和os部分。我正在尝试搜索一个目录，然后找到所有子目录。如果目录没有文件夹，请将所有文件添加到列表中。并且通过dict来组织它们。

所以例如一棵树可能看起来像这样

• 起始路径
  
  
  
  
  • 目录1
    
    
    
    
    • Subdir 1
    
    
    • Subdir 2
    
    
    • Subdir 3
      
      
      
      
      • subsubdir
        
        
        
        
        • file.jpg
        
        
        • folder1
          
          
          
          
          • file1.jpg
          
          
          • file2.jpg
          
          
        
        
        • folder2
          
          
          
          
          • file3.jpg
          
          
          • file4.jpg
          
          
        
        
      
      
    
    
  
  

即使subsubdir有一个文件，它应该被跳过，因为它有文件夹。

现在我通常可以这样做，如果我知道我要去多少目录要寻找，使用os.listdir和os.path.isdir。但是，如果我想要这样做是动态的，它将不得不补偿任何数量的文件夹和子文件夹。我一直在使用os.walk尝试，它会很容易找到的所有文件。我遇到的唯一麻烦是创造所有包含文件的路径名的类型的字典。我需要由dict组织的文件夹名称，直到起始路径。

所以最后，使用上面的例子，dict应该是这样的，其中包含文件：

  dict ['dir1'] ['subdir3'] ['subsubdir'] ['folder1'] = ['file1.jpg'，'file2.jpg'] 
 
 dict ['dir1'] ['subdir3'] ['subsubdir'] ['folder2'] = ['file3.jpg'，'file4.jpg'] 
  

感谢任何关于组织这些信息的更好的想法的帮助。谢谢。

解决方案

也许你想要的东西：

  def explore（starting_path）：
 alld = {''：{}} 
 
 for dirpath，dirnames，os.walk中的文件名（starting_path）：
d = alld 
 dirpath = dirpath [len（starting_path）：] 
 dirpath.split（os.sep）中的子目录：
 based = d 
d = d [subd] 
如果dirnames：
 for dn in dirnames：
d [dn] = {} 
 else：
 based [subd] = filenames 
 return alld [' '] 
  

例如，给定一个 / tmp / a ，以便：

  $ ls -FR / tmp / a 
b / c / d / 
 
 / tmp / a / b：
z / 
 
 / tmp / a / b / z：
 
 / tmp / a / c：
 za zu 
 
 / tmp / a / d：
  

code> print explore（'/ tmp / a'）发出： {'c'：['za'，'zu']，'b' 'z'：[]}，'d'：[]} 。

如果这是牛逼正是你以后，也许你可以明确告诉我们什么差别应该是什么？我怀疑他们也许可以很容易地固定，如果需要的话。

Hey all, this is my first time recently trying to get into the file and os part of Python. I am trying to search a directory then find all sub directories. If the directory has no folders, add all the files to a list. And organize them all by dict.

So for instance a tree could look like this

• Starting Path
  • Dir 1
    • Subdir 1
    • Subdir 2
    • Subdir 3
      • subsubdir
        • file.jpg
        • folder1
          • file1.jpg
          • file2.jpg
        • folder2
          • file3.jpg
          • file4.jpg

Even if subsubdir has a file in it, it should be skipped because it has folders in it.

Now I can normally do this if I know how many directories I am going to be looking for, using os.listdir and os.path.isdir. However if I want this to be dynamic it will have to compensate for any amount of folders and subfolders. I have tried using os.walk and it will find all the files easily. The only trouble I am having is creating all the dicts with the path names that contain file. I need the foldernames organized by dict, up until the starting path.

So in the end, using the example above, the dict should look like this with the files in it:

dict['dir1']['subdir3']['subsubdir']['folder1'] = ['file1.jpg', 'file2.jpg']

dict['dir1']['subdir3']['subsubdir']['folder2'] = ['file3.jpg', 'file4.jpg']

Would appreciate any help on this or better ideas on organizing the information. Thanks.

解决方案

Maybe you want something like:

def explore(starting_path):
  alld = {'': {}}

  for dirpath, dirnames, filenames in os.walk(starting_path):
    d = alld
    dirpath = dirpath[len(starting_path):]
    for subd in dirpath.split(os.sep):
      based = d
      d = d[subd]
    if dirnames:
      for dn in dirnames:
        d[dn] = {}
    else:
      based[subd] = filenames
  return alld['']

For example, given a /tmp/a such that:

$ ls -FR /tmp/a
b/  c/  d/

/tmp/a/b:
z/

/tmp/a/b/z:

/tmp/a/c:
za  zu

/tmp/a/d:

print explore('/tmp/a') emits: {'c': ['za', 'zu'], 'b': {'z': []}, 'd': []}.

If this isn't exactly what you're after, maybe you can show us specifically what the differences are supposed to be? I suspect they can probably be easily fixed, if need be.

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