Python:从字典的嵌套列表中创建目录树 [英] Python: Create directory tree from nested list of dictionaries

问题描述

如何从下面的python词典列表中创建目录树?子目录级别的数量必须是可变的.

How can I create a directory tree from the below list of dictionaries in python? The number of subdirectory levels has to be variable.

dirTree: 
    [{'level-1_dir-1': ''}, 
    {'level-1_dir-2': 
        [{'level-2_dir-1': 
            [{'level-3_dir-1': ''}, 
            {'level-3_dir-2': ''}, 
            {'level-3_dir-3': ''}
            ]
        }, 
        {'level-2_dir-2': ''}
        ]
    }, 
    {'level-1_dir-3': ''}, 
    {'level-1_dir-4': ''}
    ]

我想用python完成类似以下任务的事情:

I would like to accomplish something like the following tasks with python:

• 遍历1级键
  • 使用键名创建文件夹
  • 如果其值不是''
    • 创建子文件夹
    • 如果子文件夹的值不是''
      • 创建子文件夹...
      • ...等等，直到没有更深的层次为止，
      • 然后转到下一个1级键并重新开始

      推荐答案

      您可以使用与您的结构有关的适当方法来递归地创建目录

      You may recursivly create the dir, with the appropriate way regarding your structure

      def createPath(values, prefix=""):
          for item in values:
              directory, paths = list(item.items())[0]
              dir_path = join(prefix, directory)
              makedirs(dir_path, exist_ok=True)
              createPath(paths, dir_path)
      

      像 createPath(dirTree)或 createPath(dirTree，"/some/where)之类的调用来选择初始开始

      Call like createPath(dirTree) or createPath(dirTree, "/some/where) to choose the initial start

      仅包含 dict 的简单结构会更有用，因为仅用于一个映射的dict不太好

      A simplier structure with only dicts would be more usable, because a dict for one mapping only is not very nice

      res = {
          'level-1_dir-1': '',
          'level-1_dir-2':
              {
                  'level-2_dir-1':
                      {
                          'level-3_dir-1': '',
                          'level-3_dir-2': '',
                          'level-3_dir-3': ''
                      },
                  'level-2_dir-2': ''
              },
          'level-1_dir-3': '',
          'level-1_dir-4': ''
      }
      
      # with appropriate code
      def createPath(values, prefix=""):
          for directory, paths in values.items():
              dir_path = join(prefix, directory)
              makedirs(dir_path, exist_ok=True)
              if isinstance(paths, dict):
                  createPath(paths, dir_path)
      

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