遍历嵌套字典并在Python中获取路径？

Question

我有一本字典，例如：

{
   "checksum": "b884cbfb1a6697fa9b9eea9cb2054183",
   "roots": {
      "bookmark_bar": {
         "children": [ {
            "date_added": "12989159740428363",
            "id": "4",
            "name": "test2",
            "type": "url",
            "url": "chrome://bookmarks/#1"
         } ],
         "date_added": "12989159700896551",
         "date_modified": "12989159740428363",
         "id": "1",
         "name": "bookmark_bar",
         "type": "folder"
      },
      "other": {
         "children": [ {
            "date_added": "12989159740428363",
            "id": "4",
            "name": "test",
            "type": "url",
            "url": "chrome://bookmarks/#1"
         } ],
         "date_added": "12989159700896557",
         "date_modified": "0",
         "id": "2",
         "name": "aaa",
         "type": "folder"
      },
      "synced": {
         "children": [  ],
         "date_added": "12989159700896558",
         "date_modified": "0",
         "id": "3",
         "name": "bbb",
         "type": "folder"
      }
   },
   "version": 1
}

一切都始于“根”，它们有两种类型的数据：URL和文件夹，它们是字典。如果是文件夹，则必须具有键“
children”，该键的值是一个列表，我们可以在其中放置更多URL和文件夹。

现在，我想遍历此嵌套字典，以获取所有子文件夹中的URL，因此我编写了一个函数：

def traverse(dic):
    for i in dic:
        if i['type'] == 'folder':
            for j in traverse(i['children']):
                yield j
        elif i['type'] == 'url':
            yield i

我可以这样使用它：

traverse(dictionary['roots']['bookmark_bar']['children'])

完美运作。但是它只是生成URL的字典，我不知道它在哪里。我也想走这条路。我该怎么做？

Answer 1

如果我得到了您想要的东西，就不要确定，但是您可能想要这样做：

def traverse(dic, path=None):
    if not path:
        path = []
    for i in dic:
        local_path = path[:].append(i)
        if i['type'] == 'folder':
            for j in traverse(i['children'], local_path):
                yield j, local_path
        elif i['type'] == 'url':
            yield i, local_path

现在，您的函数会产生该项目和一系列键，以便在特定位置到达该项目。