blocks|key|283353|text|不是shure，如果我得到了你想要的，但你可能想要这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283354|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|code-block|syntax|javascript|283355|现在，您的函数将生成该项目和一系列键，以便在某个位置到达该项目。|283356|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Not shure if I got what you want, but you might want to do this:

<pre><code>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
</code></pre>

Now your function yields the item and a sequence of the keys to get to the item at a certain location.

blocks|key|500141|text|我有一个与您略有不同的用例:我需要将表示客户机设置的可变深度JSON结构扁平化为键-值对，以便存储在数据库中。我不能得到jsbueno的答案，因为我还需要一些可以在不显式列出或包含孩子的情况下处理情况的东西，我对其进行了修改以满足我的需求：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500142|def+traverse(dic,+path=None):
++++if+not+path:
++++++++path=[]
++++if+isinstance(dic,dict):
++++++++for+x+in+dic.keys():
++++++++++++local_path+=+path[:]
++++++++++++local_path.append(x)
++++++++++++for+b+in+traverse(dic[x],+local_path):
+++++++++++++++++yield+b
++++else:+
++++++++yield+path,dic|code-block|syntax|javascript|500143|最终结果是，我可以将这样的JSON字符串传递给我的脚本(具有可变深度)，该脚本将其转换为嵌套的dicts：|500144|{
++"servers":+{
++++"uat":+{
++++++"pkey":+true,
++++++"user":+"testval",
++++++"pass":+true
++++},
++++"dev":+{
++++++"pkey":+true,
++++++"user":+"testval",
++++++"pass":+true
++++}
++}
}|500145|对它运行上面的生成器将创建一个如下所示的列表：|500146|([u'servers',+u'uat',+u'pkey'],+True)
([u'servers',+u'uat',+u'user'],+u'testval')
([u'servers',+u'uat',+u'pass'],+True)
([u'servers',+u'dev',+u'pkey'],+True)
([u'servers',+u'dev',+u'user'],+u'testval')
([u'servers',+u'dev',+u'pass'],+True)|500147|其中，使用类似如下的内容：|500148|for+x+in+traverse(outobj):
++++pprint(('.'.join(x[0]),x[1]))|500149|然后可以转换成我想要的键值对格式，如下所示：|500150|(u'servers.uat.pkey',+True)
(u'servers.uat.user',+u'testval')
(u'servers.uat.pass',+True)
(u'servers.dev.pkey',+True)
(u'servers.dev.user',+u'testval')
(u'servers.dev.pass',+True)|500151|我知道我是在公认的答案被接受后才这样发帖的，但既然公认的答案对我不起作用，也许这个稍微更具结构不可知性的版本会对其他人有所帮助！|500152|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

I had a slightly different use case from you: I needed to flatten a variable depth JSON structure representing client settings into key-value pairs for storage in a database. I couldn't get jsbueno's answer to work, and since I also needed something that could handle cases without children being explicitly listing or contained, I modified it to suit my needs:

<pre><code>def traverse(dic, path=None):
 if not path:
 path=[]
 if isinstance(dic,dict):
 for x in dic.keys():
 local_path = path[:]
 local_path.append(x)
 for b in traverse(dic[x], local_path):
 yield b
 else: 
 yield path,dic
</code></pre>

The end result is i can pass in a JSON string like this to my script (with variable depths), which converts it to nested dicts:

<pre><code>{
 "servers": {
 "uat": {
 "pkey": true,
 "user": "testval",
 "pass": true
 },
 "dev": {
 "pkey": true,
 "user": "testval",
 "pass": true
 }
 }
}
</code></pre>

running the generator above against it creates a list that pretty-prints like this:

<pre><code>([u'servers', u'uat', u'pkey'], True)
([u'servers', u'uat', u'user'], u'testval')
([u'servers', u'uat', u'pass'], True)
([u'servers', u'dev', u'pkey'], True)
([u'servers', u'dev', u'user'], u'testval')
([u'servers', u'dev', u'pass'], True)
</code></pre>

Which, using something like:

<pre><code>for x in traverse(outobj):
 pprint(('.'.join(x[0]),x[1]))
</code></pre>

can then be transformed into the key-value pair format I want like so:

<pre><code>(u'servers.uat.pkey', True)
(u'servers.uat.user', u'testval')
(u'servers.uat.pass', True)
(u'servers.dev.pkey', True)
(u'servers.dev.user', u'testval')
(u'servers.dev.pass', True)
</code></pre>

I know I'm posting this way after the accepted answer was accepted, but since the accepted answer didn't work for me, maybe this slight more structure-agnostic version will help someone else!

I have a dictionary like:

<pre><code>{
 "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
}
</code></pre>

Everything starts at 'roots', them there are two types of data: URL and folder, they are dictionaries.
If it is a folder, it must have the key 'children', the value of the key is a list, we can put more URLs and folders in it.

Now I want to traverse this nested dictionary, to get the URL in all sub-folder, so I wrote a function:

<pre><code>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
</code></pre>

and I can use it like that:

<pre><code>traverse(dictionary['roots']['bookmark_bar']['children'])
</code></pre>

It works perfectly. But it just generate a dictionary of a URL, I don't know where is it.
I want to get the path too. How can I do it?

Traverse a nested dictionary and get the path in Python?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一本字典，比如：{   "checksum": "b884cbfb1a6697fa9b9eea9cb2054183",   "roots": {      "bookmark_bar": {         "children": [ {            "date_added": "129891597404...

问遍历嵌套字典并获取Python中的路径？
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问遍历嵌套字典并获取Python中的路径？EN

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问遍历嵌套字典并获取Python中的路径？
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