从python字典中获取所有子键的列表

我有一些字典（json输出）。我想得到基本元素，它可以是字符串列表或字符串。目前我是这样做的：-

folder="shared/"
files=os.listdir('shared')


for f in files:
    f=folder+f
    print(f)
    with open(f) as f:
        data = json.load(f)
    #data is a dict now with sub-keys
    for key,value in data.items():
        if value.keys():
            print(value)
    break

这是python代码读取的输入字典：-

{
  "shortshirt": {
    "ralphlauren": {
      "classic": [
        "That Ralph Lauren classic fit is a timeless look!",
        "Nice choice. Can’t go wrong with Ralph Lauren"
      ]
    }
  },
  "socks": {
    "": {
      "": ["Have to find the right socks to keep your feet cozy"]
    }
  }
}

这是我得到的结果：-

{'ralphlauren': {'classic': ['That Ralph Lauren classic fit is a timeless look!', 'Nice choice. Can’t go wrong with Ralph Lauren']}}
{'': {'': ['Have to find the right socks to keep your feet cozy']}}

但这正是我想要的：-

keys=[["shortshirt","ralphlauren","classic"],["socks"]]

value=[['That Ralph Lauren classic fit is a timeless look!', 'Nice choice. Can’t go wrong with Ralph Lauren'], ['Have to find the right socks to keep your feet cozy']]

---

但我不知道是有2级还是3级嵌套循环。如果我有一个内部循环，并且实际上没有嵌套键，那么我会得到值错误。我想在一个单独的列表中获取所有嵌套键，以及另一个列表中最低级别的一个或多个基值，对此的任何帮助都将不胜感激。

生成器对解决此问题非常有用。战略——

• 关键点：跟踪当前递归路径。一旦你碰到一片叶子，就产生当前路径

• 价值观：只产树叶

代码：

输入：

{
  "shortshirt": {
    "ralphlauren": {
      "classic": [
        "That Ralph Lauren classic fit is a timeless look!",
        "Nice choice. Can't go wrong with Ralph Lauren"
      ]
    }
  },
  "socks": {
    "": {
      "": ["Have to find the right socks to keep your feet cozy"]
    }
  }
}

输出：

# keys
[['classic', 'ralphlauren', 'shortshirt'], ['socks']]

# values
[['That Ralph Lauren classic fit is a timeless look!', "Nice choice. Can't go wrong with Ralph Lauren"], ['Have to find the right socks to keep your feet cozy']]

---

你的描述不清楚。为您的示例数据提供相应的预期输出。我已编辑了问题。这是因为您没有深入字典，所以只获取顶级值（即DICT）。对于您想要的内容，您必须在字典中查找，直到值不是dict。您可以递归地执行此操作。钥匙也是一样，别忘了过滤空的。如果我往下走（添加一个循环），那么我怎么知道应该走多深？我的意思是，我要先写代码，如果我写3个循环，实际有5个子键，我将如何得到它们？递归，正如@JavierPazSedano提到的，那么不管有多少子键存在

# keys
[['classic', 'ralphlauren', 'shortshirt'], ['socks']]

# values
[['That Ralph Lauren classic fit is a timeless look!', "Nice choice. Can't go wrong with Ralph Lauren"], ['Have to find the right socks to keep your feet cozy']]