Question

我有一些字典（json输出）。我想获取基本元素，它可以是字符串列表或字符串。目前，我正在这样做：-

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"]
    }
  }
}

这是我得到的输出：-

{'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']}}

但这就是我想要的：-

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']]

但是我不知道是2层还是3层嵌套循环。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，并在另一个列表中获取一个或多个基值，例如最低级别，对此的任何帮助将不胜感激。

Answer 1

生成器对于解决此问题很有用。策略–

键：跟踪当前递归路径。碰到叶子就产生当前路径。
值：仅产生叶子。

代码：

def getitems(obj):

  def getkeys(obj, stack):
    for k, v in obj.items():
      k2 = ([k] if k else []) + stack # don't return empty keys
      if v and isinstance(v, dict):
        for c in getkeys(v, k2):
          yield c
      else: # leaf
        yield k2

  def getvalues(obj):
    for v in obj.values():
      if not v: continue
      if isinstance(v, dict):
        for c in getvalues(v):
          yield c
      else: # leaf
        yield v if isinstance(v, list) else [v]

  return list(getkeys(obj,[])), list(getvalues(obj))

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']]

从python字典获取所有子键的列表

1 个答案: