难以从理解中获得关键价值

Question

我试图弄清楚如何从几乎有效的理解中获取价值。从这个数据..

{'rock': {}, 'coal1': {'gold1': {'data': ['g1']}}, 'coal2': {'gold3': {'data': ['g3']}, 'gold2': {'data': ['g2']}}}

..我正在尝试将金钥匙数据提取到简单的字典中。即 { gold : {'data' : [...]} } 项的字典基本上从数据中删除了煤炭钥匙。

也就是说从这个..

{
  "coal2": {
    "gold3": {
      "data": [
        "g3"
      ]
    },
    "gold2": {
      "data": [
        "g2"
      ]
    }
  },
  "coal1": {
    "gold1": {
      "data": [
        "g1"
      ]
    }
  },
  "rock": {}
}

为这种格式

{
  "gold3": {
    "data": [
      "g3"
    ]
  },
  "gold1": {
    "data": [
      "g1"
    ]
  },
  "gold2": {
    "data": [
      "g2"
    ]
  }
}

这几乎是可行的。这样就摆脱了岩石。

>>> {k:d for k,d in data.items() if k != 'rock'}
{'coal2': {'gold3': {'data': ['g3']}, 'gold2': {'data': ['g2']}}, 'coal1': {'gold1': {'data': ['g1']}}}

获取这些值就可以摆脱煤炭钥匙。

>>> [v for v in {k:d for k,d in data.items() if k != 'rock'}.values()]
[{'gold3': {'data': ['g3']}, 'gold2': {'data': ['g2']}}, {'gold1': {'data': ['g1']}}]

但我不知道如何从这里得到

>>> for i in [v for v in {k:d for k,d in data.items() if k != 'rock'}.values()] : print(i)
...
{'gold3': {'data': ['g3']}, 'gold2': {'data': ['g2']}}
{'gold1': {'data': ['g1']}}

到所需的结构。如果这一切都可以通过理解来完成，那就太好了。有谁知道如何做到这一点？

编辑：这两个答案都很棒，我希望我能同时接受这两个答案。我喜欢不导入任何东西，但我接受@blhsing itertools 版本只是因为它更容易理解并且性能稍微好一点。顺便说一句，岩石必须被丢弃，即使它有价值，所以我无法绕过 if k != 'rock'。所以这是结果......谢谢大家。

>>> import timeit
>>> data = {'rock': {'type':'pebble'}, 'coal1': {'gold1': {'data': ['g1']}}, 'coal2': {'gold3': {'data': ['g3']}, 'gold2': {'data': ['g2']}}}
>>> timeit.timeit( "dict(kv for x in (v for v in {k:d for k,d in data.items() if k != 'rock'}.values()) for kv in x.items())" ,  setup="from __main__ import data")            
2.6714617270044982
>>>
>>> timeit.timeit( "dict(chain.from_iterable(g.items() for g in {k:d for k,d in data.items() if k != 'rock'}.values()))" , setup="from __main__ import data; from itertools import chain")
2.22612579818815
>>>

Answer 1

只需要将 dict 的 list 变为 dict（通过添加第二行修复代码）

l=[v for v in {k:d for k,d in d.items() if k != 'rock'}.values()] # here is your own code 
newd=dict(kv for x in l for kv in x.items())
newd
Out[431]: 
{'gold1': {'data': ['g1']},
 'gold2': {'data': ['g2']},
 'gold3': {'data': ['g3']}}

带单线

dict(v for d in d.values() for v in d.items()) # d is your dict
Out[436]: 
{'gold1': {'data': ['g1']},
 'gold2': {'data': ['g2']},
 'gold3': {'data': ['g3']}}

Answer 2

您可以使用生成器表达式输出主字典值的子字典的项目，并使用 itertools.chain.from_iterable 连接这些项目，并将它们传递给 dict 构造函数：

from itertools import chain
dict(chain.from_iterable(g.items() for g in d.values()))

因此给定：

d = {'rock': {}, 'coal1': {'gold1': {'data': ['g1']}}, 'coal2': {'gold3': {'data': ['g3']}, 'gold2': {'data': ['g2']}}}

这个returns:

{'gold3': {'data': ['g3']}, 'gold2': {'data': ['g2']}, 'gold1': {'data': ['g1']}}

难以从理解中获得关键价值

stumped getting key value from a comprehension

python

dictionary

list-comprehension

dictionary-comprehension