Python - 展平词典列表

Question

词典列表:

data = [{
         'a':{'l':'Apple',
                'b':'Milk',
                'd':'Meatball'},
         'b':{'favourite':'coke',
              'dislike':'juice'}
         },
         {
         'a':{'l':'Apple1',
                'b':'Milk1',
                'd':'Meatball2'},
         'b':{'favourite':'coke2',
              'dislike':'juice3'}
         }, ...
]

我需要加入所有嵌套字典以达到预期的输出:

 [{'d': 'Meatball', 'b': 'Milk', 'l': 'Apple', 'dislike': 'juice', 'favourite': 'coke'},
  {'d': 'Meatball2', 'b': 'Milk1', 'l': 'Apple1', 'dislike': 'juice3', 'favourite': 'coke2'}]

我尝试嵌套列表理解,但不能一起加入dict:

L = [y for x in data for y in x.values()]
print (L)

[{'d': 'Meatball', 'b': 'Milk', 'l': 'Apple'}, 
 {'dislike': 'juice', 'favourite': 'coke'}, 
{'d': 'Meatball2', 'b': 'Milk1', 'l': 'Apple1'}, 
 {'dislike': 'juice3', 'favourite': 'coke2'}]

我正在寻找最快的解决方案.

Answer 1

您可以使用2个嵌套循环执行此操作,dict.update()并将内部字典添加到临时字典并在末尾添加它:

L = []
for d in data:
    temp = {}
    for key in d:
        temp.update(d[key])

    L.append(temp)

# timeit ~1.4
print(L)

哪些输出:

[{'l': 'Apple', 'b': 'Milk', 'd': 'Meatball', 'favourite': 'coke', 'dislike': 'juice'}, {'l': 'Apple1', 'b': 'Milk1', 'd': 'Meatball2', 'favourite': 'coke2', 'dislike': 'juice3'}]

Answer 2

您可以使用以下方法执行以下操作itertools.chain:

>>> from itertools import chain
# timeit: ~3.40
>>> [dict(chain(*map(dict.items, d.values()))) for d in data]
[{'l': 'Apple', 
  'b': 'Milk', 
  'd': 'Meatball', 
  'favourite': 'coke', 
  'dislike': 'juice'}, 
 {'l': 'Apple1', 
  'b': 'Milk1', 
  'dislike': 'juice3', 
  'favourite': 'coke2', 
  'd': 'Meatball2'}]

的使用chain,map,*使这个表达式实际上执行我的系统(Python的3.5.2)上的更好,是不是长得以下双重嵌套理解的缩写:

# timeit: ~2.04
[{k: v for x in d.values() for k, v in x.items()} for d in data]
# Or, not using items, but lookup by key
# timeit: ~1.67
[{k: x[k] for x in d.values() for k in x} for d in data]

注意:

RoadRunner的循环和更新方法优于这两个单行timeit: ~1.37

Answer 3

您可以使用functools.reduce简单的列表理解来展平dicts的列表

>>> from functools import reduce 

>>> data = [{'b': {'dislike': 'juice', 'favourite': 'coke'}, 'a': {'l': 'Apple', 'b': 'Milk', 'd': 'Meatball'}}, {'b': {'dislike': 'juice3', 'favourite': 'coke2'}, 'a': {'l': 'Apple1', 'b': 'Milk1', 'd': 'Meatball2'}}]
>>> [reduce(lambda x,y: {**x,**y},d.values()) for d in data]
>>> [{'dislike': 'juice', 'l': 'Apple', 'd': 'Meatball', 'b': 'Milk', 'favourite': 'coke'}, {'dislike': 'juice3', 'l': 'Apple1', 'd': 'Meatball2', 'b': 'Milk1', 'favourite': 'coke2'}]

时间基准如下:

>>> import timeit
>>> setup = """
      from functools import reduce
      data = [{'b': {'dislike': 'juice', 'favourite': 'coke'}, 'a': {'l': 'Apple', 'b': 'Milk', 'd': 'Meatball'}}, {'b': {'dislike': 'juice3', 'favourite': 'coke2'}, 'a': {'l': 'Apple1', 'b': 'Milk1', 'd': 'Meatball2'}}]
  """
>>> min(timeit.Timer("[reduce(lambda x,y: {**x,**y},d.values()) for d in data]",setup=setup).repeat(3,1000000))
>>> 1.525032774952706

我机器上其他答案的时间基准

>>> setup = """
        data = [{'b': {'dislike': 'juice', 'favourite': 'coke'}, 'a': {'l': 'Apple', 'b': 'Milk', 'd': 'Meatball'}}, {'b': {'dislike': 'juice3', 'favourite': 'coke2'}, 'a': {'l': 'Apple1', 'b': 'Milk1', 'd': 'Meatball2'}}]
    """
>>> min(timeit.Timer("[{k: v for x in d.values() for k, v in x.items()} for d in data]",setup=setup).repeat(3,1000000))
>>> 2.2488374650129117

>>> min(timeit.Timer("[{k: x[k] for x in d.values() for k in x} for d in data]",setup=setup).repeat(3,1000000))
>>> 1.8990078769857064

>>> code = """
      L = []
      for d in data:
          temp = {}
          for key in d:
              temp.update(d[key])

          L.append(temp)
    """

>>> min(timeit.Timer(code,setup=setup).repeat(3,1000000))
>>> 1.4258553800173104

>>> setup = """
      from itertools import chain
      data = [{'b': {'dislike': 'juice', 'favourite': 'coke'}, 'a': {'l': 'Apple', 'b': 'Milk', 'd': 'Meatball'}}, {'b': {'dislike': 'juice3', 'favourite': 'coke2'}, 'a': {'l': 'Apple1', 'b': 'Milk1', 'd': 'Meatball2'}}]
    """
>>> min(timeit.Timer("[dict(chain(*map(dict.items, d.values()))) for d in data]",setup=setup).repeat(3,1000000))
>>> 3.774383604992181