LoadRunner为什么URL模式解决了Javas的所有问题(LoadRunner why is URL mode solving all my issues with Javas)

 （对不起，我真的认为这是对Alex Martelli回答的评论，因为我是基于他的;但是当我最初发布的时候，我没有足够的声望发表评论） 
 
 Alex的答案实际上并不能产生预期的结果。 我并不是指拥有一系列词组清单或列表之间缺少逗号（稍后会详述）。 但是最初的问题是， 通过竞争 ，所有国家和地区的奖牌汇总。 Alex的解决方案将回答： 
 
> 'Germany': [{"Luge - Men's Singles": 'Gold'},
             {"Luge - Men's Singles": 'Silver'},
             {"Luge - Women's Singles": 'Gold'},
             {"Luge - Women's Singles": 'Bronze'},
             {'Luge - Doubles': 'Bronze'}]
 
 但我相信最初的问题实际上是要求： 
 
> 'Germany': [{"Luge - Men's Singles": ['Gold', 'Silver']},
             {"Luge - Women's Singles": ['Gold', 'Bronze'},
             {'Luge - Doubles': 'Bronze'}]
 
 问题中的数据有点混乱，我看到两种可能性： 
 
 1）显示的数据实际上是三个不同的例子，任务是分别在每个列表中合并字典条目。 就是这样 
 
[{'Germany': {"Luge - Men's Singles": 'Gold'}}, 
{'Germany': {"Luge - Men's Singles": 'Silver'}},
{'Italy': {"Luge - Men's Singles": 'Bronze'}}]
 
 你要 
 
['Germany': {"Luge - Men's Singles": ['Gold', 'Silver'],
             "Luge - Women's Singles": ['Gold', 'Bronze']},
 'Italy': {"Luge - Men's Singles": ['Bronze']}]
 
 ，给出 
 
[{'Germany': {"Luge - Women's Singles": 'Gold'}},
{'Austria': {"Luge - Women's Singles": 'Silver'}},
{'Germany': {"Luge - Women's Singles": 'Bronze'}}]
 
 你要 
 
['Germany': {"Luge - Women's Singles": ['Gold', 'Bronze']},
 'Austria': {"Luge - Women's Singles": ['Silver']}]
 
 等等。 我收集这是对问题最可能的解释。 
 
 以下代码可以做到这一点： 
 
from collections import defaultdict

merged = defaultdict(lambda: defaultdict(list))
for d in list_of_dicts:
    for k in d:
        for competition, medal in d[k].iteritems():
            merged[k][competition].append(medal)
 
 为上面显示的第一个列表运行此操作，即可获得 
 
defaultdict(<function <lambda> at 0x1907db0>,
 {'Italy': defaultdict(<type 'list'>, {"Luge - Men's Singles": ['Bronze']}),
  'Germany': defaultdict(<type 'list'>, {"Luge - Men's Singles": ['Gold', 'Silver']})})
 
 2）第二种可能性是问题中的数据是一个单独的列表，其中包含3个列表，每个列表包含字典。 我认为这不是原始问题的意思，但是，因为我已经为此编写了代码，所以它是:) 
 
from collections import defaultdict

merged = defaultdict(lambda: defaultdict(list))
for L in listoflistsofdicts:
  for d in L:
    for k in d:
      for competition, medal in d[k].iteritems():
          merged[k][competition].append(medal)
 
 运行上面的代码来显示问题中显示的列表（添加了必要的逗号，您将得到： 
 
 defaultdict(<function <lambda> at 0x1904b70>,
    {'Italy': defaultdict(<type 'list'>, {"Luge - Men's Singles": ['Bronze']}),
     'Austria': defaultdict(<type 'list'>, {'Luge - Doubles': ['Gold'],
                                            "Luge - Women's Singles": ['Silver']}),
     'Latvia': defaultdict(<type 'list'>, {'Luge - Doubles': ['Silver']}),
     'Germany': defaultdict(<type 'list'>, {'Luge - Doubles': ['Bronze'],
                                            "Luge - Men's Singles": ['Gold', 'Silver'],
                                            "Luge - Women's Singles": ['Gold', 'Bronze']})
    })
 
 请注意，这两个代码不会对奖牌类型进行排序（即，您最终可能会得到['Gold'，'Silver']或['Silver'，'Gold']）。 
 
 当然，如果在解决方案1）中使用了分隔列表，但需要将所有这些列表合并，只需将它们全部放在一个列表中，然后使用解决方案2）即可。 

(Sorry, I really meant this as a comment on Alex Martelli's answer, since mine is based on his; but when I originally posted I didn't have enough reputation to comment)
 
Alex's answer doesn't actually generate the intended result. I don't mean the finer points of having a list of lists of dicts, or the lack of commas between the lists (more on that later). But the original question wanted, as a result, a compilation of all medals by country, by competition. Alex's solution will answer:
 
> 'Germany': [{"Luge - Men's Singles": 'Gold'},
             {"Luge - Men's Singles": 'Silver'},
             {"Luge - Women's Singles": 'Gold'},
             {"Luge - Women's Singles": 'Bronze'},
             {'Luge - Doubles': 'Bronze'}]
 
But I believe the original question actually asked for:
 
> 'Germany': [{"Luge - Men's Singles": ['Gold', 'Silver']},
             {"Luge - Women's Singles": ['Gold', 'Bronze'},
             {'Luge - Doubles': 'Bronze'}]
 
The data in the question is a bit confusing, I see two possibilities:
 
1) The data shown is actually three different examples, and the task is to merge dict entries within each list, separately. That is, given
 
[{'Germany': {"Luge - Men's Singles": 'Gold'}}, 
{'Germany': {"Luge - Men's Singles": 'Silver'}},
{'Italy': {"Luge - Men's Singles": 'Bronze'}}]
 
you want
 
['Germany': {"Luge - Men's Singles": ['Gold', 'Silver'],
             "Luge - Women's Singles": ['Gold', 'Bronze']},
 'Italy': {"Luge - Men's Singles": ['Bronze']}]
 
, given 
 
[{'Germany': {"Luge - Women's Singles": 'Gold'}},
{'Austria': {"Luge - Women's Singles": 'Silver'}},
{'Germany': {"Luge - Women's Singles": 'Bronze'}}]
 
you want
 
['Germany': {"Luge - Women's Singles": ['Gold', 'Bronze']},
 'Austria': {"Luge - Women's Singles": ['Silver']}]
 
and so on. I gather this is the most likely interpretation of the question.
 
The following code does that:
 
from collections import defaultdict

merged = defaultdict(lambda: defaultdict(list))
for d in list_of_dicts:
    for k in d:
        for competition, medal in d[k].iteritems():
            merged[k][competition].append(medal)
 
Running this for the first of the lists shown above, you get
 
defaultdict(<function <lambda> at 0x1907db0>,
 {'Italy': defaultdict(<type 'list'>, {"Luge - Men's Singles": ['Bronze']}),
  'Germany': defaultdict(<type 'list'>, {"Luge - Men's Singles": ['Gold', 'Silver']})})
 
2) The second possibility is that the data in the question is one single list, containing 3 lists, each of these containing dicts. I think this is not what the original question means, but, since I'd already written the code for that, here it is :)
 
from collections import defaultdict

merged = defaultdict(lambda: defaultdict(list))
for L in listoflistsofdicts:
  for d in L:
    for k in d:
      for competition, medal in d[k].iteritems():
          merged[k][competition].append(medal)
 
Running the code above for the lists shown on the question (with the necessary commas added, you get:
 
 defaultdict(<function <lambda> at 0x1904b70>,
    {'Italy': defaultdict(<type 'list'>, {"Luge - Men's Singles": ['Bronze']}),
     'Austria': defaultdict(<type 'list'>, {'Luge - Doubles': ['Gold'],
                                            "Luge - Women's Singles": ['Silver']}),
     'Latvia': defaultdict(<type 'list'>, {'Luge - Doubles': ['Silver']}),
     'Germany': defaultdict(<type 'list'>, {'Luge - Doubles': ['Bronze'],
                                            "Luge - Men's Singles": ['Gold', 'Silver'],
                                            "Luge - Women's Singles": ['Gold', 'Bronze']})
    })
 
Please note that both of these codes don't sort medal types (i.e., you might end up with ['Gold', 'Silver'] or ['Silver', 'Gold']).
 
Of course, if you get separated lists as used in solution 1), but need a merge of all of them, simply bring them all together in a list, and use solution 2).