删除字典中的键并将值存储在列表的字典中

Question

因此，我有一本带有“项目”的字典，其中包含字典列表。 我正在尝试将其重组为“项”的字典，该字典具有包含上一本字典键的值的列表列表。

原版的：

data = { 
   "items": [ 
           { "A": 0.00, "B": 33.27, "C": "string", "D": "16122 " }, 
           { "A": 0.00, "B": 5176.66, "C": "string", "D": "21216 " } 
            ] 
       }

我想得到什么：

data = { 
    "items": [ 
           [ 0.00, 33.27, "string", "16122 " ], 
           [ 0.00, 5176.66, "string", "21216 " ] 
             ] 
         }

Answer 1

似乎operator.itemgetter几乎是您想要的：

getter = operator.itemgetter('A', 'B', 'C', 'D')
data = {'items': [getter(dct) for dct in data['items']]}

在这种情况下，你最终有一个list的tuple ，而不是一个list的list ，但在许多应用中，这可能是确定。

演示：

>>> data = { 
...    "items": [ 
...            { "A": 0.00, "B": 2184.83, "C": "string", "D": "16122 " }, 
...            { "A": 0.00, "B": 5176.66, "C": "string", "D": "21216 " } 
...             ] 
...        }
>>> import operator
>>> getter = operator.itemgetter('A', 'B', 'C', 'D')
>>> data = {'items': [getter(dct) for dct in data['items']]}
>>> data['items'][0]
(0.0, 2184.83, 'string', '16122 ')
>>> data['items'][1]
(0.0, 5176.66, 'string', '21216 ')

Answer 2

这是一种完全按照您想要的方式进行操作的方法。

#Get the column names from the first record
colNames =data['items'][0].keys()
#Get values from all records that have the same keys as in the first record
newData = { 'items' : [[record[colName] for colName in colNames] \
                   for record in data['items']] }
print newData

输出：

{'items': [[0.0, 'string', 33.27, '16122 '], [0.0, 'string', 5176.66, '21216 ']]}

Answer 3

请记住，字典是无序的-因此，当映射到列表时，需要指定键的顺序以获得值的相关顺序。 键的顺序不一定是声明的顺序，上次查看的顺序等等。

因此，更实际的示例数据是：

data = { 
   "items": [ 
           { "D": "16122 ", "A": 0.00, "B": 33.27, "C": "string" }, 
           { "B": 5176.66, "A": 0.00,  "D": "21216 ", "C": "string" } 
            ] 
       }

要将无序键映射到有序列表，您需要选择要使用的顺序。 假设您按照以下顺序排列顺序：

ordered_keys=("A", "B", "C", "D")    

然后，您可以通过简单的循环转换为结构：

for k, LoD in data.items():      # consider '.iteritems() on Py 2 and larger dicts...
    data[k]=[[di[sk] for sk in ordered_keys] for di in LoD]

>>> data
{'items': [[0.0, 33.27, 'string', '16122 '], [0.0, 5176.66, 'string', '21216 ']]}

现在，您需要决定如何处理字典列表中可能缺少的键。 除非每个字典具有完全相同的键，否则您需要一个默认值。

这是您可以执行此操作的方法：

data = { 
   "items": [ 
           { "D": "16122 ", "A": 0.00, "B": 33.27, "C": "string" }, 
           { "B": 5176.66, "A": 0.00,  "D": "21216 ", "C": "string" }, 
           {  "E": "New Key ", "C": "'A' and 'B' are missing in this dict" } 
            ] 
       }

for k, LoD in data.items():     
    keys=sorted({e for sk in LoD for e in sk})
    data[k]=[keys]+[[di.get(sk, None) for sk in keys] for di in LoD]

在这种情况下，字典列表中的所有键均被收集，排序，然后成为列表列表中的第一个元素（因此您知道哪个data键和其他data键可能具有不同的键集。）：

data = { 
   "items": [ 
           { "D": "16122 ", "A": 0.00, "B": 33.27, "C": "string" }, 
           { "B": 5176.66, "A": 0.00,  "D": "21216 ", "C": "string" }, 
           {  "E": "New Key ", "C": "'A' and 'B' are missing in this dict" } 
            ],
    "More": [
           { "D": "16122 ", "A": 0.00, "B": 33.27, "C": "string" }
            ]             
       }

for k, LoD in data.items():     
    keys=sorted({e for sk in LoD for e in sk})
    data[k]=[keys]+[[di.get(sk, None) for sk in keys] for di in LoD]

结果：

>>> for k in data:
...     print k+':'+'\n\t'+'\n\t'.join(repr(e) for e in data[k])
items:
    ['A', 'B', 'C', 'D', 'E']
    [0.0, 33.27, 'string', '16122 ', None]
    [0.0, 5176.66, 'string', '21216 ', None]
    [None, None, "'A' and 'B' are missing in this dict", None, 'New Key ']
More:
    ['A', 'B', 'C', 'D']
    [0.0, 33.27, 'string', '16122 ']