3

我想用嵌套字典和重复键创建数据结构。一个详细的例子是：

data['State1']['Landon']['abc Area'] = 'BOB'
data['State1']['Landon']['abc Area'] = 'SAM'
data['State1']['Landon']['xyz Area'] = 'John'
data['State2']['New York']['hjk Area'] = 'Ricky'

for z in data['State1'].keys() , 
# I should get list ['Landon', 'Landon', 'Landon']
for y in data['State1']['Landon'].keys() , 
# I should get list ['abc Area', 'abc Area', 'xyz Area']

目前为了存储我使用额外的计数器键的数据 

data = Autovivification()  
data[state][city][area][counter] = ID

但是在解析城市/区域的总条目（以及重复项）时，我必须使用嵌套循环直到计数器键。

for city in data['State1'].keys():
  for area in data['State1'][city].keys():
    for counter in data['State1'][city][area].keys():
     for temp in data['State1'][city][area][counter].values():
         cityList.append(city)
         areaList.append(area)

对于嵌套字典，我找到了 nosklo 发布的以下代码

class AutoVivification(dict):  
    """Implementation of perl's autovivification feature."""  
    def __getitem__(self, item):                
         try:  
            return dict.__getitem__(self, item)  
        except KeyError:   
            value = self[item] = type(self)()  
            return value

对于具有重复键的字典，我发现 Scorpil 发布的代码 

class Dictlist(dict):  
    def __setitem__(self, key, value):  
        try:   
            self[key]   
        except KeyError:   
            super(Dictlist, self).__setitem__(key, [])   
        self[key].append(value)

如何合并 Autovivification 和 Duplicate 类代码？还是有任何其他pythonic方式来处理这种情况？

4

4 回答 4

4

另一个使用defaultdict的例子：

from collections import defaultdict


data = defaultdict(  # State
    lambda: defaultdict(  # City
        lambda: defaultdict(list)  # Area
    )
)


data['State']['City']['Area'].append('area 1')
data['State']['City']['Area'].append('area 2')
data['State']['City']['Area'].append('area 2')


areas = data['State']['City']['Area']
print(areas)  # ['area 1', 'area 2', 'area 2']

total = len(areas)
print(total)  # 3

如何使用此解决方案获取您想要的项目列表：

data['State1']['Landon']['abc Area'].append('BOB')
data['State1']['Landon']['abc Area'].append('SAM')
data['State1']['Landon']['xyz Area'].append('John')
data['State2']['New York']['hjk Area'].append('Ricky')


def items_in(d):
    res = []
    if isinstance(d, list):
        res.extend(d)
    elif isinstance(d, dict):
        for k, v in d.items():
            res.extend([k] * len(items_in(v)))
    else:
        raise ValueError('Unknown data')
    return res


print(items_in(data['State1']))  # ['Landon', 'Landon', 'Landon']
print(items_in(data['State1']['Landon']))  # ['xyz Area', 'abc Area', 'abc Area']
print(items_in(data['State1']['Landon']['abc Area']))  # ['BOB', 'SAM']
print(items_in(data['State1']['Landon']['xyz Area']))  # ['John']

print(items_in(data['State2']))  # ['New York']
print(items_in(data['State2']['New York']))  # ['hjk Area']
于 2016-03-23T01:04:46.660 回答
1

一种简单的方法是将其设为列表，然后将每个新键添加到列表中：

Data['State']['City']['Area'] = []
Data['State']['City']['Area'].append( ID )
于 2016-03-22T13:54:53.447 回答
0

您可以用AutoVivication自动激活的类替换该类，Dictlists而不是dicts：

class AutoVivificationDL(Dictlist):  
    """Implementation of perl's autovivification feature."""  
    def __getitem__(self, item):                
         try:  
            return dict.__getitem__(self, item)  
        except KeyError:   
            value = self[item] = type(self)()  
            return value
于 2016-03-22T16:06:09.557 回答
0 
Data = {}


values = [
    dict(State="CA", City="San Francisco", Area="North", Id="customer1"),
    dict(State="CA", City="San Francisco", Area="Embarcadero", Id="customer1"),
    dict(State="CA", City="San Francisco", Area="North", Id="customer2"),

]

for v in values:
    #grab the existing entry.  if it doesn't exist, returns a list
    li = Data.setdefault((v["State"],v["City"],v["Area"]),[])
    li.append(v["Id"])

print "Data:%s" % (Data)

输出：

Data:{('CA', 'San Francisco', 'North'): ['customer1', 'customer2'], ('CA', 'San Francisco', 'Embarcadero'): ['customer1']}

您不仅限于一个非常简单的 Id 值，您可以将几乎任何您想要的内容添加到列表中。如果您希望在多个位置执行此操作，请查看https://docs.python.org/2/library/collections.html#collections.defaultdict，它有点内置 setdefault。

实际上，您可以将 ID 添加到字典而不是列表中，都是一样的。

于 2016-03-22T16:14:53.477 回答