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我想创建一个具有多个键和值的字典。在这一点上，我不确定我的问题是否正确。但这是我想要创建的示例：

patDct = {
           'mkey1':{'key1':'val_a1', 'key2':'val_a2', 'key3':'val_a3'},
           'mkey2':{'key1':'val_b1', 'key2':'val_b2', 'key3':'val_b3'},
           ....
          }

我有两个字典，我正在从中提取“mkey*”和“val*”的信息。'key*' 是字符串。

我有一段代码可以创建没有'mkey *'的字典，但这只会打印出最后一组值。以下是我现在所拥有的。“storedct”和“datadct”是两个给定的字典。在这里，我希望'mkey*' 来获取“item”的值。

 patDct = dict()                                                                                                                                                             
 for item in storedct :                                                                                                                                                         
     for pattern in datadct :                                                                                                                                                    
         if pattern in item :                                                                                                                                                     
             patDct['key1'] = datadct[pattern]["dpath"]                                                                                                        
             patDct['key2'] = datadct[pattern]["mask"]                                                                                                                                                                                                                                                
             patDct['key3'] = storedct[item]

感谢您的任何建议。

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2 回答 2

1

根据我从您的代码中了解到的情况，我猜想：

patDct = dict()
i = 0
for item in storedct :
    for pattern in datadct :
        if pattern in item :
            i = i + 1
            new_item = {}
            new_item['key1'] = datadct[pattern]["dpath"]
            new_item['key2'] = datadct[pattern]["mask"]
            new_item['key3'] = storedct[item]
            # I used a counter to generate the `mkey` values,
            # not sure you want it that way
            patDct['mkey{0}'.format(i)] = new_item

应该离您的需求不远...

于 2012-09-07T14:45:30.987 回答
1 
 patDct = dict()    
 n=1                                                                                                                                                        
 for item in storedct :
     patDct["mkey%s"%n] = {}
     p =  patDct["mkey%s"%n]                                                                                                                                                            
     for pattern in datadct :                                                                                                                                                    
         if pattern in item :                                                                                                                                                     
             p['key1'] = datadct[pattern]["dpath"]                                                                                                        
             p['key2'] = datadct[pattern]["mask"]                                                                                                                                                                                                                                                
             p['key3'] = storedct[item] 
     n +=1

print patDct
于 2012-09-07T14:42:29.460 回答