检查字典键是否为空

Question

我有以下字典

dict1 ={"city":"","name":"yass","region":"","zipcode":"",
       "phone":"","address":"","tehsil":"", "planet":"mars"}

我正在尝试创建一个将基于dict1的新字典，但是，

1. 它不会包含带有空字符串的键。
2. 它不会包含我不想包含的那些键。

我已经能够满足要求2，但遇到了要求1的问题。这就是我的代码。

dict1 ={"city":"","name":"yass","region":"","zipcode":"",
   "phone":"","address":"","tehsil":"", "planet":"mars"}

blacklist = set(("planet","tehsil"))    
new = {k:dict1[k] for k in dict1 if k not in blacklist} 

这给了我没有键的字典：“ tehsil”，“ planet”我也尝试了以下方法，但是没有用。

new = {k:dict1[k] for k in dict1 if k not in blacklist and dict1[k] is not None}

结果字典应如下所示：

new = {"name":"yass"}

Answer 1

这是白名单版本：

>>> dict1 ={"city":"","name":"yass","region":"","zipcode":"",
       "phone":"","address":"","tehsil":"", "planet":"mars"}
>>> whitelist = ["city","name","planet"]
>>> dict2 = dict( (k,v) for k, v in dict1.items() if v and k in whitelist )
>>> dict2
{'planet': 'mars', 'name': 'yass'}

黑名单版本：

>>> blacklist = set(("planet","tehsil"))
>>> dict2 = dict( (k,v) for k, v in dict1.items() if v and k not in blacklist )
>>> dict2
{'name': 'yass'}

两者基本上是相同的，期望一个拥有not in另一个in。如果您的python版本支持，则可以执行以下操作：

>>> dict2 = {k: v for k, v in dict1.items() if v and k in whitelist}

和

>>> dict2 = {k: v for k, v in dict1.items() if v and k not in blacklist}

Answer 2

这必须是最快的方法（使用 set Difference）：

>>> dict1 = {"city":"","name":"yass","region":"","zipcode":"",
       "phone":"","address":"","tehsil":"", "planet":"mars"}
>>> blacklist = {"planet","tehsil"}
>>> {k: dict1[k] for k in dict1.viewkeys() - blacklist if dict1[k]}
{'name': 'yass'}

白名单版本（使用集合交集）：

>>> whitelist = {'city', 'name', 'region', 'zipcode', 'phone', 'address'}
>>> {k: dict1[k] for k in dict1.viewkeys() & whitelist if dict1[k]}
{'name': 'yass'}