问题:字典和默认值
假设connectionDetails是Python字典,那么像这样的重构代码的最佳,最优雅,最“ pythonic”的方法是什么?
if "host" in connectionDetails:
    host = connectionDetails["host"]
else:
    host = someDefaultValue
 
        
        
            
            
            
                
                    
                    
Assuming connectionDetails is a Python dictionary, what’s the best, most elegant, most “pythonic” way of refactoring code like this?
if "host" in connectionDetails:
    host = connectionDetails["host"]
else:
    host = someDefaultValue
     
                 
             
            
         
        
        
回答 0
像这样:
host = connectionDetails.get('host', someDefaultValue)
 
        
        
            
            
            
                
                    
Like this:
host = connectionDetails.get('host', someDefaultValue)
     
                 
             
            
         
        
        
回答 1
您也可以这样使用defaultdict:
from collections import defaultdict
a = defaultdict(lambda: "default", key="some_value")
a["blabla"] => "default"
a["key"] => "some_value"
您可以传递任何普通函数而不是lambda:
from collections import defaultdict
def a():
  return 4
b = defaultdict(a, key="some_value")
b['absent'] => 4
b['key'] => "some_value"
 
        
        
            
            
            
                
                    
You can also use the defaultdict like so:
from collections import defaultdict
a = defaultdict(lambda: "default", key="some_value")
a["blabla"] => "default"
a["key"] => "some_value"
You can pass any ordinary function instead of lambda:
from collections import defaultdict
def a():
  return 4
b = defaultdict(a, key="some_value")
b['absent'] => 4
b['key'] => "some_value"
     
                 
             
            
         
        
        
回答 2
虽然这.get()是一个很好的习惯用法,但是它比if/else(比try/except大多数情况下可以预期字典中键的存在要慢):
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="try:\n a=d[1]\nexcept KeyError:\n a=10")
0.07691968797894333
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="try:\n a=d[2]\nexcept KeyError:\n a=10")
0.4583777282275605
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="a=d.get(1, 10)")
0.17784020746671558
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="a=d.get(2, 10)")
0.17952161730158878
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="if 1 in d:\n a=d[1]\nelse:\n a=10")
0.10071221458065338
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="if 2 in d:\n a=d[2]\nelse:\n a=10")
0.06966537335119938
 
        
        
            
            
            
                
                    
While .get() is a nice idiom, it’s slower than if/else (and slower than try/except if presence of the key in the dictionary can be expected most of the time):
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="try:\n a=d[1]\nexcept KeyError:\n a=10")
0.07691968797894333
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="try:\n a=d[2]\nexcept KeyError:\n a=10")
0.4583777282275605
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="a=d.get(1, 10)")
0.17784020746671558
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="a=d.get(2, 10)")
0.17952161730158878
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="if 1 in d:\n a=d[1]\nelse:\n a=10")
0.10071221458065338
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="if 2 in d:\n a=d[2]\nelse:\n a=10")
0.06966537335119938
     
                 
             
            
         
        
        
回答 3
对于多个不同的默认值,请尝试以下操作:
connectionDetails = { "host": "www.example.com" }
defaults = { "host": "127.0.0.1", "port": 8080 }
completeDetails = {}
completeDetails.update(defaults)
completeDetails.update(connectionDetails)
completeDetails["host"]  # ==> "www.example.com"
completeDetails["port"]  # ==> 8080
 
        
        
            
            
            
                
                    
For multiple different defaults try this:
connectionDetails = { "host": "www.example.com" }
defaults = { "host": "127.0.0.1", "port": 8080 }
completeDetails = {}
completeDetails.update(defaults)
completeDetails.update(connectionDetails)
completeDetails["host"]  # ==> "www.example.com"
completeDetails["port"]  # ==> 8080
     
                 
             
            
         
        
        
回答 4
python词典中有一个方法可以做到这一点: dict.setdefault
connectionDetails.setdefault('host',someDefaultValue)
host = connectionDetails['host']
但是,与问题所要求的不同,此方法将if 的值设置connectionDetails['host']为someDefaultValueif host尚未定义。
 
        
        
            
            
            
                
                    
There is a method in python dictionaries to do this: dict.setdefault
connectionDetails.setdefault('host',someDefaultValue)
host = connectionDetails['host']
However this method sets the value of connectionDetails['host'] to someDefaultValue if key host is not already defined, unlike what the question asked.
     
                 
             
            
         
        
        
回答 5
(这是一个很晚的答案)
一种替代方法是对类进行子dict类化并实现__missing__()方法,如下所示:
class ConnectionDetails(dict):
    def __missing__(self, key):
        if key == 'host':
            return "localhost"
        raise KeyError(key)
例子:
>>> connection_details = ConnectionDetails(port=80)
>>> connection_details['host']
'localhost'
>>> connection_details['port']
80
>>> connection_details['password']
Traceback (most recent call last):
  File "python", line 1, in <module>
  File "python", line 6, in __missing__
KeyError: 'password'
 
        
        
            
            
            
                
                    
(this is a late answer)
An alternative is to subclass the dict class and implement the __missing__() method, like this:
class ConnectionDetails(dict):
    def __missing__(self, key):
        if key == 'host':
            return "localhost"
        raise KeyError(key)
Examples:
>>> connection_details = ConnectionDetails(port=80)
>>> connection_details['host']
'localhost'
>>> connection_details['port']
80
>>> connection_details['password']
Traceback (most recent call last):
  File "python", line 1, in <module>
  File "python", line 6, in __missing__
KeyError: 'password'
     
                 
             
            
         
        
        
回答 6
测试@Tim Pietzcker对Python 3.3.5的PyPy(5.2.0-alpha0)情况的怀疑,我发现确实两者.get()和if/ else方式的执行情况相似。实际上,在if / else情况下,如果条件和赋值涉及相同的键,则似乎只有一次查找(与最后一次有两次查找的情况比较)。
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="try:\n a=d[1]\nexcept KeyError:\n a=10")
0.011889292989508249
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="try:\n a=d[2]\nexcept KeyError:\n a=10")
0.07310474599944428
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="a=d.get(1, 10)")
0.010391917996457778
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="a=d.get(2, 10)")
0.009348208011942916
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 1 in d:\n a=d[1]\nelse:\n a=10")
0.011475925013655797
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 2 in d:\n a=d[2]\nelse:\n a=10")
0.009605801998986863
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 2 in d:\n a=d[2]\nelse:\n a=d[1]")
0.017342638995614834
 
        
        
            
            
            
                
                    
Testing @Tim Pietzcker’s suspicion about the situation in PyPy (5.2.0-alpha0) for Python 3.3.5, I find that indeed both .get() and the if/else way perform similar. Actually it seems that in the if/else case there is even only a single lookup if the condition and the assignment involve the same key (compare with the last case where there is two lookups).
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="try:\n a=d[1]\nexcept KeyError:\n a=10")
0.011889292989508249
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="try:\n a=d[2]\nexcept KeyError:\n a=10")
0.07310474599944428
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="a=d.get(1, 10)")
0.010391917996457778
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="a=d.get(2, 10)")
0.009348208011942916
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 1 in d:\n a=d[1]\nelse:\n a=10")
0.011475925013655797
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 2 in d:\n a=d[2]\nelse:\n a=10")
0.009605801998986863
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 2 in d:\n a=d[2]\nelse:\n a=d[1]")
0.017342638995614834
     
                 
             
            
         
        
        
回答 7
您可以将lamba函数用作单线。制作一个connectionDetails2可以像函数一样访问的新对象      …
connectionDetails2 = lambda k: connectionDetails[k] if k in connectionDetails.keys() else "DEFAULT"
现在使用 
connectionDetails2(k)
代替 
connectionDetails[k]
如果k在键中,则返回字典值,否则返回"DEFAULT"
 
        
        
            
            
            
                
                    
You can use a lamba function for this as a one-liner. Make a new object     connectionDetails2 which is accessed like a function…
connectionDetails2 = lambda k: connectionDetails[k] if k in connectionDetails.keys() else "DEFAULT"
Now use 
connectionDetails2(k)
instead of 
connectionDetails[k]
which returns the dictionary value if k is in the keys, otherwise it returns "DEFAULT"
     
                 
             
            
         
        
        
	
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