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
回答 0
像这样:
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"
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
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
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someDefaultValue
if host
尚未定义。
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'
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
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"
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