假设connectionDetails是一个Python字典,那么像这样重构代码的最佳、最优雅、最“Python”的方式是什么呢?
if "host" in connectionDetails:
host = connectionDetails["host"]
else:
host = someDefaultValue
假设connectionDetails是一个Python字典,那么像这样重构代码的最佳、最优雅、最“Python”的方式是什么呢?
if "host" in connectionDetails:
host = connectionDetails["host"]
else:
host = someDefaultValue
当前回答
您可以使用一个lamba函数作为一行程序。创建一个新对象connectionDetails2,可以像函数一样访问它…
connectionDetails2 = lambda k: connectionDetails[k] if k in connectionDetails.keys() else "DEFAULT"
现在使用
connectionDetails2(k)
而不是
connectionDetails[k]
如果k在键中,则返回字典值,否则返回"DEFAULT"
其他回答
我相信所有这些答案都是可以的,但这表明没有“好”的方法来做到这一点。我一直使用字典而不是case语句,并添加一个默认子句,我只是调用以下函数:
def choose(choise, choises, default):
"""Choose a choice from the choises given
"""
return choises[choise] if choise in choises else default
这样我就可以使用正常的字典,没有特殊的默认子句等。
你也可以像这样使用defaultdict:
from collections import defaultdict
a = defaultdict(lambda: "default", key="some_value")
a["blabla"] => "default"
a["key"] => "some_value"
你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"
虽然.get()是一个很好的习惯用法,但它比if/else慢(并且比try/慢,除非大多数时候字典中会出现键):
>>> 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
测试@Tim Pietzcker对Python 3.3.5中PyPy (5.2.0-alpha0)情况的怀疑,我发现.get()和if/else方式确实执行相似。实际上,在if/else情况下,如果条件和赋值涉及相同的键,似乎甚至只有一次查找(与有两次查找的上一种情况相比)。
>>>> 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
(这是一个迟来的回答)
另一种方法是继承dict类的子类并实现__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'