python plt.legend用法 - 字典和默认值




legend位置 legend中文 (8)

测试@Tim Pietzcker对Python 3.3.5的PyPy(5.2.0-alpha0)情况的怀疑,我发现的确如此.get()if / else方式的表现都相似。 实际上,if / else情况下,如果条件和赋值涉及相同的键(与有两次查找的最后一种情况相比),则甚至只有一次查找。

>>>> 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

假设connectionDetails是一个Python字典,那么像这样重构代码的最好,最优雅,最“pythonic”的方式是什么?

if "host" in connectionDetails:
    host = connectionDetails["host"]
else:
    host = someDefaultValue

喜欢这个:

host = connectionDetails.get('host','someDefault')

您可以使用lamba函数作为单线程。 创建一个新的对象connectionDetails2 ,它可以像函数一样访问...

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"


虽然.get()是一个很好的习惯用法,但它比if/else慢(而且比try/except慢, try/except字典中大多数情况下可以预期键的存在):

>>> 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

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"

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"

对于多个不同的默认值,试试这个:

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

python字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,如果关键host尚未定义,则此方法将connectionDetails['host']的值设置为someDefaultValue ,与问题提出的内容不同。






python dictionary coding-style