问题描述
我对Python locals()
的奇怪行为感到困惑.基本上我想从字典理解中的locals()
字典中获取一个项目,但是失败了.这是非常基本的事情,所以:
I have been baffled by an odd behaviour of Python locals()
.
Basically I want to get an item from the dictionary of locals()
in a dictionary comprehension, yet it fails. It is an extremely basic thing, so:
>>> foo=123
>>> bar=345
>>> baz=678
>>> {k: locals()[k] for k in ('foo','bar','baz')}
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 1, in <dictcomp>
KeyError: 'foo'
>>> locals()['foo']
123
>>> locale=locals()
>>> {k: locale[k] for k in ('foo','bar','baz')}
{'foo': 123, 'bar': 345, 'baz': 678}
>>> type(locals())
<class 'dict'>
>>> def fun():
... return {'foo': 123,'bar':345}
...
>>> {k: fun()[k] for k in ('foo','bar')}
{'foo': 123, 'bar': 345}
从实用的角度来看,字典或字符串.format()
中较丑的{'foo':foo, 'bar': bar}
等都可以正常工作.
只是我想念一些东西,所以知道为什么会增加我的编码能力(到目前为止,我在编码时不发光).
On a practical side the uglier {'foo':foo, 'bar': bar}
etc. in dict or in string .format()
works fine.
It is just that I am missing something so knowing why will increase my coding chi (as of now I don't glow when coding).
推荐答案
由于Python 3中的所有理解都是使用隐藏函数实现的,因此调用locals
不会返回您期望它返回的值.
Because all comprehensions in Python 3 are implemented using a hidden function, calling locals
doesn't return the values you're expecting it to return.
您可以通过将值打印出来来查看:
You can see this by printing the values out:
>>> _ = {k: print(locals()) for k in ('foo','bar','baz')}
{'k': 'foo', '.0': <tuple_iterator object at 0x7fdf840afa90>}
{'k': 'bar', '.0': <tuple_iterator object at 0x7fdf840afa90>}
{'k': 'baz', '.0': <tuple_iterator object at 0x7fdf840afa90>}
将locals()
分配给locale
就像您所做的那样.您不是在理解内调用locals
.
Assigning locals()
to locale
, as you do, takes care of this. You aren't calling locals
inside the comprehension.
请注意,在Python 2中,情况有些模糊. dict-comps确实会以类似的方式失败,但是在dict-comps之前的list-comps可以正常工作:
Note that, in Python 2, the situation is a bit murkier. dict-comps do fail in a similar fashion but list-comps, which predate dict-comps, work just fine:
>>> _ = [locals()[k] for k in ('foo', 'bar', 'baz')]
>>> _
[20, 40, 60]
这是已解决与Py3.
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