Python vars()全局名称错误 [英] Python vars() global name error

问题描述

我在理解以下功能出了什么问题时遇到了麻烦:

I'm having a bit of trouble understanding what's going wrong with the following function:

def ness():
 pie='yum'
 vars()[pie]=4
 print vars()[pie]
 print yum

所以当我跑步时，我会得到以下结果:

So When I run that I get this result:

>>> ness()
4
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in ness
NameError: global name 'yum' is not defined

如果我不将其编写为函数，而只需一次在命令行中将其键入即可，就像这样:

If I don't write it as a function and just type it in on the command line one line at a time it works fine, like so:

>>> pie='yum'
>>> vars()[pie]=4
>>> print vars()[pie]
4
>>> print yum
4
>>> 

假设我想让事情变得比这复杂一些，而不是将yum设置为一个值并打印该值，而是定义了一些函数，并希望根据一些输入来调用其中之一:

Suppose I wanted to make things a bit more complicated than this and instead of setting yum to a value and printing that value, I define some functions, and want to call one of them based on some input:

def ness(choo):
    dic={}
    dessert=()
    dnum=[10,100]
    desserts='pie'
    dic[dessert]=str(desserts[bisect(dnum,choo)])
    vars()[dic[dessert]]()
def p():
    print 'ummmm ummm'
def i():
    print 'hooo aaaaa'
def e():
    print 'woooo'

所以当我打电话给我时，我得到一个关键错误:

So when I call ness I get a key error:

>>> ness(3)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 7, in ness
KeyError: 'p'

现在我知道我可以使用一些elif语句来做类似的事情，但是我想知道这是否也可以工作，以及是否使用像这样的bisect会更有效(比如我是否需要检查1000个choo值)比起使用Elifs.

Now I know I can do things like this with some elif statements, but I'm wondering if this would work too, and if using bisect like this would be more efficient (say if i need to check 1000 values of choo) than using elifs.

非常感谢您的帮助.

推荐答案

可以通过exec来做到这一点

There is way to do it with exec

>>> def ness():
...  pie='yum'
...  exec pie+"=4"
...  print vars()[pie]
...  print yum
...
>>>
>>> ness()
4
4

但是，与其这样做，不如使用一个新的dict更好，更安全

But Instead of doing that, using a new dict is better and safe

>>> def ness():
...  dic={}
...  pie='yum'
...  dic[pie]=4
...  print dic[pie]
...  print dic['yum']
...
>>> ness()
4
4
>>>

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