[英]Command exec in a function, name error - python3
I'm begginer in Python and I'm in front of a problem that I can't understand.我是 Python 的初学者,我遇到了一个我无法理解的问题。 I tried to just define a variable with a exec() and then just print it.
我试图用 exec() 定义一个变量,然后打印它。 And it's working well.
它运作良好。 BUT when I do the same code in a function, it's not working...
但是当我在 function 中执行相同的代码时,它不起作用......
Example:例子:
def fonct():
possibilite = [0,1,2,3]
testc = 0
testl = 1
commande = "proba"+str(testc+1)+str(testl)+" = possibilite"
exec(commande)
print(commande)
print(proba11)
The same thing but not in a function has for result, that the command print(proba11) returns [0,1,2,3] so it works.同样的事情,但不是在 function 有结果,命令 print(proba11) 返回 [0,1,2,3] 所以它工作。 But for the example I got this:
但对于这个例子,我得到了这个:
proba11 = possibilite
NameError: name 'proba11' is not defined
There is no stories about globals or locals, everything is local...没有关于全球人或本地人的故事,一切都是本地人......
Updating the local variables with exec()
in Python 3 is tricky due to the way local variables are stored.由于局部变量的存储方式,在 Python 3 中使用
exec()
更新局部变量很棘手。 It used to work in Python 2.7 and earlier.它曾经在 Python 2.7 及更早版本中工作。
To workaround this, you need to要解决此问题,您需要
locals
dictionary to exec
locals
字典传递给exec
Like so:像这样:
def fonct():
possibilite = [0, 1, 2, 3]
testc = 0
testl = 1
varname = "proba" + str(testc + 1) + str(testl)
commande = varname + " = possibilite"
_locals = locals()
exec(commande, globals(), _locals)
proba11 = _locals[varname]
print(proba11)
Which works as expected.哪个按预期工作。
You can read more about it here:你可以在这里读更多关于它的内容:
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