The following code doesn't work, I assume because the locals() variable inside the comprehension will refer to the nested block where comprehension is evaluated:
def f():
a = 1
b = 2
list_ = ['a', 'b']
dict_ = {x : locals()[x] for x in list_}
I could use globals()
instead, and it seems to work, but that may come with some additional problems (eg, if there was a variable from a surrounding scope that happens to have the same name).
Is there anything that would make the dictionary using the variables precisely in the scope of function f
?
Note: I am doing this because I have many variables that I'd like to put in a dictionary later, but don't want to complicate the code by writing dict_['a']
instead of a
in the meantime.
You could perhaps do this:
def f():
a = 1
b = 2
list_ = ['a', 'b']
locals_ = locals()
dict_ = dict((x, locals_[x]) for x in list_)
However, I would strongly discourage the use of locals()
for this purpose.
I believe that you're right: the locals()
inside the dict comprehension will refer to the comprehension's namespace.
One possible solution (if it hasn't already occurred to you):
f_locals = locals()
dict_ = {x : f_locals[x] for x in list_}
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