Can't access variable created by altering locals() inside function

Question

I need to assign some variables when calling a funcion with the name of the variable as an argument.

Therefore, I loop through a tuple with the desired names, assigning them via the locals() dict.

It works, but then I can't access them by name - even inside the function itself.

def function():
    var_names = ("foo","bar","foobar")
    for var_name in var_names:
        locals()[var_name] = len(var_name)
    print foo

Throws:

Traceback (most recent call last):
  File "error_test.py", line 8, in <module>
    function()
  File "error_test.py", line 5, in function
    print foo
NameError: global name 'foo' is not defined

With the following code it works well:

def function():
    var_names = ("foo","bar","foobar")
    for var_name in var_names:
        locals()[var_name] = len(var_name)
    print locals()["foo"]

Isn't it that the locals() dict contains just the normal function variables? Why isn't it working?

Don't expect to reference `foo` by changing a key-value pair in the `locals` dictionary. — Malik Brahimi, Dec 28 '15 at 20:57
You will run into the same problem if you try to extend the functionality of built-in type classes like `str` — Malik Brahimi, Dec 28 '15 at 20:58
If you looked at the [documentation](https://docs.python.org/2/library/functions.html#locals) of the `locals` function, you'd see that it explicitly warns you about this. — user2357112, Dec 28 '15 at 21:02
a useful post: http://stackoverflow.com/questions/5036700/how-can-you-dynamically-create-variables-in-python-via-a-while-loop — newtover, Dec 28 '15 at 21:28
and another one: http://stackoverflow.com/questions/8028708/dynamically-set-local-variable-in-python — newtover, Dec 28 '15 at 21:30

score 1 · Accepted Answer · edited Jun 20 '20 at 09:12

When you write:

for var_name in var_names:
    locals()[var_name] = len(var_name)

You modify the locals() dictionary:

as @user2357112 aptly linked in the docs:

Note

The contents of this dictionary should not be modified; changes may not affect the values of local and free variables used by the interpreter.

So modifying locals() deletes the local variables, hence the NameError: global name 'foo' is not defined error.

Ulf Aslak · Answer 2 · 2016-12-08T22:04:59.370

Quoting this post, the explanation seems to be that inside the function you're only retrieving a copy of the namespace and not the namespace itself:

locals does not actually return the local namespace, it returns a copy. So changing it does nothing to the value of the variables in the local namespace.

I'm still looking for a way to change that namespace inside a function, so I'd love to see someone answer this question with a better solution than this.

Can't access variable created by altering locals() inside function

2 Answers2