Jonathan Jonathan - 1 month ago 18
Python Question

Dynamically set local variable

How do you dynamically set local variable in Python?

(where the variable name is dynamic)

UPDATE: I'm aware this isn't good practice, and the remarks are legit, but this doesn't make it a bad question, just a more theoretical one - I don't see why this justifies downvotes.

Answer

Contrary to other answers already posted you cannot modify locals() directly and expect it to work.

>>> def foo():
    lcl = locals()
    lcl['xyz'] = 42
    print(xyz)


>>> foo()

Traceback (most recent call last):
  File "<pyshell#6>", line 1, in <module>
    foo()
  File "<pyshell#5>", line 4, in foo
    print(xyz)
NameError: global name 'xyz' is not defined

Modifying locals() is undefined. Outside a function when locals() and globals() are the same it will work; inside a function is will usually not work.

Use a dictionary, or set an attribute on an object:

d = {}
d['xyz'] = 42
print(d['xyz'])

or if you prefer, use a class:

class C: pass

obj = C()
setattr(obj, 'xyz', 42)
print(obj.xyz)

Edit: Access to variables in namespaces that aren't functions (so modules, class definitions, instances) are usually done by dictionary lookups (as Sven points out in the comments there are exceptions, for example classes that define __slots__). Function locals can be optimised for speed because the compiler (usually) knows all the names in advance, so there isn't a dictionary until you call locals().

In the C implementation of Python locals() (called from inside a function) creates an ordinary dictionary initialised from the current values of the local variables. Within each function any number of calls to locals() will return the same dictionary, but every call to locals() will update it with the current values of the local variables. This can give the impression that assignment to elements of the dictionary are ignored (I originally wrote that this was the case). Modifications to existing keys within the dictionary returned from locals() therefore only last until the next call to locals() in the same scope.

In IronPython things work a bit differently. Any function that calls locals() inside it uses a dictionary for its local variables so assignments to local variables change the dictionary and assignments to the dictionary change the variables BUT that's only if you explicitly call locals() under that name. If you bind a different name to the locals function in IronPython then calling it gives you the local variables for the scope where the name was bound and there's no way to access the function locals through it:

>>> def foo():
...     abc = 123
...     lcl = zzz()
...     lcl['abc'] = 456
...     deF = 789
...     print(abc)
...     print(zzz())
...     print(lcl)
...
>>> zzz =locals
>>> foo()
123
{'__doc__': None, '__builtins__': <module '__builtin__' (built-in)>, 'zzz': <built-in function locals>, 'foo': <function foo at 0x000000000000002B>, '__name__': '__main__', 'abc': 456}
{'__doc__': None, '__builtins__': <module '__builtin__' (built-in)>, 'zzz': <built-in function locals>, 'foo': <function foo at 0x000000000000002B>, '__name__': '__main__', 'abc': 456}
>>>

This could all change at any time. The only thing guaranteed is that you cannot depend on the results of assigning to the dictionary returned by locals().