Eval scope in Python 2 vs. 3

Question

I came across bizarre eval behavior in Python 3 - local variables aren't picked up when eval is called in a list comprehension.

def apply_op():     x, y, z = [0.5, 0.25, 0.75]     op = "x,y,z"     return [eval(o) for o in op.split(",")] print(apply_op()) 

It errors in Python 3:

▶ python --version Python 3.4.3 ▶ python eval.py Traceback (most recent call last):   File "eval.py", line 7, in <module>     print(apply_op())   File "eval.py", line 5, in apply_op     return [eval(o) % 1 for o in op.split(",")]   File "eval.py", line 5, in <listcomp>     return [eval(o) % 1 for o in op.split(",")]   File "<string>", line 1, in <module> NameError: name 'x' is not defined 

And it works fine in Python 2:

▶ python --version Python 2.7.8 ▶ python eval.py [0.5, 0.25, 0.75] 

Moving it outside of the list comprehension removes the problem.

def apply_op():     x, y, z = [0.5, 0.25, 0.75]     return [eval("x"), eval("y"), eval("z")] 

Is this intended behavior, or is it a bug?

Answer 1

There is a closed issue in the bug tracker for this: Issue 5242.

The resolution for this bug is won't fix.

Some comments from the Issue read:

This is expected, and won't easily fix. The reason is that list comprehensions in 3.x use a function namespace "under the hood" (in 2.x, they were implemented like a simple for loop). Because inner functions need to know what names to get from what enclosing namespace, the names referenced in eval() can't come from enclosing functions. They must either be locals or globals.

eval() is probably already an hack, there's no need to add another hack to make it work. It's better to just get rid of eval() and find a better way to do what you want to do.