python - Python 2 与 3 中的评估范围

Question

我在 Python 3 中遇到了奇怪的 eval 行为 - 在列表推导中调用 eval 时，不会拾取局部变量。

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())

Python 3 中的错误:

▶ 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

它在 Python 2 中运行良好:

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

将它移到列表推导之外可以解决问题。

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

这是预期的行为，还是错误？

Answer 1

在错误跟踪器中有一个 已关闭 问题:Issue 5242 .

此错误的解决方案是无法修复。

问题中的一些评论如下:

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.