为什么Python生成器在exec'd脚本中将其范围与全局混淆？

Question

好的,所以我在一个工具的配置脚本是exec'python脚本' 的环境中工作.exec调用是这样的:

outer.py:

exec(open("inner.py").read(), globals(), {})

现在,我想在exec'd脚本中做一些相对基本的迭代.在这种情况下,当某些值不在白名单中时执行工作:

inner.py:

items = (
  'foo/bar',
  'foo/baz',
  'foof',
  'barf/fizz',
)

whitelist = (
  'foo/',
)

for key in items:
  try:
    # Not terribly efficient, but who cares; computers are fast.
    next(True for prefix in whitelist if key.startswith(prefix))

  # Do some work here when the item doesn't match the whitelist.
  except StopIteration:
    print("%10s isn't in the whitelist!" % key)

---

运行python inner.py产生预期结果:

      foof isn't in the whitelist!
 barf/fizz isn't in the whitelist!

这是一个奇怪的部分:运行python outer.py似乎表明解释器对生成器的范围感到困惑:

Traceback (most recent call last):
  File "outer.py", line 1, in <module>
    exec(open("inner.py").read(), globals(), {})
  File "<string>", line 15, in <module>
  File "<string>", line 15, in <genexpr>
NameError: global name 'key' is not defined

---

其他一些说明:

• 您可以print(key)在for循环内部(在生成器运行之前)完成.

• locals()在解决问题的exec行中替换空字典outer.py,但该代码不受我的控制.

• 我正在运行OS X内置的Python 2.7.2(Mountain Lion).

Answer 1

确实很狡猾.

来自doc:

如果给出两个单独的对象作为全局变量和局部变量,则代码将被执行,就像它嵌入在类定义中一样.

(请注意,当您exec(..., globals(), locals())在模块级别运行时,这不适用,因为globals() is locals(),即不是两个单独的对象).

这表明您可以通过运行此脚本来简单地重现该问题:

class A(object):

  items = (
    'foo/bar',
    'foo/baz',
    'foof',
    'barf/fizz',
  )

  whitelist = (
    'foo/',
  )

  for key in items:
    try:
      # Not terribly efficient, but who cares; computers are fast.
      next(True for prefix in whitelist if key.startswith(prefix))
      # Found!
      print(key)
    except StopIteration:
      pass

"好的,但为什么我在这里得到错误？"

很高兴你问.答案就在这里.