How to __enter__ n context managers?

Question

Using the with statement, we can enter many context handlers using only one level of indentation/nesting:

>>> from contextlib import contextmanager
>>> @contextmanager
... def frobnicate(n):
...     print('frobbing {}'.format(n))
...     yield
... 
>>> frob1 = frobnicate(1)
>>> frob2 = frobnicate(2)
>>> with frob1, frob2:
...     pass
... 
frobbing 1
frobbing 2

But this doesn't seem to work:

>>> frobs = [frobnicate(1), frobnicate(2)]
>>> with *frobs:
...     pass
# SyntaxError: invalid syntax

How can we enter n context managers without having to manually write out each one?

Answer 1

python2.7 had contextlib.nested to do exactly that, however, it was deprecated due to error-prone quirks.

This function has two major quirks that have led to it being deprecated. Firstly, as the context managers are all constructed before the function is invoked, the __new__() and __init__() methods of the inner context managers are not actually covered by the scope of the outer context managers. That means, for example, that using nested() to open two files is a programming error as the first file will not be closed promptly if an exception is thrown when opening the second file.

Secondly, if the __enter__() method of one of the inner context managers raises an exception that is caught and suppressed by the __exit__() method of one of the outer context managers, this construct will raise RuntimeError rather than skipping the body of the with statement.

python3.3 does a better job with contextlib.ExitStack which would look like:

from contextlib import ExitStack

with ExitStack() as stack:
    contexts = [stack.enter_context(frobnicate(i)) for i in range(2)]
    ...

See contextlib2 for a backport to python2.x code.