Create a "with" block on several context managers?

Question

Suppose you have three objects you acquire via context manager, for instance A lock, a db connection and an ip socket.
You can acquire them by:

with lock:
   with db_con:
       with socket:
            #do stuff

But is there a way to do it in one block? something like

with lock,db_con,socket:
   #do stuff

Furthermore, is it possible, given an array of unknown length of objects that have context managers, is it possible to somehow do:

a=[lock1, lock2, lock3, db_con1, socket, db_con2]
with a as res:
    #now all objects in array are acquired

If the answer is “no”, is it because the need for such a feature implies bad design, or maybe I should suggest it in a pep? 😛

Asked By: olamundo

||

Source

Answer 1

The first part of your question is possible in Python 3.1.

With more than one item, the context managers are processed as if multiple with statements were nested:
with A() as a, B() as b:
    suite
is equivalent to
with A() as a:
    with B() as b:
        suite
Changed in version 3.1: Support for multiple context expressions

Answered By: Mark Byers

Answer 2

In Python 2.7 and 3.1 and above, you can write:

with A() as X, B() as Y, C() as Z:
    do_something()

This is normally the best method to use, but if you have an unknown-length list of context managers you’ll need one of the below methods.

In Python 3.3, you can enter an unknown-length list of context managers by using contextlib.ExitStack:

with ExitStack() as stack:
    for mgr in ctx_managers:
        stack.enter_context(mgr)
    # ...

This allows you to create the context managers as you are adding them to the ExitStack, which prevents the possible problem with contextlib.nested (mentioned below).

contextlib2 provides a backport of ExitStack for Python 2.6 and 2.7.

In Python 2.6 and below, you can use contextlib.nested:

from contextlib import nested

with nested(A(), B(), C()) as (X, Y, Z):
    do_something()

is equivalent to:

m1, m2, m3 = A(), B(), C()
with m1 as X:
    with m2 as Y:
        with m3 as Z:
            do_something()

Note that this isn’t exactly the same as normally using nested with, because A(), B(), and C() will all be called initially, before entering the context managers. This will not work correctly if one of these functions raises an exception.

contextlib.nested is deprecated in newer Python versions in favor of the above methods.

Answered By: interjay

Answer 3

The second part of your question is solved with contextlib.ExitStack in Python 3.3.

Answered By: Neil G

Answer 4

UPDATE: Starting in python 3.10, you’ll be able to use parenthesized context managers! Thanks @iforapsy!

with (
    mock.patch('aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa') as a,
    mock.patch('bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb') as b,
    mock.patch('cccccccccccccccccccccccccccccccccccccccccc') as c,
):
    do_something()

Original Answer For python versions < 3.10

@interjay’s Answer is correct. However, if you need to do this for long context managers, for example mock.patch context managers, then you quickly realize you want to break this across lines. Turns out you can’t wrap them in parens, so you have to use backslashes. Here’s what that looks like:

with mock.patch('aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa') as a, 
        mock.patch('bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb') as b, 
        mock.patch('cccccccccccccccccccccccccccccccccccccccccc') as c:
    do_something()

Answered By: sage88

Answer 5

Following on from @sage88’s response you can always assign those patches to have meaningful variable names before you enter into them.

You could create those patches in multiple lines

a_patch = mock.patch('aaaaaaa') 
b_patch = mock.patch('bbbbbbb')
c_patch = mock.patch('ccccccc') 
with a_patch as a, b_patch as b, c_patch as c:    
    do_something()

Answered By: poulter7

Create a "with" block on several context managers?

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