For completeness, the itertools solution (which returns an iterator) is
>>> os = ["Linux","Windows"]
>>> region = ["us-east-1", "us-east-2"]
>>> import itertools
>>> itertools.product(os,region)
<itertools.product object at 0x0000011C543D3340>
>>> list(itertools.product(os,region))
[('Linux', 'us-east-1'), ('Linux', 'us-east-2'), ('Windows', 'us-east-1'),
('Windows', 'us-east-2')]
There are probably use cases where you want the iterator and others where you
want the list comprehension. I think the itertools looks nice and it's easier
to generalize it to N=3,4,5,... lists than writing out `for a in list1 for b in
list2 for c in list3` etc -- also, you can do things like
itertools.product(*list_of_lists) to get the product if you have a variable
number of lists. Not exactly sure about speed comparison but my instinct is
that a double/triple list comprehension is going to be slower than letting
itertools do its magic.
---- On Tue, 01 Mar 2022 18:20:16 -0600 <2qdxy4rzwzuui...@potatochowder.com>
wrote ----
> On 2022-03-01 at 19:12:10 -0500,
> Larry Martell <larry.mart...@gmail.com> wrote:
>
> > If I have 2 lists, e.g.:
> >
> > os = ["Linux","Windows"]
> > region = ["us-east-1", "us-east-2"]
> >
> > How can I get a list of tuples with all possible permutations?
> >
> > So for this example I'd want:
> >
> > [("Linux", "us-east-1"), ("Linux", "us-east-2"), ("Windows",
> > "us-east-1"), "Windows", "us-east-2')]
> >
> > The lists can be different lengths or can be 0 length. Tried a few
> > different things with itertools but have not got just what I need.
>
> [(o, r) for o in os for r in region]
>
> Feel free to import itertools, but it's not really necessary. ;-)
> --
> https://mail.python.org/mailman/listinfo/python-list
>
--
https://mail.python.org/mailman/listinfo/python-list
