All permutations from 2 lists

[Larry Martell]

On Wed, Mar 2, 2022 at 9:42 PM Avi Gross via Python-list
<python-list at python.org> wrote:
>
> Larry,
>
> i waited patiently to see what others will write and perhaps see if you explain better what you need. You seem to gleefully swat down anything offered. So I am not tempted to engage.

But then you gave in to the temptation.

> And it is hard to guess as it is not clear what you will do with this.

In the interests of presenting a minimal example I clearly
oversimplified. This is my use case: I get a dict from an outside
source. The dict contains key/value pairs that I need to use to query
a mongodb database. When the values in the dict are all scalar I can
pass the dict directly into the query, e.g.:
self._db_conn[collection_name].find(query). But if any of the values
are lists that does not work. I need to query with something like the
cross product of all the lists. It's not a true product since if a
list is empty it means no filtering on that field, not no filtering on
all the fields.  Originally I did not know I could generate a single
query that did that. So I was trying to come up with a way to generate
a list of all the permutations and was going to issue a query for each
individually.  Clearly that would become very inefficient if the lists
were long or there were a lot of lists. I then found that I could
specify a list with the "$in" clause, hence my solution.

> def query_lfixer(query):
>     for k, v in query.items():
>         if type(v)==list:
>             query[k] = {"$in": v}
>     return query
>
> self._db_conn[collection_name].find(query_lfixer(query))
>
>
> So why did so many of us bother?

Indeed - so why did you bother?