How to set local variable in list comprehension?

Solution 1:

Use nested list comprehension:

[x for x in [map_to_obj(v) for v in v_list] if x]

or better still, a list comprehension around a generator expression:

[x for x in (map_to_obj(v) for v in v_list) if x]

Solution 2:

Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), it's possible to use a local variable within a list comprehension in order to avoid calling twice the same function:

In our case, we can name the evaluation of map_to_obj(v) as a variable o while using the result of the expression to filter the list; and thus use o as the mapped value:

[o for v in [v1, v2, v3, v4] if (o := map_to_obj(v))]

Solution 3:

A variable assignment is just a singular binding:

[x   for v in l   for x in [v]]

This is a more general answer and also closer to what you proposed. So for your problem you can write:

[x   for v in v_list   for x in [map_to_obj(v)]   if x]

Solution 4:

You can avoid re-calculation by using python built-in filter:

list(filter(lambda t: t is not None, map(map_to_obj, v_list)))

Solution 5:

A local variable can be set within a comprehension by cheating a bit and using an extra 'for' which "iterates" through a 1-element tuple containing the desired value for the local variable. Here's a solution to the OP's problem using this approach:

[o for v in v_list for o in (map_to_obj(v),) if o]

Here, o is the local variable being set equal to map_to_obj(v) for each v.

In my tests this is slightly faster than Lying Dog's nested generator expression (and also faster than the OP's double-call to map_to_obj(v), which, surprisingly, can be faster than the nested generator expression if the map_to_obj function isn't too slow).