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[ Python: How to set local variable in list comprehension? ]

I have a method that take a list and return an object

# input a list, returns an object
def map_to_obj(lst):
    a_list = f(lst)
    return a_list[0] if a_list else None

I want to get a list that contains all the mapped elements that isn't None.

Like this:

v_list = [v1, v2, v3, v4]

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

But it seems not good to call the map_to_obj method twice in the list comprehension.

Is there a way to have local variables in list comprehension so that it can have better performance?

Or does the compiler optimize it automatically?

Here is what I want:

(sml like)
[let mapped = map_to_obj(v) in for v in v_list if mapped end] 

Answer 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]

Answer 2

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)))

Answer 3

List comprehensions are fine for the simple cases, but sometimes a plain old for loop is the simplest solution:

other_list = []
for v in v_list:
    obj = map_to_obj(v)
    if obj:

Now if you really want a list comp and dont want to build an tmp list, you can use the iterator versions of filter and map:

import itertools as it
result = list(it.ifilter(None, it.imap(map_to_obj, v_list)))

or more simply :

import itertools as it
result = filter(None, it.imap(map_to_obj, v_list)))

The iterator versions don't build a temporary list, they use lazy evaluation.

Answer 4

I have figured out a way of using reduce:

def map_and_append(lst, v):
    mapped = map_to_obj(v)
    if mapped is not None:
    return lst

reduce(map_and_append, v_list, [])

How about the performance of this?