Issue
I have the following toy function:
def foo(a):
return [a+5]
And I am running the following code:
my_lst = [foo(x) for x in range(10) if x%2 == 0]
Getting:
[[5], [7], [9], [11], [13]]
But need:
[5,7,9,11,13]
But I want to get a plain list, not a list of lists.
How can I do it without itertools.chain.from_iterable(my_lst)
but with list comprehension?
What is the best practice? itertools or list comprehension in this case?
Please advice.
I have tried:
[j[0] for j in [foo(x) for x in range(10) if x%2 == 0]]
Should I do it like this or there is a better way?
Solution
With list comprehension, it's done using two for
loops:
my_lst = [subx for x in range(10) if x%2 == 0 for subx in foo(x)]
Also if you have a list of lists of one elements, you can "transpose" after the fact the list using zip
:
bad_lst = [foo(x) for x in range(10) if x%2 == 0]
[my_lst] = zip(*bad_lst)
Answered By - Lærne
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