Double variable list comprehension in python

Question

Can we , in list comprehension , abstract two variables ?
like let's say we have a list of lists,A, and we need to create a list which contains the odd indexed members of all the lists in A .
I tried the following :
new_list = [ [ j[i] for j in matrix] for i in range(len(j)) and i % 2 == 0 ]
but I am getting an error stating "j" is not defined .
Can we have a solution in the above fashion where list comprehension with two variables is being used ?

Answer 1

Kanishk Viman You're welcome!
enumerate() creates a tuple for each element in a list which containts the element's index and the element itself:
print(*enumerate(a)) # (0, [1, 9, 2, 6]) (1, [5, 1, 10, 7]) (2, [9, 1, 2, 4, 5]) (3, [9, 8, 9, 8, 3]) (4, [7, 4, 5, 8, 4]) (5, [7, 3, 5, 9, 6])
In a for loop, you can assign the index to the variable i and the element to the variable l like this:
for i, l in enumerate(a):
print(i, l)
output:
0 [1, 9, 2, 6]
1 [5, 1, 10, 7]
2 [9, 1, 2, 4, 5]
(...)
I did the same thing in my list comprehension:
b = [l for i, l in enumerate(a) if i%2]
=> for every element in a, assign the variable i to its index and l to the element and only take the element (l) if its index is odd (i%2).
Maybe if the tuple is in parentheses it's a little more obvious what it does:
b = [l for (i, l) in enumerate(a) if i%2]

Answer 2

You can, but that would overcomplicate what you're trying to do:
a = [[1, 9, 2, 6], [5, 1, 10, 7], [9, 1, 2, 4, 5], [9, 8, 9, 8, 3], [7, 4, 5, 8, 4], [7, 3, 5, 9, 6]]
b = [a[i] for i in range(len(a)) if i%2]
print(b) # [[5, 1, 10, 7], [9, 8, 9, 8, 3], [7, 3, 5, 9, 6]]
I'd probably do this:
b = [l for i, l in enumerate(a) if i%2]
Here's an example of an actually nested list comprehension:
b = [[i, j] for i in range(2) for j in range(3)]
print(b) # [[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2]]
You can use as many variables and nest as many lists as you need.

Answer 3

This might be hard to understand but does the job (?). Let's wait for Anna answer.
a = [[1,2,3] ,[4,5,6] , [7,8,9]]
b = [i for x in ([c[i] for i in range(0, len(c), 2)] for c in a) for i in x]
print(b) # [1, 3, 4, 6, 7, 9]

Answer 4

Kanishk Viman Did you see my last post? The output is exactly the same as the expected output in your post... 🤔

Answer 5

@Anna : Thanks a ton :) But I need help with the syntax , [I for i ,...] ...how is it deduced that I for i will mean I[i] ?

Answer 6

for a = [ [1,2,3] ,[4,5,6] , [7,8,9]]
the expected output is :
[1,3,4,6,7,9]
but I am getting [4,5,6] with b = [l for i, l in enumerate(a) if i%2]

Answer 7

Oh, okay. I understood "we need to create a list which contains the odd indexed members of all the lists in A" that the new list should only contain a[1], a[3] etc. Wait a minute...

Answer 8

I'm playing around with something like this, but it looks pretty terrible:
b = [m for i, m in enumerate([e for l in a for e in l]) if i%2]

Answer 9

I misunderstood the question yet again 😅
Here you go:
a = [[1,2,3], [4,5,6], [7,8,9]]
b = [e for l in a for i, e in enumerate(l) if i%2 == 0]
print(b) # [1, 3, 4, 6, 7, 9]

Answer 10

@Anna : Thanks a lot again :) ..I understood the logic now , afraid that it might not make a list out of all the odd elements of the lists contained in the initial list

Answer 11

@Diego : This solution worked . But I see you have created a generator there... any reason why a generator but not a list ?

Answer 12

@diego : Works with lists as well , thanks a lot :)

Answer 13

Kanishk Viman I'm more used to generators.

Answer 14

@Anna : Sorry Anna ..missed it somehow ... Thanks a lot ...Apologies again