An elegant way to split specific items in a list (array)?

Question

For example, I have a list which only contains zeroes, ones and twos

ls = [
[1, 1, 0, 2, 2],
[1, 1, 1, 2, 2, 2],
[0, 1, 0, 0, 0, 2, 0]
]

I want to split this list it into two list,

ls1 contains the ones, and ls2 contains the twos. I would like to keep the shape and use 0 to replace 2's in ls1 and 1's in ls2 . The expected result is:

ls1 = [
[1, 1, 0, 0, 0],
[1, 1, 1, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0]
]

ls2 = [
[0, 0, 0, 2, 2],
[0, 0, 0, 2, 2, 2],
[0, 0, 0, 0, 0, 2, 0]
]

I know I can use a for loop to handle it, but is there an elegant way?

Answer 1

ls1, ls2 = ([[b & x for b in a] for a in ls] for x in (1, 2))

Answer 2

Using nested list comprehension:

ls1 = [[1 if e == 1 else 0 for e in l] for l in ls]
ls2 = [[2 if e == 2 else 0 for e in l] for l in ls]

# ls1
[[1, 1, 0, 0, 0], 
 [1, 1, 1, 0, 0, 0], 
 [0, 1, 0, 0, 0, 0, 0]]

# ls2
[[0, 0, 0, 2, 2], 
 [0, 0, 0, 2, 2, 2], 
 [0, 0, 0, 0, 0, 2, 0]]

Answer 3

Nothing especially "elegant" about it, but you could simply use a list comprehension:

ls = [
[1, 1, 0, 2, 2],
[1, 1, 1, 2, 2, 2],
[0, 1, 0, 0, 0, 2, 0]
]

def keep_only(val, lst):
    return [[v if v==val else 0 for v in sublist] for sublist in lst]

ls1 = keep_only(1, ls)
ls2 = keep_only(2, ls)

Output:

print(ls1)
print(ls2)

# [[1, 1, 0, 0, 0], [1, 1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0, 0]]
# [[0, 0, 0, 2, 2], [0, 0, 0, 2, 2, 2], [0, 0, 0, 0, 0, 2, 0]]