Merge list into sparse list efficiently

Question

I have two lists:

a = [None, None, 1, None, 4, None, None, 5, None]
b = [7,8,2,3,6,9]

I want to merge them, either to create a new list or just update a, by filling in the Nones with the values from b, so

a = [7,8,1,2,4,3,6,5,9]

What's the most efficient way of doing this?

For extension, I'll be wanting to do this with every permutation of b. Does this allow shortcutting the technique at all?

Answer 1

This is one approach. Using a list comprehension and converting b to a iterator object.

Demo:

a = [None, None, 1, None, 4, None, None, 5, None]
b = [7,8,2,3,6,9]
b = iter(b)

print( [next(b) if i is None else i for i in a] )

Output:

[7, 8, 1, 2, 4, 3, 6, 5, 9]

Answer 2

With list.pop() method on reversed 2nd b list:

a = [None, None, 1, None, 4, None, None, 5, None]
b = [7,8,2,3,6,9]
tmp = b[::-1]

result = [tmp.pop() if i is None else i for i in a]
print(result)    # [7, 8, 1, 2, 4, 3, 6, 5, 9]

Answer 3

for reversed_index in range(len(a) - 1, -1, -1):
    if a[reversed_index] is None:
        a[reversed_index] = b.pop()

It should be efficient as it modifies a in place and right popping a list is also efficient. As far as I think, it is an O(len(a)) solution.