python - Merge two or more lists with given order of merging

Question

On start I have 2 lists and 1 list that says in what order I should merge those two lists. For example I have first list equal to [a, b, c] and second list equal to [d, e] and 'merging' list equal to [0, 1, 0, 0, 1].

That means: to make merged list first I need to take element from first list, then second, then first, then first, then second... And I end up with [a, d, b, c, e]. To solve this I just used for loop and two "pointers", but I was wondering if I can do this task more pythonic... I tried to find some functions that could help me, but no real result.

Answer 1

You could create iterators from those lists, loop through the ordering list, and call next on one of the iterators:

i1 = iter(['a', 'b', 'c'])
i2 = iter(['d', 'e'])
# Select the iterator to advance: `i2` if `x` == 1, `i1` otherwise
print([next(i2 if x else i1) for x in [0, 1, 0, 0, 1]]) # ['a', 'd', 'b', 'c', 'e']

It's possible to generalize this solution to any number of lists as shown below

def ordered_merge(lists, selector):
    its = [iter(l) for l in lists]
    for i in selector:
        yield next(its[i])

In [4]: list(ordered_merge([[3, 4], [1, 5], [2, 6]], [1, 2, 0, 0, 1, 2]))
Out[4]: [1, 2, 3, 4, 5, 6]

If the ordering list contains strings, floats, or any other objects that can't be used as list indexes, use a dictionary:

def ordered_merge(mapping, selector):
    its = {k: iter(v) for k, v in mapping.items()}
    for i in selector:
        yield next(its[i])

In [6]: mapping = {'A': [3, 4], 'B': [1, 5], 'C': [2, 6]}

In [7]: list(ordered_merge(mapping, ['B', 'C', 'A', 'A', 'B', 'C']))
Out[7]: [1, 2, 3, 4, 5, 6]

Of course, you can use integers as dictionary keys as well.

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Alternatively, you could remove elements from the left side of each of the original lists one by one and add them to the resulting list. Quick example:

In [8]: A = ['a', 'b', 'c']
   ...: B = ['d', 'e']
   ...: selector = [0, 1, 0, 0, 1]
   ...: 

In [9]: [B.pop(0) if x else A.pop(0) for x in selector]
Out[9]: ['a', 'd', 'b', 'c', 'e']

I would expect the first approach to be more efficient (list.pop(0) is slow).