Combine two or more lists with the specified merge order

Question

At the beginning, I have 2 lists and 1 list, which says in what order I should combine these two lists. For example, I have the first list equal to [a, b, c] , and the second list is [d, e] and 'merging' equal to [0, 1, 0, 0, 1] .

This means: first create a combined list, I need to take an element from the first list, then the second, then first, then first, then the second ... And I get [a, d, b, c, e] . To solve this problem, I just used two “pointers” for the loop, but I was wondering if I could make this task more pythonic ... I tried to find some functions that could help me, but no real result.

Answer 1

You can create iterators from these lists, iterate over the list of orders 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'] 

You can 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 order list contains strings, floats or any other objects that cannot be used as list indices, use the 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 also use integers as dictionary keys.

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Alternatively, you can remove the items on the left side of each of the source 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 expect the first approach will be more efficient ( list.pop(0) slow).

Answer 2

How about this

 list1 = ['a', 'b', 'c'] list2 = ['d', 'e'] options = [0,1,0,0,1] list1_iterator = iter(list1) list2_iterator = iter(list2) new_list = [next(list2_iterator) if option else next(list1_iterator) for option in options] print(new_list) # Output ['a', 'd', 'b', 'c', 'e']