Merge elements in list based on given indices

Question

I want to merge element in the list based on given start and stop index of tuple (non-overlap for tuple). I'll leave the indices that don't mention as it is. This is my example

ls = ['1', '2', '3', '4', '5', '6', '7']
merge = [(1, 3), (5, 7)]

Here, I want to merge index from [1:3] together and [5:7] together so the output should look something like following

['1', '23', '4', '5', '67']

I tried to loop using range(len(ls)) but it doesn't seem to be the right way to tackle this problem. Let me know if someone has simple way to solve this problem.

Are the merge tuples guaranteed to not overlap? Are they guaranteed to be in any ordering? — Matthew Cole, Apr 21 '17 at 19:09
Yeah, no overlap! I'll fix the question right now. Also, I did make a typo. — titipata, Apr 21 '17 at 19:29
What about ordering of the tuples in `merge`? See my comment on Vadim's solution below. — Matthew Cole, Apr 21 '17 at 19:30
It would be best if we can consider any ordering. However, i can sort the the `merge` by start position before hand in this case. — titipata, Apr 21 '17 at 19:37

score 7 · Answer 1 · answered Apr 21 '17 at 19:23

7

A quick-and-dirty solution would be:

ls = ['1', '2', '3', '4', '5', '6', '7']
merge = [(1, 3), (5, 7)]

result = []
index = 0

for start, end in merge:
    result += ls[index:start]
    result.append("".join(ls[start:end]))
    index = end

print result # ['1', '23', '4', '5', '67']

answered Apr 21 '17 at 19:23

Vadim Landa

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Ah, yes, this is certainly a reasonable solution, especially compared to the monstrosity I had posted originally. – juanpa.arrivillaga Apr 21 '17 at 19:25
1

This solution only works if `merge` is ordered and there are no overlaps. (Consider two such examples: `merge = [(5,7),(1,3)]` and `merge = [(1,3),(2,4)]`). I asked a question about this on the OP. – Matthew Cole Apr 21 '17 at 19:29
Thanks @Vadim! I can't accept both solution at once so I gave you thumbs up instead! – titipata Apr 21 '17 at 19:47
No worries, the other solution is definitely nice. Cheers! – Vadim Landa Apr 21 '17 at 19:50

RomanPerekhrest · Accepted Answer · 2017-04-21T19:40:03.470

3

Short "trick" with reversed merge list:

ls = ['1', '2', '3', '4', '5', '6', '7']
merge = [(1, 3), (5, 7)]

for t in merge[::-1]:
    merged = ''.join(ls[t[0]:t[1]])  # merging values within a range
    ls[t[0]:t[1]] = [merged]         # slice replacement

print(ls)

The output:

['1', '23', '4', '5', '67']

edited Apr 21 '17 at 19:40

answered Apr 21 '17 at 19:31

RomanPerekhrest

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1

This solution has the added benefit that it requires no ordering of the tuples in the `merge` list to work correctly. – Matthew Cole Apr 21 '17 at 19:33
@RomanPerekhrest, thanks Roman. Both solutions (from you and Vadim) are great. It just opens my way of solving this kind of problem :) – titipata Apr 21 '17 at 20:06

score 1 · Answer 3 · answered Apr 21 '17 at 20:47

For the fun of it, because I've been learning Haskell, a recursive solution:

def recursive(ls, merge):
    if merge == []:
        return ls
    else:
        x, xs = merge[0], merge[1:]
        return ls[:x[0]] + [''.join(ls[x[0]:x[1]])] + recursive(ls, xs)[x[1]:]

Only works if there are no overlapping intervals, however.

Merge elements in list based on given indices

3 Answers3

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