convert list of sequential number into interval or range

Question

I have a list like this:

a = [1, 2, 3, 4, 5, 8, 9, 10, 11, 15, 16, 17]

I would like to get the result as:

b = [1-5, 8-11, 15-17]

since in my list a I have sequential number from 1 to 5, 8 to 11 and then 15 to 17.

How should I do this in Python?

What have you tried so far. Give us your [Minimal, Complete, and Verifiable Code](https://stackoverflow.com/help/mcve) — Partho63, Feb 04 '19 at 05:21

score 4 · Accepted Answer · answered Feb 04 '19 at 05:27

Using itertools.groupby as described here but with a little change by @luca:

import itertools

def to_ranges(iterable):
    iterable = sorted(set(iterable))
    for key, group in itertools.groupby(enumerate(iterable),
                                        lambda t: t[1] - t[0]):
        group = list(group)
        yield group[0][1], group[-1][1]


x = [1, 2, 3, 4, 5, 8, 9, 10, 11, 15, 16, 17]
print( list(to_ranges(x)))

OUTPUT:

[(1, 5), (8, 11), (15, 17)]

score 0 · Answer 2 · answered Feb 04 '19 at 05:42

    def range_extract(lst):
            'Yield 2-tuple ranges or 1-tuple single elements from list of increasing ints'
            lenlst = len(lst)
            i = 0
            while i< lenlst:
                low = lst[i]
                while i <lenlst-1 and lst[i]+1 == lst[i+1]: i +=1
                hi = lst[i]
                if   hi - low >= 2:
                    yield (low, hi)
                elif hi - low == 1:
                    yield (low,)
                    yield (hi,)
                else:
                    yield (low,)
                i += 1

    def printr(ranges):
            print( ','.join( (('%i-%i' % r) if len(r) == 2 else '%i' % r)
                             for r in ranges ) )

    if __name__ == '__main__':
            lst = [1, 2, 3, 4, 5, 8, 9, 10, 11, 15, 16, 17]
            x=range_extract(lst)
            print(list(x))

    # output : [(1, 5), (8, 11), (15, 17)]

convert list of sequential number into interval or range

2 Answers2