Combine consecutive numbers into range tuples

Question

I have a list like

[38, 98, 110, 111, 112, 120, 121, 898]

How can I merge consecutive numbers into a pair representing a range?

desired output:

['38',
 '98', 
 '110,112',
 '120,121',
 '898']

@hexparrot he has 110, 111, 112; I think he means to include all numbers in the 'range' until he reaches a number which is no longer consecutive. — 2rs2ts, Jul 01 '13 at 23:14
ah, such a silly oversight. now, i recall a codegolf that does this...brb. — hexparrot, Jul 01 '13 at 23:14

John La Rooy · Accepted Answer · 2013-07-01T23:25:54.110

You can do this nicely with itertools.groupby by noticing that the difference between the items in your list, and a counter are constant while the numbers are consecutive

eg

38 - 0 = 38
98 - 1 = 97
110 - 2 = 108
111 - 3 = 108
112 - 4 = 108
120 - 5 = 115
121 - 6 = 115
898 - 7 = 891

in this example, the 108's are grouped together and the 115's are grouped together. Now some code

>>> from itertools import groupby, count
>>> L = [38, 98, 110, 111, 112, 120, 121, 898]
>>> groups = groupby(L, key=lambda item, c=count():item-next(c))
>>> tmp = [list(g) for k, g in groups]

see what we have so far

>>> tmp
[[38], [98], [110, 111, 112], [120, 121], [898]]

convert to the desired result

>>> [str(x[0]) if len(x) == 1 else "{},{}".format(x[0],x[-1]) for x in tmp]
['38', '98', '110,112', '120,121', '898']

Elazar · Answer 2 · 2013-07-01T23:38:22.547

4

A straight-forward solution:

def ranger(lst):
    fr, to = lst[0], lst[0]
    for x in lst[1:]:
        if x == to+1:
            to = x
        else:
            yield fr, to
            fr, to = x, x
    yield fr, to

res = [','.join(map(str, sorted(set(x)))) for x in ranger(lst)]

edited Jul 01 '13 at 23:38

answered Jul 01 '13 at 23:19

Elazar

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the output doesnt match it makes a duplicate of the single numbers and pairs them – Serial Jul 01 '13 at 23:22

Combine consecutive numbers into range tuples

2 Answers2