Identify groups of varying continuous numbers in a list

Question

In this other SO post, a Python user asked how to group continuous numbers such that any sequences could just be represented by its start/end and any stragglers would be displayed as single items. The accepted answer works brilliantly for continuous sequences.

I need to be able to adapt a similar solution but for a sequence of numbers that have potentially (not always) varying increments. Ideally, how I represent that will also include the increment (so they'll know if it was every 3, 4, 5, nth)

Referencing the original question, the user asked for the following input/output

[2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]  # input
[(2,5), (12,17), 20]

What I would like is the following (Note: I wrote a tuple as the output for clarity but xrange would be preferred using its step variable):

[2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]  # input
[(2,5,1), (12,17,1), 20]  # note, the last element in the tuple would be the step value

And it could also handle the following input

[2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20]  # input
[(2,8,2), (12,17,1), 20]  # note, the last element in the tuple would be the increment

I know that xrange() supports a step so it may be possible to even use a variant of the other user's answer. I tried making some edits based on what they wrote in the explanation but I wasn't able to get the result I was looking for.

For anyone that doesn't want to click the original link, the code that was originally posted by Nadia Alramli is:

ranges = []
for key, group in groupby(enumerate(data), lambda (index, item): index - item):
    group = map(itemgetter(1), group)
    if len(group) > 1:
        ranges.append(xrange(group[0], group[-1]))
    else:
        ranges.append(group[0])

Answer 1

The itertools pairwise recipe is one way to solve the problem. Applied with itertools.groupby, groups of pairs whose mathematical difference are equivalent can be created. The first and last items of each group are then selected for multi-item groups or the last item is selected for singleton groups:

from itertools import groupby, tee, izip


def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

def grouper(lst):
    result = []
    for k, g in groupby(pairwise(lst), key=lambda x: x[1] - x[0]):
        g  = list(g)
        if len(g) > 1:
            try:
                if g[0][0] == result[-1]:
                    del result[-1]
                elif g[0][0] == result[-1][1]:
                    g = g[1:] # patch for duplicate start and/or end
            except (IndexError, TypeError):
                pass
            result.append((g[0][0], g[-1][-1], k))
        else:
            result.append(g[0][-1]) if result else result.append(g[0])
    return result

---

Trial: input -> grouper(lst) -> output

Input: [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
Output: [(2, 5, 1), (12, 17, 1), 20]

Input: [2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20]
Output: [(2, 8, 2), (12, 17, 1), 20]

Input: [2, 4, 6, 8, 12, 12.4, 12.9, 13, 14, 15, 16, 17, 20]
Output: [(2, 8, 2), 12, 12.4, 12.9, (13, 17, 1), 20] # 12 does not appear in the second group

---

Update: (patch for duplicate start and/or end values)

s1 = [i + 10 for i in xrange(0, 11, 2)]; s2 = [30]; s3 = [i + 40 for i in xrange(45)]

Input: s1+s2+s3
Output: [(10, 20, 2), (30, 40, 10), (41, 84, 1)]

# to make 30 appear as an entry instead of a group change main if condition to len(g) > 2
Input: s1+s2+s3
Output: [(10, 20, 2), 30, (41, 84, 1)]

Input: [2, 4, 6, 8, 10, 12, 13, 14, 15, 16, 17, 20]
Output: [(2, 12, 2), (13, 17, 1), 20]

Answer 2

You can create an iterator to help grouping and try to pull the next element from the following group which will be the end of the previous group:

def ranges(lst):
    it = iter(lst)
    next(it)  # move to second element for comparison
    grps = groupby(lst, key=lambda x: (x - next(it, -float("inf"))))
    for k, v in grps:
        i = next(v)
        try:
            step = next(v) - i  # catches single element v or gives us a step
            nxt = list(next(grps)[1])
            yield xrange(i, nxt.pop(0), step)
            # outliers or another group
            if nxt:
                yield nxt[0] if len(nxt) == 1 else xrange(nxt[0], next(next(grps)[1]), nxt[1] - nxt[0])
        except StopIteration:
            yield i  # no seq

which give you:

In [2]: l1 = [2, 3, 4, 5, 8, 10, 12, 14, 13, 14, 15, 16, 17, 20, 21]

In [3]: l2 = [2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20]

In [4]: l3 = [13, 14, 15, 16, 17, 18]

In [5]: s1 = [i + 10 for i in xrange(0, 11, 2)]

In [6]: s2 = [30]

In [7]: s3 = [i + 40 for i in xrange(45)]

In [8]: l4 = s1 + s2 + s3

In [9]: l5 = [1, 2, 5, 6, 9, 10]

In [10]: l6 = {1, 2, 3, 5, 6, 9, 10, 13, 19, 21, 22, 23, 24}

In [11]: 

In [11]: for l in (l1, l2, l3, l4, l5, l6):
   ....:         print(list(ranges(l)))
   ....:     
[xrange(2, 5), xrange(8, 14, 2), xrange(13, 17), 20, 21]
[xrange(2, 8, 2), xrange(12, 17), 20]
[xrange(13, 18)]
[xrange(10, 20, 2), 30, xrange(40, 84)]
[1, 2, 5, 6, 9, 10]
[xrange(1, 3), 5, 6, 9, 10, 13, 19, xrange(21, 24)]

When the step is 1 it is not included in the xrange output.