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 dis

Answer 1

Here is a quickly written (and extremely ugly) answer:

def test(inArr):
    arr=inArr[:] #copy, unnecessary if we use index in a smart way
    result = []
    while len(arr)>1: #as long as there can be an arithmetic progression
        x=[arr[0],arr[1]] #take first two
        arr=arr[2:] #remove from array
        step=x[1]-x[0]
        while len(arr)>0 and x[1]+step==arr[0]: #check if the next value in array is part of progression too
            x[1]+=step #add it
            arr=arr[1:]
        result.append((x[0],x[1],step)) #append progression to result
    if len(arr)==1:
        result.append(arr[0])
    return result

print test([2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20])

This returns [(2, 8, 2), (12, 17, 1), 20]

Slow, as it copies a list and removes elements from it

It only finds complete progressions, and only in sorted arrays.

In short, it is shitty, but should work ;)

There are other (cooler, more pythonic) ways to do this, for example you could convert your list to a set, keep removing two elements, calculate their arithmetic progression and intersect with the set.

You could also reuse the answer you provided to check for certain step sizes. e.g.:

ranges = []
step_size=2
for key, group in groupby(enumerate(data), lambda (index, item): step_size*index - item):
    group = map(itemgetter(1), group)
    if len(group) > 1:
        ranges.append(xrange(group[0], group[-1]))
    else:
        ranges.append(group[0])

Which finds every group with step size of 2, but only those.

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.

Answer 3

I came across such a case once. Here it goes.

import more_itertools as mit
iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]  # input
x = [list(group) for group in mit.consecutive_groups(iterable)]
output = [(i[0],i[-1]) if len(i)>1 else i[0] for i in x]
print(output)

Answer 4

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]