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.