Finding clusters of numbers in a list

Not strictly necessary if your list is small, but I'd probably approach this in a "stream-processing" fashion: define a generator that takes your input iterable, and yields the elements grouped into runs of numbers differing by <= 15. Then you can use that to generate your dictionary easily.

def grouper(iterable):
    prev = None
    group = []
    for item in iterable:
        if prev is None or item - prev <= 15:
            group.append(item)
        else:
            yield group
            group = [item]
        prev = item
    if group:
        yield group

numbers = [123, 124, 128, 160, 167, 213, 215, 230, 245, 255, 257, 400, 401, 402, 430]
dict(enumerate(grouper(numbers), 1))

prints:

{1: [123, 124, 128],
 2: [160, 167],
 3: [213, 215, 230, 245, 255, 257],
 4: [400, 401, 402],
 5: [430]}

As a bonus, this lets you even group your runs for potentially-infinite lists (as long as they're sorted, of course). You could also stick the index generation part into the generator itself (instead of using enumerate) as a minor enhancement.

import itertools
import numpy as np

numbers = np.array([123, 124, 128, 160, 167, 213, 215, 230, 245, 255, 257, 400, 401, 402, 430])
nd = [0] + list(np.where(np.diff(numbers) > 15)[0] + 1) + [len(numbers)]

a, b = itertools.tee(nd)
next(b, None)
res = {}
for j, (f, b) in enumerate(itertools.izip(a, b)):
    res[j] = numbers[f:b]

If you can use itertools and numpy. Adapted pairwise for the iterator tricks. The +1 is needed to shift the index, adding the 0 and len(numbers) onto the list makes sure the first and last entries are included correctly.

You can obviously do this with out itertools, but I like tee.