Python: Uniquefying a list with a twist

Question

Lets say I have a list:

L = [15,16,57,59,14]

The list contains mesurements, that are not very accurate: that is the real value of an element is +-2 of the recorded value. So 14,15 and 16 can have the same value. What I want to do is to uniquefy that list, taking into account the mesurement errors. The output should therefor be:

l_out = [15,57]

or

l_out = [(14,15,16),(57,59)]

I have no problem producing either result with a for loop. However, I am curious if there could be a more elegant solution. Ideas much appriciated.

Answer 1

As lazyr pointed out in the comments, a similar problem has been posted here. Using the cluster module the solution to my problem would be:

>>> from cluster import *
>>> L = [15,16,57,59,14]
>>> cl = HierarchicalClustering(L, lambda x,y: abs(x-y))
>>> cl.getlevel(2)
[[14, 15, 16], [57, 59]]

or (to get unique list with mean values of each group):

>>> [mean(cluster) for cluster in cl.getlevel(2)]
[15, 58]

Answer 2

If you want standard lib python, itertool's groupby is your friend:

from itertools import groupby

L = [15,16,57,59,14]

# Stash state outside key function. (a little hacky).
# Better way would be to create stateful class with a __call__ key fn.
state = {'group': 0, 'prev': None}
thresh = 2

def _group(cur):
    """Group if within threshold."""
    if state["prev"] is not None and abs(state["prev"] - cur) > thresh:
        state["group"] += 1 # Advance group
    state["prev"] = cur
    return state["group"]

# Group, then drop the group key and inflate the final tuples.
l_out = [tuple(g) for _, g in groupby(sorted(L), key=_group)]

print l_out
# -> [(14, 15, 16), (57, 59)]