Detecting consecutive integers in a list [duplicate]

Question

I have a list containing data as such:

[1, 2, 3, 4, 7, 8, 10, 11, 12, 13, 14] 

I'd like to print out the ranges of consecutive integers:

1-4, 7-8, 10-14 

Is there a built-in/fast/efficient way of doing this?

Answer 1

From the docs:

>>> from itertools import groupby >>> from operator import itemgetter >>> data = [ 1, 4,5,6, 10, 15,16,17,18, 22, 25,26,27,28] >>> for k, g in groupby(enumerate(data), lambda (i, x): i-x): ...     print map(itemgetter(1), g) ... [1] [4, 5, 6] [10] [15, 16, 17, 18] [22] [25, 26, 27, 28] 

You can adapt this fairly easily to get a printed set of ranges.

Answer 2

A short solution that works without additional imports. It accepts any iterable, sorts unsorted inputs, and removes duplicate items:

def ranges(nums):     nums = sorted(set(nums))     gaps = [[s, e] for s, e in zip(nums, nums[1:]) if s+1 < e]     edges = iter(nums[:1] + sum(gaps, []) + nums[-1:])     return list(zip(edges, edges)) 

Example:

>>> ranges([2, 3, 4, 7, 8, 9, 15]) [(2, 4), (7, 9), (15, 15)]  >>> ranges([-1, 0, 1, 2, 3, 12, 13, 15, 100]) [(-1, 3), (12, 13), (15, 15), (100, 100)]  >>> ranges(range(100)) [(0, 99)]  >>> ranges([0]) [(0, 0)]  >>> ranges([]) [] 

This is the same as @dansalmo's solution which I found amazing, albeit a bit hard to read and apply (as it's not given as a function).

Note that it could easily be modified to spit out "traditional" open ranges [start, end), by e.g. altering the return statement:

    return [(s, e+1) for s, e in zip(edges, edges)]