How to do an inverse `range`, i.e. create a compact range based on a set of numbers?

Question

Python has a range method, which allows for stuff like:

>>> range(1, 6)
[1, 2, 3, 4, 5]

What I’m looking for is kind of the opposite: take a list of numbers, and return the start and end.

>>> magic([1, 2, 3, 4, 5])
[1, 5] # note: 5, not 6; this differs from `range()`

This is easy enough to do for the above example, but is it possible to allow for gaps or multiple ranges as well, returning the range in a PCRE-like string format? Something like this:

>>> magic([1, 2, 4, 5])
['1-2', '4-5']
>>> magic([1, 2, 3, 4, 5])
['1-5']

Edit: I’m looking for a Python solution, but I welcome working examples in other languages as well. It’s more about figuring out an elegant, efficient algorithm. Bonus question: is there any programming language that has a built-in method for this?

Answer 1

A nice trick to simplify the code is to look at the difference of each element of the sorted list and its index:

a = [4, 2, 1, 5]
a.sort()
print [x - i for i, x in enumerate(a)]

prints

[1, 1, 2, 2]

Each run of the same number corresponds to a run of consecutive numbers in a. We can now use itertools.groupby() to extract these runs. Here's the complete code:

from itertools import groupby

def sub(x):
    return x[1] - x[0]

a = [5, 3, 7, 4, 1, 2, 9, 10]
ranges = []
for k, iterable in groupby(enumerate(sorted(a)), sub):
     rng = list(iterable)
     if len(rng) == 1:
         s = str(rng[0][1])
     else:
         s = "%s-%s" % (rng[0][1], rng[-1][1])
     ranges.append(s)
print ranges

printing

['1-5', '7', '9-10']

Answer 2

Sort numbers, find consecutive ranges (remember RLE compression?).

Something like this:

input = [5,7,9,8,6, 21,20, 3,2,1, 22,23, 50]

output = []
first = last = None # first and last number of current consecutive range
for item in sorted(input):
  if first is None:
    first = last = item # bootstrap
  elif item == last + 1: # consecutive
    last = item # extend the range
  else: # not consecutive
    output.append((first, last)) # pack up the range
    first = last = item
# the last range ended by iteration end
output.append((first, last))

print output

Result: [(1, 3), (5, 9), (20, 23), (50, 50)]. You figure out the rest :)

Answer 3

I thought you might like my generalised clojure solution.

(def r [1 2 3 9 10])

(defn successive? [a b]
  (= a (dec b)))

(defn break-on [pred s]
  (reduce (fn [memo n]
            (if (empty? memo)
              [[n]]
              (if (pred (last (last memo)) n)
                (conj (vec (butlast memo))
                      (conj (last memo) n))
                (conj memo [n]))))
          []
          s))

(break-on successive? r)