converting a list of integers into range in python

Question

Is there something existing in python that can convert an increasing list of integers into a range list

E.g. given the set {0, 1, 2, 3, 4, 7, 8, 9, 11} I want to get { {0,4}, {7,9}, {11,11} }.

I can write a program to do this, but want to know if there is an inbuilt function in python

Answer 1

Using itertools.groupby() produces a concise but tricky implementation:

import itertools

def ranges(i):
    for a, b in itertools.groupby(enumerate(i), lambda (x, y): y - x):
        b = list(b)
        yield b[0][1], b[-1][1]

print list(ranges([0, 1, 2, 3, 4, 7, 8, 9, 11]))

Output:

[(0, 4), (7, 9), (11, 11)]

Answer 2

You can use a list comprehension with a generator expression and a combination of enumerate() and itertools.groupby():

>>> import itertools
>>> l = [0, 1, 2, 3, 4, 7, 8, 9, 11]
>>> [[t[0][1], t[-1][1]] for t in
... (tuple(g[1]) for g in itertools.groupby(enumerate(l), lambda (i, x): i - x))]
[[0, 4], [7, 9], [11, 11]]

---

First, enumerate() will build tuples from the list items and their respective index:

>>> [t for t in enumerate(l)]
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 7), (6, 8), (7, 9), (8, 11)]

Then groupby() will group those tuples using the difference between their index and their value (which will be equal for consecutive values):

>>> [tuple(g[1]) for g in itertools.groupby(enumerate(l), lambda (i, x): i - x)]
[((0, 0), (1, 1), (2, 2), (3, 3), (4, 4)), ((5, 7), (6, 8), (7, 9)), ((8, 11),)]

From there, we only need to build lists from the values of the first and last tuples of each group (which will be the same if the group only contains one item).

You can also use [(t[0][1], t[-1][1]) ...] to build a list of range tuples instead of nested lists, or even ((t[0][1], t[-1][1]) ...) to turn the whole expression into a iterable generator that will lazily build the range tuples on the fly.

Answer 3

This is an improvement over the very elegant @juanchopanza answer. This one covers non-unique and non-sorted input and is python3 compatible too:

import itertools

def to_ranges(iterable):
    iterable = sorted(set(iterable))
    for key, group in itertools.groupby(enumerate(iterable),
                                        lambda t: t[1] - t[0]):
        group = list(group)
        yield group[0][1], group[-1][1]

Example:

>>> x
[44, 45, 2, 56, 23, 11, 3, 4, 7, 9, 1, 2, 2, 11, 12, 13, 45]

>>> print( list(to_ranges(x))) 
[(1, 4), (7, 7), (9, 9), (11, 13), (23, 23), (44, 45), (56, 56)]

Answer 4

This generator:

def ranges(p):
    q = sorted(p)
    i = 0
    for j in xrange(1,len(q)):
        if q[j] > 1+q[j-1]:
            yield (q[i],q[j-1])
            i = j
    yield (q[i], q[-1])

sample = [0, 1, 2, 3, 4, 7, 8, 9, 11]
print list(ranges(sample))
print list(ranges(reversed(sample)))
print list(ranges([1]))
print list(ranges([2,3,4]))
print list(ranges([0,2,3,4]))
print list(ranges(5*[1]))

Produces these results:

[(0, 4), (7, 9), (11, 11)]
[(0, 4), (7, 9), (11, 11)]
[(1, 1)]
[(2, 4)]
[(0, 0), (2, 4)]
[(1, 1)]

Note that runs of repeated numbers get compressed. I don't know if that's what you want. If not, change the > to a !=.

I understand your question. I looked into itertools and tried to think of a solution that could be done in a couple of lines of Python, which would have qualified as "almost a built in", but I couldn't come up with anything.

Answer 5

Generating range pairs:

def ranges(lst):
    s = e = None
    r = []
    for i in sorted(lst):
        if s is None:
            s = e = i
        elif i == e or i == e + 1:
            e = i
        else:
            r.append((s, e))
            s = e = i
    if s is not None:
        r.append((s, e))
    return r

Example:

>>> lst = [1, 5, 6, 7, 12, 15, 16, 17, 18, 30]
>>> print repr(ranges(lst))
[(1, 1), (5, 7), (12, 12), (15, 18), (30, 30)]

As a generator:

def gen_ranges(lst):
    s = e = None
    for i in sorted(lst):
        if s is None:
            s = e = i
        elif i == e or i == e + 1:
            e = i
        else:
            yield (s, e)
            s = e = i
    if s is not None:
        yield (s, e)

Example:

>>> lst = [1, 5, 6, 7, 12, 15, 16, 17, 18, 30]
>>> print repr(','.join(['%d' % s if s == e else '%d-%d' % (s, e) for (s, e) in gen_ranges(lst)]))
'1,5-7,12,15-18,30'

Answer 6

Nothing built-in, or in any libraries that I know of. Not very helpful, I know, but I've never come across anything like what you want.

Here are some ideas for your program atleast (in C++, but it can give you some other ideas):

Converting sets of integers into ranges

Answer 7

In the case there is no such feature in python, here is an implementation

p = []
last = -2                                                            
start = -1

for item in list:
    if item != last+1:                        
        if start != -1:
            p.append([start, last])
        start = item
    last = item

p.append([start, last])

Answer 8

Put it shorter:

ranges=lambda l:map(lambda x:(x[0][1],x[-1][1]),map(lambda (x,y):list(y),itertools.groupby(enumerate(l),lambda (x,y):x-y)))

Answer 9

I think the other answers are hard to understand, and probably inefficient. Hope this is easier and faster.

def ranges(ints):
    ints = sorted(set(ints))
    range_start = previous_number = ints[0]
    for number in ints[1:]:
        if number == previous_number + 1:
            previous_number = number
        else:
            yield range_start, previous_number
            range_start = previous_number = number
    yield range_start, previous_number