Formatting consecutive numbers

Question

I\'m trying to format a list of integers with Python and I\'m having a few difficulties achieving what I\'d like.

Input is a sorted list of Integers:

Answer 1

You can use groupby and count from itertools module like this way:

Edit:

Thanks to @asongtoruin comment. For removing duplicates from the input you can use: sorted(set(a)).

from itertools import groupby, count

a = [1, 2, 3, 6, 8, 9]
clustered = [list(v) for _,v in groupby(sorted(a), lambda n, c = count(): n-next(c))]

for k in clustered:
    if len(k) > 1:
        print("{0}-{1}".format(k[0], k[-1]))
    else:
        print("{0}".format(k[0]))

Output:

1-3
6
8-9

Or maybe you can do something like this in order to have a pretty output:

from itertools import groupby, count

a = [1, 2, 3, 6, 8, 9]
clustered = [list(v) for _,v in groupby(sorted(a), lambda n, c = count(): n-next(c))]
out = ", ".join(["{0}-{1}".format(k[0], k[-1]) if len(k) > 1 else "{0}".format(k[0]) for k in clustered ])

print(out)

Output:

1-3, 6, 8-9

Update:

I'm guessing, using itertools modules may confuse many Python's new developers. This is why i decided to rewrite the same solution without importing any package and trying to show what groupby and count are doing behind the scenes:

def count(n=0, step=1):
    """Return an infinite generator of numbers"""
    while True:
        n += step
        yield n


def concat(lst):
    """Group lst elements based on the result of elm - next(_count)"""
    _count, out = count(), {}
    for elm in sorted(lst):
        c = elm - next(_count)
        if c in out:
            out[c].append(elm)
        else:
            out[c] = [elm]
    return out


def pretty_format(dct):
    for _, value in dct.items():
        if len(value) > 1:
            yield '{}-{}'.format(value[0], value[-1])
        else:
            yield '{}'.format(value[0])


lst = [1, 2, 3, 6, 8, 9]
dct = concat(lst)
formatted = list(pretty_format(dct))
print(formatted)

Output:

['1-3', '6', '8-9']