Identify groups of continuous numbers in a list

EDIT 2: To answer the OP new requirement

ranges = []for key, group in groupby(enumerate(data), lambda (index, item): index - item):    group = map(itemgetter(1), group)    if len(group) > 1:        ranges.append(xrange(group[0], group[-1]))    else:        ranges.append(group[0])

Output:

[xrange(2, 5), xrange(12, 17), 20]

You can replace xrange with range or any other custom class.

Python docs have a very neat recipe for this:

from operator import itemgetterfrom itertools import groupbydata = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]for k, g in groupby(enumerate(data), lambda (i,x):i-x):    print map(itemgetter(1), g)

Output:

[2, 3, 4, 5][12, 13, 14, 15, 16, 17]

If you want to get the exact same output, you can do this:

ranges = []for k, g in groupby(enumerate(data), lambda (i,x):i-x):    group = map(itemgetter(1), g)    ranges.append((group[0], group[-1]))

output:

[(2, 5), (12, 17)]

EDIT: The example is already explained in the documentation but maybe I should explain it more:

The key to the solution is differencing with a range so that consecutive numbers all appear in same group.

If the data was: [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]Then groupby(enumerate(data), lambda (i,x):i-x) is equivalent of the following:

groupby(    [(0, 2), (1, 3), (2, 4), (3, 5), (4, 12),    (5, 13), (6, 14), (7, 15), (8, 16), (9, 17)],    lambda (i,x):i-x)

The lambda function subtracts the element index from the element value. So when you apply the lambda on each item. You'll get the following keys for groupby:

[-2, -2, -2, -2, -8, -8, -8, -8, -8, -8]

groupby groups elements by equal key value, so the first 4 elements will be grouped together and so forth.

I hope this makes it more readable.

python 3 version may be helpful for beginners

import the libraries required first

from itertools import groupbyfrom operator import itemgetterranges =[]for k,g in groupby(enumerate(data),lambda x:x[0]-x[1]):    group = (map(itemgetter(1),g))    group = list(map(int,group))    ranges.append((group[0],group[-1]))

more_itertools.consecutive_groups was added in version 4.0.

Demo

import more_itertools as mititerable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20][list(group) for group in mit.consecutive_groups(iterable)]# [[2, 3, 4, 5], [12, 13, 14, 15, 16, 17], [20]]

Code

Applying this tool, we make a generator function that finds ranges of consecutive numbers.

def find_ranges(iterable):    """Yield range of consecutive numbers."""    for group in mit.consecutive_groups(iterable):        group = list(group)        if len(group) == 1:            yield group[0]        else:            yield group[0], group[-1]iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]list(find_ranges(iterable))# [(2, 5), (12, 17), 20]

The source implementation emulates a classic recipe (as demonstrated by @Nadia Alramli).

Note: more_itertools is a third-party package installable via pip install more_itertools.

The "naive" solution which I find somewhat readable atleast.

x = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 22, 25, 26, 28, 51, 52, 57]def group(L):    first = last = L[0]    for n in L[1:]:        if n - 1 == last: # Part of the group, bump the end            last = n        else: # Not part of the group, yield current group and start a new            yield first, last            first = last = n    yield first, last # Yield the last group>>>print list(group(x))[(2, 5), (12, 17), (22, 22), (25, 26), (28, 28), (51, 52), (57, 57)]