Detecting Consecutive Integers in a List

Detecting consecutive integers in a list

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.

Find consecutive integers in a list

This is because you only check the next number. When you want the second number (like 9 or 3), you have to include a check for the previous number too. This will make the if a bit longer, but it'll work.

num=[8,9,4,1,2,3]

for i in range(len(num)):
    if (
        (  # check for the next number
            i + 1 != len (num) and  # don't check the end of the list
            num[i]+1==num[i+1] 
        ) or (  # check for the previous number
            i != 0 and  # don't check before the list
            num [i-1] == num [i] - 1
        )
    ): print('Con',num[i])

Also, I had to remove the -1 in your range, because I already do a manual check, and as pointed out, this prvented 3 from being shown.

Identify groups of continuous numbers in a list

more_itertools.consecutive_groups was added in version 4.0.

Demo

import more_itertools as mit

iterable = [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.

Detecting Consecutive Integers in a list [Python 3]

Turns out I didn't read the original post enough, a solution for python 3 was posted in a comment to the solution:

"Change the lambda to lambda ix : ix[0] - ix[1] and it works in both Python 3 and Python 2 (well, not counting the print statement). – Kevin May 20 '15 at 4:17"

Detecting consecutive integers and collapse to string

int[] numbers = { 1, 2, 3, 4, 7, 8, 10, 11, 12, 13, 14 };

return string.Join(", ",
    numbers
        .Select((n, i) => new { value = n, group = n - i })
        .GroupBy(o => o.group)
        .Select(g => g.First().value + "-" + g.Last().value)
);

Find pairs of consecutive integers in array PYTHON

In response to the more recent clarifications:

def find_pairs(xs):
    result = []
    i = 0
    
    while i < len(xs) - 1:
        if xs[i] == xs[i + 1]:
            result.append(xs[i])
            i += 1
        i += 1

    return result

Testing:

>>> xs = [3, 4, 4, 4, 5, 6, 6, 5, 4, 4]
>>> find_pairs(xs)
[4, 6, 4]

---

Update: Minor off-by-one bug fix.

Identify if list has consecutive elements that are equal

You can use itertools.groupby() and a generator expression within any()
^*:

>>> from itertools import groupby
>>> any(sum(1 for _ in g) > 1 for _, g in groupby(lst))
True

Or as a more Pythonic way you can use zip(), in order to check if at least there are two equal consecutive items in your list:

>>> any(i==j for i,j in zip(lst, lst[1:])) # In python-2.x,in order to avoid creating a 'list' of all pairs instead of an iterator use itertools.izip()
True

Note: The first approach is good when you want to check if there are more than 2 consecutive equal items, otherwise, in this case the second one takes the cake!

---

^{* Using sum(1 for _ in g) instead of len(list(g)) is very optimized in terms of memory use (not reading the whole list in memory at once) but the latter is slightly faster.}

Find consecutive strings representing numbers in a Python list

Try something like this:

l = ['1133', '1300', '1418', '1443', '1473', '1600', '1601', '1988', '1990', '1991', '1992', '1993', '1994', '1995', '1996', '1997', '1998', '2000', '2003', '2004', '2005', '2006', '2007', '2008', '2009', '2010', '2012', '2013', '2014', '2015', '2153', '2600', '3000', '3714', '3785', '3896', '3995', '4001', '4436', '5094', '5346', '8000']

size = len(l)
i = 0
while i < size:
   j = i
   while j + 1 < size and int(l[j]) + 1 == int(l[j + 1]):
      j += 1
   if j != i:
       print (l[i:j+1])
   i = j + 1

Output:

['1600', '1601']
['1990', '1991', '1992', '1993', '1994', '1995', '1996', '1997', '1998']
['2003', '2004', '2005', '2006', '2007', '2008', '2009', '2010']
['2012', '2013', '2014', '2015']

---

Related Topics