Splitting list based on missing numbers in a sequence

Question

I am looking for the most pythonic way of splitting a list of numbers into smaller lists based on a number missing in the sequence. For example, if the initial list was:

seq1 = [1, 2, 3, 4, 6, 7, 8, 9, 10]

the function would yield:

[[1, 2, 3, 4], [6, 7, 8, 9, 10]]

or

seq2 = [1, 2, 4, 5, 6, 8, 9, 10]

would result in:

[[1, 2], [4, 5, 6], [8, 9, 10]]

Asked By: billybandicoot

||

Source

Answer 1

Python 3 version of the code from the old Python documentation:

>>> # Find runs of consecutive numbers using groupby.  The key to the solution
>>> # is differencing with a range so that consecutive numbers all appear in
>>> # same group.
>>> 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[0] - i_x[1]):
...     print(list(map(itemgetter(1), g)))
...
[1]
[4, 5, 6]
[10]
[15, 16, 17, 18]
[22]
[25, 26, 27, 28]

The groupby function from the itertools module generates a break every time the key function changes its return value. The trick is that the return value is the number in the list minus the position of the element in the list. This difference changes when there is a gap in the numbers.

The itemgetter function is from the operator module, you’ll have to import this and the itertools module for this example to work.

Alternatively, as a list comprehension:

>>> [map(itemgetter(1), g) for k, g in groupby(enumerate(seq2), lambda i_x: i_x[0] - i_x[1])]
[[1, 2], [4, 5, 6], [8, 9, 10]]

Answered By: Mad Scientist

Answer 2

Another option which doesn’t need itertools etc.:

>>> data = [1, 4, 5, 6, 10, 15, 16, 17, 18, 22, 25, 26, 27, 28]
>>> spl = [0]+[i for i in range(1,len(data)) if data[i]-data[i-1]>1]+[None]
>>> [data[b:e] for (b, e) in [(spl[i-1],spl[i]) for i in range(1,len(spl))]]
... [[1], [4, 5, 6], [10], [15, 16, 17, 18], [22], [25, 26, 27, 28]]

Answered By: muksie

Answer 3

This is a solution that works in Python 3 (based on previous answers that work in python 2 only).

>>> from operator import itemgetter
>>> from itertools import *
>>> groups = []
>>> for k, g in groupby(enumerate(seq2), lambda x: x[0]-x[1]):
>>>     groups.append(list(map(itemgetter(1), g)))
... 
>>> print(groups)
[[1, 2], [4, 5, 6], [8, 9, 10]]

or as a list comprehension

>>> [list(map(itemgetter(1), g)) for k, g in groupby(enumerate(seq2), lambda x: x[0]-x[1])]
[[1, 2], [4, 5, 6], [8, 9, 10]]

Changes were needed because

Removal of tuple parameter unpacking PEP 3113
map returning an iterator instead of a list

Answered By: johnnyB

Answer 4

My way

alist = [1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 15, 16, 17, 18, 20, 21, 22]
newlist = []
start = 0
end = 0
for index,value in enumerate(alist):
    if index < len(alist)-1:
        if alist[index+1]> value+1:
            end = index +1
            newlist.append(alist[start:end])
            start = end
    else:
            newlist.append(alist[start: len(alist)])
print(newlist)

Result

[[1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12], [15, 16, 17, 18], [20, 21, 22]]

Answered By: Dave B

Answer 5

I like this one better because it doesn’t require any extra libraries or special treatment for first case:

a = [1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 15, 16, 17, 18, 20, 21, 22]
b = []
subList = []
prev_n = -1

for n in a:
    if prev_n+1 != n:            # end of previous subList and beginning of next
        if subList:              # if subList already has elements
            b.append(subList)
            subList = []
    subList.append(n)
    prev_n = n

if subList:
    b.append(subList)

print a
print b

Output:

[1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 15, 16, 17, 18, 20, 21, 22]

[[1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12], [15, 16, 17, 18], [20, 21, 22]]

Answered By: Paulo Zappi

Splitting list based on missing numbers in a sequence

Question:

Answers: