Finding groups of increasing numbers in a list

Question

The aim is to find groups of increasing/monotonic numbers given a list of integers. Each item in the resulting group must be of a +1 increment from the previous item

Answer 1

def igroups(L):
    R=[[]]
    [R[-1].append(L[i]) for i in range(len(L)) if (L[i-1]+1==L[i] if L[i-1]+1==L[i] else R.append([L[i]]))]
    return [P for P in R if len(P)>1]


tests=[[],
    [0, 0, 0],
    [7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5],
    [8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6],
    [9, 1, 2, 3, 1, 1, 2, 3, 5],
    [4,3,2,1,1,2,3,3,4,3],
    [1, 4, 3],
    [1],
    [1,2],
    [2,1]
    ]
for L in tests:
    print(L)
    print(igroups(L))
    print("-"*10)

outputting the following:

[]
[]
----------
[0, 0, 0]
[]
----------
[7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5]
[[7, 8, 9, 10], [0, 1, 2, 3, 4, 5]]
----------
[8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6]
[[8, 9, 10, 11], [1, 2, 3, 4, 5, 6]]
----------
[9, 1, 2, 3, 1, 1, 2, 3, 5]
[[1, 2, 3], [1, 2, 3]]
----------
[4, 3, 2, 1, 1, 2, 3, 3, 4, 3]
[[1, 2, 3], [3, 4]]
----------
[1, 4, 3]
[]
----------
[1]
[]
----------
[1, 2]
[[1, 2]]
----------
[2, 1]
[]
----------

EDIT My first attemp using itertools.groupby was a fail, sorry for that.

Answer 2

If two consecutive numbers are increasing by one I form a list (group) of tuples of those numbers.

When non-increasing and if the list (group) is non-empty, I unpack it and zip again to rebuild the pair of sequence which were broken by the zip. I use set comprehension to eliminate duplicate numbers.

  def extract_increasing_groups(seq):
    seq = tuple(seq)

    def is_increasing(a,b):
        return a + 1 == b

    def unzip(seq):
        return tuple(sorted({ y for x in zip(*seq) for y in x}))

    group = []
    for a,b in zip(seq[:-1],seq[1:]):
        if is_increasing(a,b):
            group.append((a,b))
        elif group:
            yield unzip(group)
            group = []

    if group:
        yield unzip(group)

if __name__ == '__main__':

    x = [17, 17, 19, 20, 21, 22, 0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12,

         13, 14, 14, 14, 28, 29, 30, 31, 32, 33, 34, 35, 36, 40]

    for group in extract_increasing_groups(x):
        print(group)

Simpler one using set;

from collections import namedtuple
from itertools import islice, tee

def extract_increasing_groups(iterable):

    iter1, iter2 = tee(iterable)
    iter2 = islice(iter2,1,None)

    is_increasing = lambda a,b: a + 1 == b
    Igroup = namedtuple('Igroup','group, len')

    group = set()
    for pair in zip(iter1, iter2):
        if is_increasing(*pair):
            group.update(pair)
        elif group:
            yield Igroup(tuple(sorted(group)),len(group))
            group = set()

    if group:
        yield Igroup(tuple(sorted(group)), len(group))


if __name__ == '__main__':

    x = [17, 17, 19, 20, 21, 22, 0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 14, 14, 28, 29, 30, 31, 32, 33, 34, 35, 36, 40]
    total = 0
    for group in extract_increasing_groups(x):
        total += group.len
        print('Group: {}\nLength: {}'.format(group.group, group.len))
    print('Total: {}'.format(total))

Answer 3

A couple of different ways using itertools and numpy:

from itertools import groupby, tee, cycle

x = [17, 17, 19, 20, 21, 22, 0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 14, 14, 28, 29, 30, 31, 32, 33, 34, 35,
     36, 1, 2, 3, 4,34,54]


def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, key=lambda j: j + 1 == next(x2))
    for k, v in grps:
        if k:
            yield tuple(v) + (next((next(grps)[1])),)


print(list(sequences(x)))

[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), (28, 29, 30, 31, 32, 33, 34, 35, 36), (1, 2, 3, 4)]

Or using python3 and yield from:

def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, key=lambda j: j + 1 == next(x2))
    yield from (tuple(v) + (next((next(grps)[1])),) for k,v in grps if k)

print(list(sequences(x)))

Using a variation of my answer here with numpy.split :

out = [tuple(arr) for arr in np.split(x, np.where(np.diff(x) != 1)[0] + 1) if arr.size > 1]

print(out)

[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), (28, 29, 30, 31, 32, 33, 34, 35, 36), (1, 2, 3, 4)]

And similar to ekhumoro's answer:

def sequences(x):
    it = iter(x)
    prev, temp = next(it), []
    while prev is not None:
        start = next(it, None)
        if prev + 1 == start:
            temp.append(prev)
        elif temp:
            yield tuple(temp + [prev])
            temp = []
        prev = start

To get the length and the tuple:

def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, key=lambda j: j + 1 == next(x2))
    for k, v in grps:
        if k:
            t = tuple(v) + (next(next(grps)[1]),)
            yield t, len(t)


def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, lambda j: j + 1 == next(x2))
    yield from ((t, len(t)) for t in (tuple(v) + (next(next(grps)[1]),)
                                      for k, v in grps if k))



def sequences(x):
        it = iter(x)
        prev, temp = next(it), []
        while prev is not None:
            start = next(it, None)
            if prev + 1 == start:
                temp.append(prev)
            elif temp:
                yield tuple(temp + [prev]), len(temp) + 1
                temp = []
            prev = start

Output will be the same for all three:

[((19, 20, 21, 22), 4), ((0, 1, 2), 3), ((4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), 11)
, ((28, 29, 30, 31, 32, 33, 34, 35, 36), 9), ((1, 2, 3, 4), 4)]

Answer 4

With itertools.groupby, the problem of partionning a list of integers L in sublists of adjacent and increasing consecutive items from L can be done with a one-liner. Nevertheless I don't know how pythonic it can be considered ;)

Here is the code with some simple tests:

[EDIT : now subsequences are increasing by 1, I missed this point the first time.]

from itertools import groupby

def f(i):
    return  L[i-1]+1==L[i]


def igroups(L):
    return [[L[I[0]-1]]+[L[i] for i in I] for I in [I for (v,I) in [(k,[i for i in list(g)]) for (k, g) in groupby(range(1, len(L)), f)] if v]]

outputting:

tests=[
    [0, 0, 0, 0],
    [7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5],
    [8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6],
    [9, 1, 2, 3, 1, 1, 2, 3, 5],
    [4,3,2,1,1,2,3,3,4,3],
    [1, 4, 3],
    [1],
    [1,2, 2],
    [2,1],
    [0, 0, 0, 0, 2, 5, 5, 8],
    ]
for L in tests:
    print(L)
    print(igroups(L))
    print('-'*10)


[0, 0, 0, 0]
[]
----------
[7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5]
[[7, 8, 9, 10], [0, 1, 2, 3, 4, 5]]
----------
[8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6]
[[8, 9, 10, 11], [1, 2, 3, 4, 5, 6]]
----------
[9, 1, 2, 3, 1, 1, 2, 3, 5]
[[1, 2, 3], [1, 2, 3]]
----------
[4, 3, 2, 1, 1, 2, 3, 3, 4, 3]
[[1, 2, 3], [3, 4]]
----------
[1, 4, 3]
[]
----------
[1]
[]
----------
[1, 2, 2]
[[1, 2]]
----------
[2, 1]
[]
----------
[0, 0, 0, 0, 2, 5, 5, 8]
[]
----------

Some explanation. If you "unroll" the code, the logic is more apparant :

from itertools import groupby

def f(i):
    return L[i]==L[i-1]+1

def igroups(L):
    monotonic_states = [(k,list(g)) for (k, g) in groupby(range(1, len(L)), f)]
    increasing_items_indices = [I for (v,I) in monotonic_states if v]
    print("\nincreasing_items_indices ->", increasing_items_indices, '\n')
    full_increasing_items= [[L[I[0]-1]]+[L[i] for i in I] for I in increasing_items_indices]
    return full_increasing_items

L= [2, 8, 4, 5, 6, 7, 8, 5, 9, 10, 11, 12, 25, 26, 27, 42, 41]
print(L)
print(igroups(L))

outputting :

[2, 8, 4, 5, 6, 7, 8, 5, 9, 10, 11, 12, 25, 26, 27, 42, 41]

increasing_items_indices -> [[3, 4, 5, 6], [9, 10, 11], [13, 14]]

[[4, 5, 6, 7, 8], [9, 10, 11, 12], [25, 26, 27]]

We need a key function f that compares an item with the preceding one in the given list. Now, the important point is that the groupby function with the key function f provides a tuple (k, S) where S represents adjacent indices from the initial list and where the state of f is constant, the state being given by the value of k: if k is True, then S represents increasing (by 1) items indices else non-increasing items indices. (in fact, as the example above shows, the list S is incomplete and lacks the first item).

I also made some random tests with one million items lists : igroups function returns always the correct response but is 4 times slower than a naive implementation! Simpler is easier and faster ;)

Thanks alvas for your question, it gives me a lot of fun!

Answer 5

I think this works. It's not fancy but it's simple. It constructs a start list sl and an end list el, which should always be the same length, then uses them to index into x:

def igroups(x):
    sl = [i for i in range(len(x)-1)
          if (x == 0 or x[i] != x[i-1]+1) and x[i+1] == x[i]+1]

    el = [i for i in range(1, len(x))
          if x[i] == x[i-1]+1 and (i == len(x)-1 or x[i+1] != x[i]+1)]

    return [x[sl[i]:el[i]+1] for i in range(len(sl))]

Answer 6

EDIT:

Here's a code-golf solution (142 characters):

def f(x):s=[0]+[i for i in range(1,len(x)) if x[i]!=x[i-1]+1]+[len(x)];return [x[j:k] for j,k in [s[i:i+2] for i in range(len(s)-1)] if k-j>1]

Expanded version:

def igroups(x):
    s = [0] + [i for i in range(1, len(x)) if x[i] != x[i-1] + 1] + [len(x)]
    return [x[j:k] for j, k in [s[i:i+2] for i in range(len(s)-1)] if k - j > 1]

Commented version:

def igroups(x):
    # find the boundaries where numbers are not consecutive
    boundaries = [i for i in range(1, len(x)) if x[i] != x[i-1] + 1]
    # add the start and end boundaries
    boundaries = [0] + boundaries + [len(x)]
    # take the boundaries as pairwise slices
    slices = [boundaries[i:i + 2] for i in range(len(boundaries) - 1)]
    # extract all sequences with length greater than one
    return [x[start:end] for start, end in slices if end - start > 1]

Original solution:

Not sure whether this counts as "pythonic" or "not too verbose":

def igroups(iterable):
    items = iter(iterable)
    a, b = None, next(items, None)
    result = [b]
    while b is not None:
        a, b = b, next(items, None)
        if b is not None and a + 1 == b:
            result.append(b)
        else:
            if len(result) > 1:
                yield tuple(result)
            result = [b]

print(list(igroups([])))
print(list(igroups([0, 0, 0])))
print(list(igroups([7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5])))
print(list(igroups([8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6])))
print(list(igroups([9, 1, 2, 3, 1, 1, 2, 3, 5])))

Output:

[]
[]
[(7, 8, 9, 10), (0, 1, 2, 3, 4, 5)]
[(8, 9, 10, 11), (1, 2, 3, 4, 5, 6)]
[(1, 2, 3), (1, 2, 3)]