Find all consecutive sub-sequences of length n in a sequence

Question

I want to find all consecutive sub-sequences of length n in a sequence.

E.g. say n was 3 and the sequence was:

[0,1,7,3,4,5,10]

I w

Answer 1

>>> x = [0,1,7,3,4,5,10]
>>> [x[n:n+3] for n in range(len(x)-2)]
[[0, 1, 7], [1, 7, 3], [7, 3, 4], [3, 4, 5], [4, 5, 10]]

Answer 2

The following might probably be suit for you:

def subseqs(xs, n):
  all_seqs = (xs[i:j+1] for i, _ in enumerate(xs) for j, _ in enumerate(xs))
  return filter(lambda seq: len(seq) == n, all_seqs)

>>> xs = [1, 2, 3, 4, 5, 6] # can be also range(1, 7) or list(range(1, 7)) 
>>> list(subseqs(xs, 3))
[[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6]]

Or simply, for getting all of the sequences of a list named 'xs':

[xs[i:j+1] for i, _ in enumerate(xs) for j, _ in enumerate(xs)]

For getting the sequences of a list named 'xs' that are only from length n:

[xs[i:j+1] for i, _ in enumerate(xs) for j, _ in enumerate(xs) if len(xs[i:j+1]) == n]

Answer 3

def subseqs(seq, length):
    for i in xrange(len(seq) - length + 1):
        yield seq[i:i+length]

Use it ike this:

>>> for each in subseqs("hello", 3):
...     print each
...
hel
ell
llo

Of course it works also with lists:

>>> list(subseqs([1, 2, 3, 4, 5, 6, 7, 8], 3))
[[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [5, 6, 7], [6, 7, 8]]

Answer 4

>>> x = [0,1,7,3,4,5,10]
>>> n = 3
>>> zip(*(x[i:] for i in range(n)))
[(0, 1, 7), (1, 7, 3), (7, 3, 4), (3, 4, 5), (4, 5, 10)]

If you want the result to be a list of lists instead of list of tuples, use map(list, zip(...)).