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 want a function that would produce as output:
[[0,1,7],[1,7,3],[7,3,4],[3,4,5],[4,5,10]]
Thanks in advance!
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4 Answers
19
>>> 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(...)).
Andrew Clark
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Beautiful. Can you explain me what the * operator in the zip function does? – Raghavendra Kumar Oct 30 '16 at 14:25
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It unpacks a list .Useful if you are passing a list of arguments to a function , yet it expects a individual arguments – mossplix Jun 23 '17 at 06:49
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I know comments are supposed to be more constructive, but I simply found this beautiful. – kevin_theinfinityfund Feb 28 '20 at 05:35
14
>>> 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]]
JBernardo
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6To make this general for any subsequence size: `[x[i:i+n] for i in range(len(x)-n+1)]` where `n` is the desired length of the subsequence. – Steven Rumbalski Jul 12 '11 at 20:57
2
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]]
pillmuncher
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1
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]
Shaikovsky
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