我希望能够在一个序列中找到所有长度为n的连续子序列。
例如,假设n为3,序列为:
[0,1,7,3,4,5,10]
我希望有一个函数能够产生以下输出:
[[0,1,7],[1,7,3],[7,3,4],[3,4,5],[4,5,10]]
提前感谢您!
>>> 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)]
如果你想要的结果是一个列表的列表而不是元组的列表,可以使用map(list, zip(...))
。
>>> 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]]
[x[i:i+n] for i in range(len(x)-n+1)]
,其中 n
是所需子序列的长度。 - Steven Rumbalskidef subseqs(seq, length):
for i in xrange(len(seq) - length + 1):
yield seq[i:i+length]
使用方法如下:
>>> for each in subseqs("hello", 3):
... print each
...
hel
ell
llo
>>> 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]]
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]]
或者,仅需获取名为'xs'的列表的所有序列:
[xs[i:j+1] for i, _ in enumerate(xs) for j, _ in enumerate(xs)]
[xs[i:j+1] for i, _ in enumerate(xs) for j, _ in enumerate(xs) if len(xs[i:j+1]) == n]