Pythonic : Find all consecutive sub-sequences of certain length

Question

我有一个整数列表，我想在这个列表中找到所有长度为 n 的连续子序列。例如:

>>> int_list = [1,4,6,7,8,9]
>>> conseq_sequences(int_list, length=3)
[[6,7,8], [7,8,9]]

我能想到的最好的是:

def conseq_sequences(self, li, length):
    return [li[n:n+length]
            for n in xrange(len(li)-length+1)
            if li[n:n+length] == range(li[n], li[n]+length)]

这并不过分可读。是否有任何可读的 pythonic 方式来执行此操作？

Answer 1

这是一个更通用的解决方案，适用于任意输入可迭代对象(不仅仅是序列):

from itertools import groupby, islice, tee
from operator import itemgetter

def consecutive_subseq(iterable, length):
    for _, consec_run in groupby(enumerate(iterable), lambda x: x[0] - x[1]):
        k_wise = tee(map(itemgetter(1), consec_run), length)
        for n, it in enumerate(k_wise):
            next(islice(it, n, n), None) # consume n items from it
        yield from zip(*k_wise)

• itertools.groupby在输入中查找连续的子字符串，例如 6, 7, 8, 9。它基于 the example from the docs that shows how to find runs of consecutive numbers :

  The key to the solution is differencing with a range generated by enumerate() so that consecutive integers all appear in same group (run).

• itertools.tee + zip 允许按 k 方向迭代子字符串——pairwise recipe from the itertools docs 的概括.

• next(islice(iterator, n, n), None) 来自 the consume recipe there .

例子:

print(*consecutive_subseq([1,4,6,7,8,9], 3))
# -> (6, 7, 8) (7, 8, 9)

代码使用 Python 3 语法，如果需要可以针对 Python 2 进行调整。

另请参阅，What is the most pythonic way to sort dates sequences?