Question

如果还没有被要求，我会感到惊讶。

我们说我有一个数组[5,6,7,29,34]，我想检查序列5,6,7是否出现在它（它做了）。订单很重要。

我该怎么做？

Answer 1

只是为了好玩，这是一个快速（非常快）和脏（非常脏）的解决方案（有点缺陷，所以不要真的使用它）：

>>> str([5,6,7]).strip('[]') in str([5,6,7,29,34])
True

RightWay™可能会使用list.index()来查找第一个元素的候选匹配项，然后验证与切片和列表相等的完全匹配：

>>> def issubsequence(sub, seq):
        i = -1
        while True:
            try:
                i = seq.index(sub[0], i+1)  # locate first character
            except ValueError:
                return False
            if seq[i : i+len(sub)] == sub:  # verify full match
                return True         

>>> issubsequence([5, 6, 7], [5,6,7,29,34])
True
>>> issubsequence([5, 20, 7], [5,6,7,29,34])
False

编辑：OP在评论中澄清，子序列必须按顺序排列，但不必处于连续位置。这有一个不同的，更复杂的解决方案，已在这里得到解答：How do you check if one array is a subsequence of another?

Answer 2

这是一个很好的解决方案：

def is_sublist(a, b):
    if not a: return True
    if not b: return False
    return b[:len(a)] == a or is_sublist(a, b[1:])

如Stefan Pochmann所述，这可以改写为：

def is_sublist(a, b):
    return b[:len(a)] == a or bool(b) and is_sublist(a, b[1:])

Answer 3

这是一个对任何可迭代对象都有效（高效！）的解决方案：

import collections
import itertools

def consume(iterator, n=None):
    """Advance the iterator n-steps ahead. If n is none, consume entirely."""
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)

def is_slice(seq, subseq):
    """Returns whether subseq is a contiguous subsequence of seq."""
    subseq = tuple(subseq)  # len(subseq) is needed so we make it a tuple.
    seq_window = itertools.tee(seq, n=len(subseq))
    for steps, it in enumerate(seq_window):
        # advance each iterator to point to subsequent values in seq.
        consume(it, n=steps)
    return any(subseq == seq_slice for seq_slice in izip(*seq_window))

consume来自itertools recipes。

如何检查有序的非连续子序列是否在数组中？蟒蛇

3 个答案: