python - 在列表中查找连续值

Question

我有一个值列表:

a = [1,3,4,5,2]

我现在想要以下功能:

does_segment_exist(a, [1,3,4]) #True
does_segment_exist(a, [3,4,5]) #True
does_segment_exist(a, [4,5,2]) #True
does_segment_exist(a, [1,4,5]) #False
does_segment_exist(a, [1,3]) #True
does_segment_exist(a, [1,4]) #False

因此必须按连续顺序找到这些值。

我有在 Python 3 中执行此操作的巧妙方法吗？

Answer 1

您可以使用滚动窗口迭代器，在本例中是旧版本的 itertools 文档:

from itertools import islice

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result
    for elem in it:
        result = result[1:] + (elem,)
        yield result

def does_segment_exist(iterable, sublist):
    return tuple(sublist) in window(iterable, len(sublist))

print(does_segment_exist([1,3,4,5,2], [3,4,5]))

如果你只需要它在列表上工作，而不是任何可迭代的，你可以使用:

def does_segment_exist(seq, sublist):
    # seq and sublist must both be lists
    n = len(sublist)
    return sublist in (seq[i:i+n] for i in range(len(seq) + 1 - n))

雷蒙德提到的方法的基本实现:

def does_segment_exist(seq, sublist):
    first = sublist[0]
    i = 0
    n = len(sublist)
    while True:
        try:
            i = seq.index(first, i)
        except ValueError:
            return False
        if sublist == seq[i:i+n]:
            return True
        i += 1

print(does_segment_exist([1,3,4,5,2], [3,4,5]))

此方法的优点是它不必对第一个匹配项之前的每个索引进行切片，只需对与段中第一个值的匹配项对应的索引进行切片。