如何检查列表中是否存在子序列? [英] How to check subsequence exists in a list?

问题描述

在python中，可以使用is关键字来检查包含内容，例如

In python, it's possible to use the is keyword to check for contains, e.g.

>>> 3 in [1,2,3,4,5]
True

但是，如果要检查单个整数列表是否在引用列表[1,2,3,4,5]中，则不会产生相同的输出:

But this doesn't yield the same output if it's checking whether a list of a single integer is inside the reference list [1,2,3,4,5]:

>>> [3] in [1,2,3,4,5]
False

此外，使用以下方法无法检查参考列表中的子序列:

Also, checking a subsequence in the reference list cannot be achieved with:

>>> [3,4,5] in [1,2,3,4,5]
False

是否有一种方法可以检查子序列，以使以下结果返回true?，例如函数调用x_in_y():

>>> x_in_y([3,4,5], [1,2,3,4,5])
True
>>> x_in_y([3], [1,2,3,4,5])
True
>>> x_in_y(3, [1,2,3,4,5])
True
>>> x_in_y([2,3], [1,2,3,4,5])
True
>>> x_in_y([2,4], [1,2,3,4,5])
False
>>> x_in_y([1,5], [1,2,3,4,5])
False

也许来自itertools或operator?

Maybe something from itertools or operator?

(注意，输入列表可以是唯一的)

(Note, the input lists can be non-unique)

推荐答案

x_in_y()可以通过切片原始列表并将切片与输入列表进行比较来实现:

x_in_y() can be implemented by slicing the original list and comparing the slices to the input list:

def x_in_y(query, base):
    try:
        l = len(query)
    except TypeError:
        l = 1
        query = type(base)((query,))

    for i in range(len(base)):
        if base[i:i+l] == query:
            return True
    return False

如果使用的是Python2，请将range更改为xrange.

Change range to xrange if you are using Python2.

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