Python：从具有约束的列表中采样索引

Question

我想知道如何从具有如下约束的列表中采样索引：

               1111
     01234567890123
a = "ABAAABABBBABAB"
b = "ABABBB"

问题是如何找到a的有序采样字符等于b

所以，问题的答案应该如下所示，

[0, 1, 2, 5, 7, 8],
[0, 1, 2, 5, 7, 9],
[0, 1, 2, 5, 7, 11],
...,
[2, 5, 6, 7, 8, 9],
...

因为b 等于a[0] + a[1] + a[2] + a[5] + a[7] + a[8]。 所以[0, 1, 2, 5, 7, 8] 是问题的一个答案。 但是答案的指标应该满足有序的条件。 不能是[2, 1, 3, 5, 7, 8]。 a[2] + a[1] + a[3] + a[5] + a[7] + a[8] 等于b，但不能满足有序状态条件。

a 的长度为 22，b 的长度总是小于a 的长度。

我实施了蛮力解决这个问题，但它需要很大的时间复杂度。

有没有比蛮力更好的算法？ 如果有，应该如何实现？

Answer 1

如果我理解正确，这是一个global sequence alignment 问题，您可以使用BioPython 来解决：

from Bio import pairwise2

a = "ABAAABABBBABAB"
b = "ABABBB"

for alignment in pairwise2.align.globalxx(a, b):
    positions = [i for i, v in enumerate(alignment.seqB) if v != "-"]
    print(positions)

输出 （部分）

[0, 1, 6, 7, 8, 13]
[0, 1, 4, 7, 8, 13]
[0, 1, 3, 7, 8, 13]
[0, 1, 2, 7, 8, 13]
[0, 1, 4, 5, 8, 13]
[0, 1, 3, 5, 8, 13]
[0, 1, 2, 5, 8, 13]
[0, 1, 4, 5, 7, 13]
[0, 1, 3, 5, 7, 13]
[0, 1, 2, 5, 7, 13]

Answer 2

不能在恒定时间内完成，但可以以大致线性的时间复杂度完成（至少，相对于 A - 我认为复杂度的上限是 O(A * B)）如下：

def indices(A, B):
    # store partial patterns and complete patterns as we go
    # partial patterns will be stored in a list, where each element
    # is a 2-tuple (next_char_of_B_to_look_for, current_index_list)
    partial_patterns = []
    complete_patterns = []
    for (idx, char) in enumerate(A):
        # look for the next element, for any partial patterns we have
        # if we find it, then mark the index and increment the char to look for
        for p in partial_patterns:
            if p[0] < len(B) and char == B[p[0]]:
                p[1].append(idx)
                p[0] += 1
                # p[0] == len(B) indicates a completed pattern, and also 
                # will stop this `if` from being triggered again for this 
                # now-completed pattern (which will be ignored hereafter)
                if p[0] == len(B):
                    complete_patterns.append(p[1])
        # always look for the first element of B to start a new partial pattern
        if char == B[0]:
            partial_patterns.append([1, [idx]])
    return complete_patterns

a = "ABAAABABBBABAB"
b = "ABABBB"
indices(a, b)
# [[0, 1, 2, 5, 7, 8], 
#  [2, 5, 6, 7, 8, 9], 
#  [3, 5, 6, 7, 8, 9], 
#  [4, 5, 6, 7, 8, 9]]

我相信这是给定输入的正确输出（因为索引 6 是最后一个 A，后面至少有三个 Bs）

Answer 3

建立一个类似于Longest common subsequence problem的表，然后使用Reading out all LCSs部分提取所有LCS（这里LCS是B序列）

Answer 4

简单的解决方案是：

a = "ABAAABABBBABAB"
b = "ABABBB"

total_res = []
def find_indices(index):
    b_index = 0
    res = []
    for c in range(index, len(a)):
        if b_index > len(b) - 1:
            break
        if a[c] == b[b_index]:
            res.append(c)
            b_index += 1
    if res not in total_res and res:
        print(res)
        total_res.append(res)
for c in range(len(a)):
    find_indices(c)

输出：

[0, 1, 2, 5, 7, 8]
[2, 5, 6, 7, 8, 9]
[3, 5, 6, 7, 8, 9]
[4, 5, 6, 7, 8, 9]
[6, 7, 10, 11, 13]
[10, 11, 12, 13]
[12, 13]