在两个字符串之间找到最长的完美匹配

Question

我正在尝试编写一个脚本，当提供两个字符串时，它将执行两个功能：

1。查找从pos[0] 开始在两个字符串中相同的最长字符序列

Seq1 = 'ATCCTTAGC'
Seq2 = 'ATCCAGCAATTC'
        ^^^^ Match from pos[0] to pos[3]
Pos: 0:3
Length: 4
Seq: ATCC

2。查找两个字符串中存在的最长的连续字符

Seq1 = 'TAGCTCCTTAGC' # Contains 'TCCTT'
Seq2 = 'GCAGCCATCCTTA' # Contains 'TCCTT'
        ^ No match at pos[0]
Pos1: 4:8
Pos2  7:11
Length: 5
Seq: TCCTT

要完成问题 1，我有以下几点：

#!/usr/bin/python

upstream_seq = 'ATACATTGGCCTTGGCTTAGACTTAGATCTAGACCTGAAAATAACCTGCCGAAAAGACCCGCCCGACTGTTAATACTTTACGCGAGGCTCACCTTTTTGTTGTGCTCCC'
downstream_seq = 'ATACACGAAAAGCGTTCTTTTTTTGCCACTTTTTTTTTATGTTTCAAAACGGAAAATGTCGCCGTCGTCGGGAGAGTGCCTCCTCTTAGTTTATCAAATAAAGCTTTCG'

print("Upstream:   %s\nDownstream: %s\n") % (upstream_seq, downstream_seq)

mh = 0
pos_count = 0
seq = ""
position =""
longest_hom=""
for i in range(len(upstream_seq)):
    pos_count += 1
    if upstream_seq[i] == downstream_seq[i]:
        mh += 1
        seq += upstream_seq[i]
        position = pos_count
        longest_hom = mh

    else:
        mh = 0
        break

print("Pos: 0:%s\nLength: %s\nSeq: %s\n") % (position , longest_hom, seq)

---

Upstream:   ATACATTGGCCTTGGCTTAGACTTAGATCTAGACCTGAAAATAACCTGCCGAAAAGACCCGCCCGACTGTTAATACTTTACGCGAGGCTCACCTTTTTGTTGTGCTCCC
Downstream: ATACACGAAAAGCGTTCTTTTTTTGCCACTTTTTTTTTATGTTTCAAAACGGAAAATGTCGCCGTCGTCGGGAGAGTGCCTCCTCTTAGTTTATCAAATAAAGCTTTCG

Pos: 0:5
Length: 5
Seq: ATACA

---

我遇到了问题 2。到目前为止，我已经使用BioPython's pairwise2 考虑了两个序列之间的比对。但是，在这种情况下，我只想要完美匹配（没有间隙，没有扩展），我只想看到最长的序列，而不是我似乎得到的共识：

from Bio import pairwise2 as pw2

global_align = pw2.align.globalms(upstream_seq, downstream_seq, 3, -1, -.5, -.5)

print(global_align[0])

---

('ATACATT-G----GCC-TTGGCTTA-----G--ACTTAGATCTAG-----ACCTGAA----AATAACCTGCCGAAAA-GACC-CGCCCGACTGTTAATACTT-TACGCG-AG-GCT-CAC-C-T-TT--TTGT-TG----T---GCTCC--C-', 'ATACA--CGAAAAG-CGTT--CTT-TTTTTGCCACTT---T-T--TTTTTA--TG--TTTCAA-AA-C-G--GAAAATG---TCG--C--C-G----T-C--GT-CG-GGAGAG-TGC-CTCCTCTTAGTT-TAT-CAAATAAAGCT--TTCG', 151.0, 0, 153)

---

问题：如何找到两个字符串中存在的最长的连续字符？

Answer 1

以下是问题 1 的简短代码：

upstream_seq = 'ATACATTGGCCTTGGCTTAGACTTAGATCTAGACCTGAAAATAACCTGCCGAAAAGACCCGCCCGACTGTTAATACTTTACGCGAGGCTCACCTTTTTGTTGTGCTCCC'
downstream_seq = 'ATACACGAAAAGCGTTCTTTTTTTGCCACTTTTTTTTTATGTTTCAAAACGGAAAATGTCGCCGTCGTCGGGAGAGTGCCTCCTCTTAGTTTATCAAATAAAGCTTTCG'

common_prefix = ''

for x,y in zip(upstream_seq, downstream_seq):
    if x == y:
        common_prefix += x
    else:
        break
print(common_prefix)
# ATACA

问题 2 的简单方法是简单地为每个字符串生成一组每个子字符串，计算它们的交集并按长度排序：

upstream_seq = 'ATACATTGGCCTTGGCTTAGACTTAGATCTAGACCTGAAAATAACCTGCCGAAAAGACCCGCCCGACTGTTAATACTTTACGCGAGGCTCACCTTTTTGTTGTGCTCCC'
downstream_seq = 'ATACACGAAAAGCGTTCTTTTTTTGCCACTTTTTTTTTATGTTTCAAAACGGAAAATGTCGCCGTCGTCGGGAGAGTGCCTCCTCTTAGTTTATCAAATAAAGCTTTCG'

def all_substrings(string):
    n = len(string)
    return {string[i:j+1] for i in range(n) for j in range(i,n)}

print(all_substrings('ABCA'))
# {'CA', 'BC', 'ABC', 'C', 'BCA', 'AB', 'A', 'B', 'ABCA'}
print(all_substrings(upstream_seq) & all_substrings(downstream_seq))
# {'AAAG', 'CA', 'A', 'AAC', 'TGTT', 'ACT', 'CTTAG', 'GCT', 'ATAC', 'AAAA', 'TTTA', 'AAT', 'GTGC', 'CTT', 'AAAAG', 'TTTG', 'CGAA', 'AA', 'CGAAAAG', 'GCC', 'ACA', 'TGCC', 'AAATAA', 'CTCC', 'TTTTT', 'CGCC', 'CAC', 'GAG', 'CTC', 'CGAAAA', 'ATC', 'TCA', 'GA', 'CGC', 'TGT', 'GT', 'GC', 'GAAA', 'ACTTT', 'AAG', 'TTTT', 'CT', 'AATA', 'TCC', 'CGAAA', 'GAA', 'GAAAAG', 'GTT', 'AG', 'TC', 'AAAAT', 'CC', 'TTT', 'AATAA', 'CTTTT', 'ACTT', 'TTA', 'CTTT', 'GCTT', 'GCCG', 'GTG', 'TACA', 'TT', 'GCG', 'TTTTTG', 'TAG', 'TTG', 'TTAG', 'AAATA', 'CTTTTT', 'AAAT', 'TAA', 'ACG', 'TG', 'GCCT', 'G', 'TAC', 'CCT', 'TCT', 'ATA', 'CTTA', 'CCG', 'CG', 'ATAA', 'GG', 'ATACA', 'AGA', 'TGC', 'C', 'T', 'AT', 'GAAAA', 'CGA', 'GAAAAT', 'TA', 'AC', 'AAA', 'TTTTG'}
print(max(all_substrings(upstream_seq) & all_substrings(downstream_seq), key=len))
# CGAAAAG

如果您想要更有效的方法，您应该使用suffix tree。

如果你不想重新发明轮子，你可以简单地使用difflib.SequenceMatcher.find_longest_match

Answer 2

longest common substring problem 可以通过多种方式处理，其中一些方式比其他方式更有效。一种非常有效的解决方案涉及动态编程，它在 python 2 和 3 中的实现可以在wikibooks 中找到。一个天真的解决方案，更简单，更容易理解，但效率较低，是这样的：

def longest_common_substring(s1, s2):
    current_match_start = -1
    current_match_end = -1

    best_match_start = current_match_start
    best_match_end = current_match_end

    min_len = min(len(s1), len(s2))
    for i in range(min_len):
        if s1[i] == s2[i]:
            current_match_start = current_match_end = i
            j = 0
            while s1[i+j] == s2[i+j] and i+j < min_len:
               j += 1
            current_match_end = current_match_start + j

            if current_match_end - current_match_start > best_match_end - best_match_start:
                best_match_start = current_match_start
                best_match_end = current_match_end

    return s1[best_match_start:best_match_end]

upstream_seq = 'ATACATTGGCCTTGGCTTAGACTTAGATCTAGACCTGAAAATAACCTGCCGAAAAGACCCGCCCGACTGTTAATACTTTACGCGAGGCTCACCTTTTTGTTGTGCTCCC'
downstream_seq = 'ATACACGAAAAGCGTTCTTTTTTTGCCACTTTTTTTTTATGTTTCAAAACGGAAAATGTCGCCGTCGTCGGGAGAGTGCCTCCTCTTAGTTTATCAAATAAAGCTTTCG'

print(longest_common_substring(upstream_seq, downstream_seq))

Answer 3

正如 Eric Duminil 的回答中提到的，解决此问题的一种方法是使用difflib.SequenceMatcher.find_longest_match：

from difflib import SequenceMatcher

upstream_seq = 'ATACATTGGCCTTGGCTTAGACTTAGATCTAGACCTGAAAATAACCTGCCGAAAAGACCCGCCCGACTGTTAATACTTTACGCGAGGCTCACCTTTTTGTTGTGCTCCC'
downstream_seq = 'ATACACGAAAAGCGTTCTTTTTTTGCCACTTTTTTTTTATGTTTCAAAACGGAAAATGTCGCCGTCGTCGGGAGAGTGCCTCCTCTTAGTTTATCAAATAAAGCTTTCG'

s = SequenceMatcher(None, upstream_seq, downstream_seq)
match = s.find_longest_match(0, len(upstream_seq), 0, len(downstream_seq))

print(match)

upstream_start = match[0]
upstream_end = match[0]+match[2]
seq = upstream_seq[match[0]:(match[0]+match[2])]
downstream_start = match[1]
downstream_end = match[1]+match[2]

print("Upstream seq: %s\nstart-stop: %s-%s\n") % (seq, upstream_start, upstream_end)
print("Downstream seq: %s\nstart-stop: %s-%s\n") % (seq, downstream_start, downstream_end)

---

Match(a=49, b=5, size=7)
Upstream seq: CGAAAAG
start-stop: 49-56

Downstream seq: CGAAAAG
start-stop: 5-12