改进 DNA 比对去间隙的代码设计

Question

这是一个关于更有效的代码设计的问题：

假设三个对齐的 DNA 序列（seq1、seq2 和 seq3；它们都是字符串）代表两个基因（gene1 和gene2）。这些基因的起始和终止位置相对于比对的 DNA 序列是已知的。

# Input
align = {"seq1":"ATGCATGC", # In seq1, gene1 and gene2 are of equal length
         "seq2":"AT----GC",
         "seq3":"A--CA--C"}
annos = {"seq1":{"gene1":[0,3], "gene2":[4,7]},
         "seq2":{"gene1":[0,3], "gene2":[4,7]},
         "seq3":{"gene1":[0,3], "gene2":[4,7]}}

我希望从比对中删除间隙（即破折号）并保持基因开始和停止位置的相对关联。

# Desired output
align = {"seq1":"ATGCATGC", 
         "seq2":"ATGC",
         "seq3":"ACAC"}
annos = {"seq1":{"gene1":[0,3], "gene2":[4,7]},
         "seq2":{"gene1":[0,1], "gene2":[2,3]},
         "seq3":{"gene1":[0,1], "gene2":[2,3]}}

获得所需的输出并不像看起来那么简单。下面我为这个问题写了一些（行号）伪代码，但肯定有更优雅的设计。

1  measure length of any aligned gene  # take any seq, since all seqs aligned
2  list_lengths = list of gene lengths  # order is important
3  for seq in alignment
4      outseq = ""
5      for each num in range(0, length(seq))  # weird for-loop is intentional
6          if seq[num] == "-"
7              current_gene = gene whose start/stop positions include num
8              subtract 1 from length of current_gene
9              subtract 1 from lengths of all genes following current_gene in list_lengths
10         else
11             append seq[num] to outseq
12     append outseq to new variable
13     convert gene lengths into start/stop positions and append ordered to new variable

谁能给我建议/示例，让我设计更短、更直接的代码？

Answer 1

此答案处理您更新的 annos 字典，从评论到 cdlanes 答案。该答案使annos 字典的seq2 gene2 的索引为[2,1] 不正确。如果序列包含该区域中的所有间隙，我提出的解决方案将从字典中删除 gene 条目。还要注意的是，如果一个基因在最后的align 中只包含一个字母，那么anno[geneX] 将具有相同的开始和停止索引--> 请参阅seq3 gene1 来自您的评论annos。

align = {"seq1":"ATGCATGC",
         "seq2":"AT----GC",
         "seq3":"A--CA--C"}

annos = {"seq1":{"gene1":[0,3], "gene2":[4,7]},
         "seq2":{"gene1":[0,3], "gene2":[4,7]},
         "seq3":{"gene1":[0,3], "gene2":[4,7]}}


annos3 = {"seq1":{"gene1":[0,2], "gene2":[3,4], "gene3":[5,7]}, 
          "seq2":{"gene1":[0,2], "gene2":[3,4], "gene3":[5,7]}, 
          "seq3":{"gene1":[0,2], "gene2":[3,4], "gene3":[5,7]}}

import re
for name,anno in annos.items():
    # indices of gaps removed usinig re
    removed = [(m.start(0)) for m in re.finditer(r'-', align[name])]

    # removes gaps from align dictionary
    align[name] = re.sub(r'-', '', align[name])

    build_dna = ''
    for gene,inds in anno.items():

        start_ind = len(build_dna)+1

        #generator to sum the num '-' removed from gene
        num_gaps = sum(1 for i in removed if i >= inds[0] and i <= inds[1])

        # build the de-gapped string
        build_dna+= align[name][inds[0]:inds[1]+1].replace("-", "")

        end_ind = len(build_dna)

        if num_gaps == len(align[name][inds[0]:inds[1]+1]): #gene is all gaps
            del annos[name][gene] #remove the gene entry
            continue
        #update the values in the annos dictionary
        annos[name][gene][0] = start_ind-1
        annos[name][gene][1] = end_ind-1

结果：

In [3]: annos
Out[3]:  {'seq1': {'gene1': [0, 3], 'gene2': [4, 7]},
          'seq2': {'gene1': [0, 1], 'gene2': [2, 3]},
          'seq3': {'gene1': [0, 1], 'gene2': [2, 3]}}

来自上述 3 基因annos 的结果。只需替换 annos 变量：

In [5]: annos3
Out[5]:  {'seq1': {'gene1': [0, 2], 'gene2': [3, 4], 'gene3': [5, 7]},
          'seq2': {'gene1': [0, 1], 'gene3': [2, 3]},
          'seq3': {'gene1': [0, 0], 'gene2': [1, 2], 'gene3': [3, 3]}}

Answer 2

以下匹配两个测试用例的示例程序的输出：

align = {"seq1":"ATGCATGC",
         "seq2":"AT----GC",
         "seq3":"A--CA--C"}

annos = {"seq1":{"gene1":[0,3], "gene2":[4,7]},
         "seq2":{"gene1":[0,3], "gene2":[4,7]},
         "seq3":{"gene1":[0,3], "gene2":[4,7]}}

(START, STOP) = (0, 1)

for alignment, sequence in align.items():
    new_sequence = ''
    gap = 0

    for position, codon in enumerate(sequence):
        if '-' == codon:
            for gene in annos[alignment].values():
                if gene[START] > (position - gap):
                    gene[START] -= 1
                if gene[STOP] >= (position - gap):
                    gene[STOP] -= 1
            gap += 1
        else:
            new_sequence += codon

    align[alignment] = new_sequence

运行结果：

python3 -i test.py
>>> align
{'seq2': 'ATGC', 'seq1': 'ATGCATGC', 'seq3': 'ACAC'}
>>> 
>>> annos
{'seq1': {'gene1': [0, 3], 'gene2': [4, 7]}, 'seq2': {'gene1': [0, 1], 'gene2': [2, 3]}, 'seq3': {'gene1': [0, 1], 'gene2': [2, 3]}}
>>> 

我希望这仍然足够优雅、直接、简短和 Pythonic！

Answer 3

我自己对上述问题的解决方案既不是优雅的也不是 Pythonic，而是达到了所需的输出。非常欢迎任何改进建议！

import collections
import operator
# measure length of any aligned gene  # take any seq, since all seqs aligned
align_len = len(align.itervalues().next())
# initialize output
align_out, annos_out = {}, {}
# loop through annos
for seqname, anno in annos.items():
# operate on ordered sequence lengths instead on ranges
    ordseqlens = collections.OrderedDict()
# generate ordered sequence lengths
    for k,v in sorted(anno.items(), key=operator.itemgetter(1)):
        ordseqlens[k] = v[1]-v[0]+1
# start (and later append to) sequence output
    align_out[seqname] = ""
# generate R-style for-loop
    for pos in range(0, len(align[seqname])):
        if align[seqname][pos] == "-":
            try:
                current_gene = next(key for key, a in anno.items() if a[0] <= pos <= a[1])
            except StopIteration:
                print("No annotation provided for position", pos, "in sequence", seqname)
# subtract 1 from lengths of current_gene
            ordseqlens[current_gene] = ordseqlens[current_gene]-1
# append nucleotide unless a gap
        else:
            align_out[seqname] += align[seqname][pos]
# convert modified ordered sequence lengths back into start/stop positions
    summ = 0
    tmp_dict = {}
    for k,v in ordseqlens.items():
        tmp_dict[k] = [summ, v+summ-1]
        summ = v+summ
# save start/stop positions to correct annos
    annos_out[seqname] = tmp_dict

这段代码的输出是：

>>> align_out
{'seq3': 'ACAC',
 'seq2': 'ATGC',
 'seq1': 'ATGCATGC'}

>>> annos_out
{'seq3': {'gene1': [0, 1], 'gene2': [2, 3]},
 'seq2': {'gene1': [0, 1], 'gene2': [2, 3]},
 'seq1': {'gene1': [0, 3], 'gene2': [4, 7]}}

Answer 4

所以，我认为尝试将每个序列分解为基因然后删除破折号的方法会导致大量不必要的簿记。相反，直接查看破折号然后根据它们的相对位置更新所有索引可能更容易。这是我编写的一个似乎运行正常的函数：

from copy import copy

def rewriteGenes(align, annos):
    alignments = copy(align)
    annotations = copy(annos)

    for sequence, alignment in alignments.items():
        while alignment.find('-') > -1:
            index = alignment.find('-')
            for gene, (start, end) in annotations[sequence].items():
                if index < start: 
                    annotations[sequence][gene][0] -= 1
                if index <= end: 
                    annotations[sequence][gene][1] -= 1
            alignment = alignment[:index] + alignment[index+1:]
        alignments[sequence] = alignment

    return (alignments, annotations)

这会遍历每个比对中的破折号，并在删除基因索引时更新它们。

请注意，这会为以下测试用例生成索引为 [2,1] 的基因：

align = {"seq1":"ATGCATGC",
         "seq2":"AT----GC",
         "seq3":"A--CA--C"}
annos = {"seq1":{"gene1":[0,2], "gene2":[3,4], "gene3":[5,7]}, 
         "seq2":{"gene1":[0,2], "gene2":[3,4], "gene3":[5,7]}, 
         "seq3":{"gene1":[0,2], "gene2":[3,4], "gene3":[5,7]}}

这是必要的，因为您的索引设置方式不允许空基因。例如，索引[2,2] 将是从索引 2 开始的长度为 1 的序列。