python - 改进DNA比对解盖的代码设计

这是有关更有效的代码设计的问题:

假定代表两个基因(gene1和gene2)的三个对齐的DNA序列(seq1，seq2和seq3；它们都是字符串)。相对于比对的DNA序列，这些基因的起始和终止位置是已知的。

# 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]}}

我希望从比对中消除缺口(即破折号)，并保持基因起始和终止位置的相对关联。

# 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]}}

获得所需的输出看起来并不那么琐碎。下面，我针对此问题编写了一些(行编号)伪代码，但是肯定有一个更优雅的设计。

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

谁能给我一些更简短，直接的代码设计建议/示例？

最佳答案

此答案处理您从注释到cdlanes答案的更新的annos字典。该答案使annos字典的seq2 gene2的索引错误[2,1]。如果序列在该区域中包含所有空位，则我建议的解决方案将从字典中删除gene条目。还需要注意的是，如果一个基因在最后的align中仅包含一个字母，则anno[geneX]的起始和终止索引将相等–>请从注释的seq3中查看gene1 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个基因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]}}

关于python - 改进DNA比对解盖的代码设计，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/34816513/