Question

所以我正在尝试编写一个可用于分析DNA的程序，现在我正试图将这些基因分成“链”。为了实现这一点，我需要分析链并将其分成三个STOP密码子之一（三个碱基对的组）。我现在的代码是这样的：

class Strand:

  def __init__(self, code):
    self.code = [code]
    self.endCodons = []
    self.genes = []

  def getGenes(self):
    for codon in self.endCodons:
      for code in self.code:
        code = code.split(codon)


strand = Strand("ATCATGCACATAGAAACTGATACACACCACAGTGATCACATGAAGTACACATG")
strand.getGenes()
print(strand.genes)

但是，当我运行它时，它会返回一个空列表。我可以使用一些建议。

Answer 1

通过每个终止密码子运行循环并按此分裂将导致错误的输出，因为我认为这些终止密码子可以以序列中的任何顺序出现，并且对终止密码子列表的迭代将要求停止在那里同样的顺序。

所以，如果我理解正确，你需要从左到右扫描你的字符串，然后搜索密码子：

class Strand:
  def __init__(self, code):
    self.code = code
    self.endCodons = ["TAG", "TAA", "TGA"]
    self.genes = []

  def getGenes(self):
    if (len(self.code) % 3 != 0):
      print("Input sequence is not divisible by 3?")

    # In this, we assume each stop codon is always 3 characters.
    iteration = 0
    lastGeneEnd = 0
    while (iteration < len(self.code)):
      # What is our current 3 character sequence? (Unless it's at the end)
      currentSequence = self.code[iteration:iteration + 3]

      # Check if our current 3 character sequence is an end codon
      if (currentSequence in self.endCodons):
        # What will our gene length be?
        geneLength = (iteration + 3) - lastGeneEnd

        # Make sure we only break into multiples of 3
        overlap = 3 - (geneLength % 3)
        # There is no overlap if our length is already a multiple of 3
        if (overlap == 3): overlap = 0

        # Modify the gene length to reflect our overlap into a multiple of 3
        geneLength = geneLength + overlap

        # Update the iteration so we don't process any more than we need
        iteration = iteration + overlap + 3

        # Grab the entire gene sequence, including the stop codon
        gene = self.code[lastGeneEnd:iteration]

        # If we have a 3-length gene and there's nothing left, just append to the last gene retrieved as it has
        # got to be part of the last sequence
        if (len(gene) == 3 and iteration >= len(self.code)):
          lastIndex = len(self.genes) - 1
          self.genes[lastIndex] = self.genes[lastIndex] + gene
          break

        # Make sure we update the last end index so we don't include portions of previous positives
        lastGeneEnd = iteration

        # Append the result to our genes and continue
        self.genes.append(gene)

        continue

      iteration = iteration + 1

strand = Strand("ATCATGCACATAGAAACTGATACACACCACAGTGATCACATGAAGTACACATG")
strand.getGenes()
print("Got Genes: ")
print(strand.genes)

for gene in strand.genes:
  print("Sequence '%s' is a multiple of 3: %u" % (gene, len(gene) % 3 == 0))

我不是真正的生物学家，所以我可能做了一些不正确的假设。

<强>编辑：

代码保证打破三倍的数量，但我似乎仍然不太理解所需的逻辑。它在给定的示例中有效，但我不确定它是否在其他情况下正常工作。

容器Python拆分字符串

1 个答案: