Python. 用BioPython将genbank文件中三个最长的基因核苷酸序列排序并写入fasta文件
我对Python还比较陌生,所以请原谅我这个问题的愚蠢之处。我有一个genbank文件,写了一段代码,可以找出最长的三个基因,并把它们放到一个新生成的fasta文件里。
from Bio import SeqIO
file="sequence.gb"
output=open("Top3.faa", "w")
record=SeqIO.parse(file, "genbank")
rec=next(record)
print('The genes with the top 3 longest lengths have beens saved in Top3.faa')
for f in rec.features:
end=f.location.end.position
start=f.location.start.position
length=end-start
bug=(rec.seq)
if f.type=='CDS':
if 'gene' in f.qualifiers:
if length>7000:
geneName=f.qualifiers['gene']
name=str(geneName)
lenth=str(length)
seq=str(bug[start:end])
output.write('>')
output.write(lenth)
output.write('\n')
output.write(seq)
output.write('\n')
output.close()
我想做的是,不再手动检查基因长度是否超过7kb,而是让代码自己完成这个检查,自动找到最长的三个基因。如果能给我一些建议,告诉我该往哪个方向去做,我会非常感激。谢谢!
1 个回答
1
你可以保存一个包含N个最大元素及其大小的列表。
可以像这样做(虽然我不能测试,所以可能会出错,但这个思路是对的):
from Bio import SeqIO
file="sequence.gb"
output=open("Top3.faa", "w")
record=SeqIO.parse(file, "genbank")
rec=next(record)
print('The genes with the top 3 longest lengths have beens saved in Top3.faa')
# Largest genes and their size, sorted from the shortest to the longest.
# size first, gene name next, then seq.
largest_genes = [ (0, None, None) ] * 3; # initialize with the number of genes you need.
for f in rec.features:
end = f.location.end.position
start = f.location.start.position
length = end-start
bug = (rec.seq)
if f.type=='CDS' and 'gene' in f.qualifiers:
if length > largest_genes[0][0]: # [0] gives the first, [0] gives the length.
# This one is larger than the smallest one we have.
largest_genes = largest_genes[1:] # drop the smallest one.
# add this one
largest_genes.append((length, f.qualifiers['gene'], str(bug[start:end])))
largest_genes.sort() # re-sort.
for length, name, seq in largest_genes:
# name is not used but available.
output.write('>')
output.write(str(lenth))
output.write('\n')
output.write(seq)
output.write('\n')
output.close()