如何使用 python 获取 fasta 文件中重复序列的计数

How to get the count of duplicated sequences in fasta file using python

我有一个这样的fasta文件: test_fasta.fasta

>XXKHH_1
AAAAATTTCTGGGCCCC
>YYYXXKHH_1
TTAAAAATTTCTGGGCCCCGGGAAAAAA
>TTDTT_11
TTTGGGAATTAAACCCT
>ID_2SS
TTTGGGAATTAAACCCT
>YKHH_1
TTAAAAATTTCTGGGCCCCGGGAAAAAA
>YKHSH_1S
TTAAAAATTTCTGGGCCCCGGGAAAAAA

我想获取重复序列的计数并追加文件中每个序列的总计数(从多到少排序)并得到如下所示的结果:

>YYYXXKHH_1_counts3
TTAAAAATTTCTGGGCCCCGGGAAAAAA
>TTDTT_11_counts2
TTTGGGAATTAAACCCT
>XXKHH_1_counts1
AAAAATTTCTGGGCCCC

我有这段代码可以找到重复的序列并将它们的 ID 连接在一起,但我没有将它们连接在一起,我只想在结果中将重复项的计数附加在 ID 中,如上所示。

from Bio import SeqIO
from collections import defaultdict

dedup_records = defaultdict(list)
for record in SeqIO.parse("test_fasta.fasta", "fasta"):
    # Use the sequence as the key and then have a list of id's as the value
    dedup_records[str(record.seq)].append(record.id)
with open("Output.fasta", 'w') as output:
    for seq, ids in dedup_records.items():
        # Join the ids and write them out as the fasta
        output.write(">{}\n".format('|'.join(ids)))
        output.write(seq + "\n")

因为您已经在输出循环中的 ids 列表中拥有每个重复记录的 ID,您可以简单地输出第一个 ID(根据您的预期输出,您显然想要),然后是ids 列表的长度:

for seq, ids in sorted(dedup_records.items(), key=lambda t: len(t[1]), reverse=True):
    output.write(">{}_counts{}\n".format(ids[0], len(ids)))
    output.write(seq + "\n")