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python - 如何在不使用 Biopython 的情况下从 FASTA 文件中获取此输出？

问题描述

我需要从 FASTA 文件中获取下面显示的输出，但不使用 BioPython。有人有想法吗？

这是使用 BioPython 的代码：

from Bio import SeqIO
records = SeqIO.parse("data/assembledSeqs.fa", "fasta")
for i, seq_record in enumerate(records):
    print("Sequence %d:" % i)
    print("Number of A's: %d" % seq_record.seq.count("A"))
    print("Number of C's: %d" % seq_record.seq.count("C"))
    print("Number of G's: %d" % seq_record.seq.count("G"))
    print("Number of T's: %d" % seq_record.seq.count("T"))
    print()

FASTA 文件如下所示：

>chr12_9180206_+:chr12_118582391_+:a1;2 total_counts: 115 Seed: 4 K:    20 length: 79
TTGGTTTCGTGGTTTTGCAAAGTATTGGCCTCCACCGCTATGTCTGGCTGGTTTACGAGC
AGGACAGGCCGCTAAAGTG
>chr12_9180206_+:chr12_118582391_+:a2;2 total_counts: 135 Seed: 4 K: 20 length: 80
CTAACCCCCTACTTCCCAGACAGCTGCTCGTACAGTTTGGGCACATAGTCATCCCACTCG
GCCTGGTAACACGTGCCAGC
>chr1_8969882_-:chr1_568670_-:a1;113 total_counts: 7600 Seed: 225 K: 20 length: 86
CACTCATGAGCTGTCCCCACATTAGGCTTAAAAACAGATGCAATTCCCGGACGTCTAAAC
CAAACCACTTTCACCGCCACACGACC
>chr1_8969882_-:chr1_568670_-:a2;69 total_counts: 6987 Seed: 197 K: 20   length: 120
TGAACCTACGACTACACCGACTACGGCGGACTAATCTTCAACTCCTACATACTTCCCCCA
TTATTCCTAGAACCAGGCGACCTGCGACTCCTTGACGTTGACAATCGAGTAGTACTCCCG

我需要获得以下输出：

Sequence 0:
Number of A's: 14
Number of C's: 17
Number of G's: 24
Number of T's: 24

Sequence 1:
Number of A's: 17
Number of C's: 30
Number of G's: 16
Number of T's: 17

Sequence 2:
Number of A's: 27
Number of C's: 31
Number of G's: 12
Number of T's: 16

Sequence 3:
Number of A's: 31
Number of C's: 41
Number of G's: 20
Number of T's: 28

我已经尝试过了，但我无法获得相同的输出。

def count_bases (fasta_file_name):
    with open(fasta_file_name) as file_content:
        for seqs in file_content:
            if seqs.startswith('>'):
                for i, seq in enumerate('>'):
                    print("Sequence %d:" % i)
            else:
                print("Number of A's: %d" % seqs.count("A"))
                print("Number of C's: %d" % seqs.count("C"))
                print("Number of G's: %d" % seqs.count("G"))
                print("Number of T's: %d" % seqs.count("T"))
                print()
    return bases

result = count_bases('data/assembledSeqs.fa')

标签： pythonbioinformaticsfasta

解决方案

这似乎有效。它使用正则表达式re模块来识别标题行并获得以下核苷酸序列的总长度。此值用于确定要读取并连接在一起的后续数据行数。

因为它实际上解析了序列头中的信息，所以它应该能够处理任何长度的序列。

我不知道每行的官方碱基总数是多少，但在示例 fasta 文件中似乎是 60，因此已将其硬编码到代码中。

import re

pattern = r""">.+?length:\s(\d+)"""
regex = re.compile(pattern)

MAX_PER_LINE = 60

def count_bases (fasta_file_name):
    bases = None  # Undefined.

    with open(fasta_file_name) as inp:
        i = 0  # Sequence counter.
        line = next(inp, None)  # Read first line.
        while line:
            match = regex.search(line)
            if match:
                length = int(match.group(1))
                nlines = (length + MAX_PER_LINE-1) // MAX_PER_LINE
                # Read and concatenate data from the required number of lines.
                seqs = ''.join(next(inp).rstrip() for _ in range(nlines))

                print("Sequence %d:" % i)
                print("Number of A's: %d" % seqs.count("A"))
                print("Number of C's: %d" % seqs.count("C"))
                print("Number of G's: %d" % seqs.count("G"))
                print("Number of T's: %d" % seqs.count("T"))
                print()
                i += 1

            line = next(inp, None)  # Read next line.

    return bases

fasta_file = 'assembledSeqs.fa'
result = count_bases(fasta_file)

这是一个变体，它使用 a 一次collections.Counter计算所有碱基的数量。这可能比count对每个单独使用序列方法更快，因此如果您有大量要处理的数据，这很有用。

def count_bases (fasta_file_name):
    bases = None  # Undefined.

    with open(fasta_file_name) as inp:
        i = 0  # Sequence counter.
        line = next(inp, None)  # Read first line.
        while line:
            match = regex.search(line)
            if match:
                length = int(match.group(1))
                nlines = (length + MAX_PER_LINE-1) // MAX_PER_LINE
                # Read and concatenate data from the required number of lines.
                seqs = ''.join(next(inp).rstrip() for _ in range(nlines))

                print("Sequence %d:" % i)
                counter = Counter(seqs)  # Count the number of each base.
                for base in 'ACGT':
                    if base in counter:
                        print("Number of {}'s: {}".format(base, counter[base]))
                print()
                i += 1

            line = next(inp, None)  # Read next line.

    return bases

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