无需使用BioPython即可读取FASTA中的核苷酸

时间:2018-12-26 20:08:50

标签: python python-3.x biopython fasta

我需要获得通过以下代码获得的相同输出,但不使用BioPython。我被困住了...有人可以帮助我吗?谢谢!!!

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

我已经尝试过,但是根本不起作用

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')

3 个答案:

答案 0 :(得分:0)

This will work. First we want to parse your fasta file into headers and sequences, then flatten the list of lists of sequences to a list of strings, then count the number of each nucleotide in the string, then print:

import sys

fasta = sys.argv[1]

def fastaParser(infile):
    seqs = []
    headers = []
    with open(infile) as f:
        sequence = ""
        header = None
        for line in f:
            if line.startswith('>'):
                headers.append(line[1:-1])
                if header:
                    seqs.append([sequence])
                sequence = ""
                header = line[1:]
            else:
                sequence += line.rstrip()
        seqs.append([sequence])
    return headers, seqs

headers, seqs = fastaParser(fasta)

flat_seqs = [item for sublist in seqs for item in sublist]

def countNucs(instring):
    # will count upper and lower case sequences, if do not want lower case remove .upper()
    g = instring.upper().count('G') 
    c = instring.upper().count('C')
    a = instring.upper().count('A')
    t = instring.upper().count('T')
    return 'G = {}, C = {}, A = {}, T = {}'.format(g, c, a, t)

for header, seq in zip(headers, flat_seqs):
    print(header, countNucs(seq))

using your example fasta to run on the command-line:

[me]$ python3.5 fasta.py infasta.fa

output:

chr12_9180206_+:chr12_118582391_+:a1;2 total_counts: 115 Seed: 4 K:    20 length: 79 G = 24, C = 17, A = 14, T = 24
chr12_9180206_+:chr12_118582391_+:a2;2 total_counts: 135 Seed: 4 K: 20 length: 80 G = 16, C = 30, A = 17, T = 17
chr1_8969882_-:chr1_568670_-:a1;113 total_counts: 7600 Seed: 225 K: 20 length: 86 G = 12, C = 31, A = 27, T = 16
chr1_8969882_-:chr1_568670_-:a2;69 total_counts: 6987 Seed: 197 K: 20   length: 120 G = 20, C = 41, A = 31, T = 28

答案 1 :(得分:0)

我将使用Counter字典,如下所示:

from collections import Counter

with open('temp.fasta', 'r') as f:
    seq_name, d = '', ''
    for line in f:
        if line.startswith('>'):
            print(seq_name, d)
            seq_name = line.rstrip('\n')
            d = Counter()
        else:
            for i in line.rstrip('\n'):
                d[i] += 1

我以您提供的示例为例,并将其另存为temp.fasta,运行上面显示的代码后,输出如下:

>chr1_8969882_-:chr1_568670_-:a2;69 total_counts: 6987 Seed: 197 K: 20   length: 120 Counter({'C': 41, 'A': 31, 'T': 28, 'G': 20})
>chr12_9180206_+:chr12_118582391_+:a1;2 total_counts: 115 Seed: 4 K:    20 length: 79 Counter({'T': 24, 'G': 24, 'C': 17, 'A': 14})
>chr12_9180206_+:chr12_118582391_+:a2;2 total_counts: 135 Seed: 4 K: 20 length: 80 Counter({'C': 30, 'T': 17, 'A': 17, 'G': 16})
>chr1_8969882_-:chr1_568670_-:a1;113 total_counts: 7600 Seed: 225 K: 20 length: 86 Counter({'C': 31, 'A': 27, 'T': 16, 'G': 12})

我将把格式留给您。

答案 2 :(得分:0)

我觉得@Kayvee从 collections 库中对Counter的建议是正确的,但是该建议附带的示例代码在许多方面都存在不足,即无法运行,并且不能很好地使用Counter。这是解决该问题的方法:

from collections import Counter

def display(index, counter):
    print("Sequence {}:".format(index))
    for nucleotide in 'ACGT':
        print("Number of {}'s: {}".format(nucleotide, counter[nucleotide]))
    print()

with open('assembledSeqs.fa') as file:
    index = 0
    counter = Counter()

    for line in file:
        string = line.rstrip('\n')

        if string.startswith('>'):

            if index > 0:
                display(index, counter)

            index += 1
            counter.clear()
        else:
            counter.update(string)

    if counter:  # pop out the last one in the now closed file
        display(index, counter)

此输出应该与您从BioPython获得的输出相匹配:

> python3 test.py
Sequence 1:
Number of A's: 14
Number of C's: 17
Number of G's: 24
Number of T's: 24

Sequence 2:
Number of A's: 17
Number of C's: 30
Number of G's: 16
Number of T's: 17

Sequence 3:
Number of A's: 27
Number of C's: 31
Number of G's: 12
Number of T's: 16

Sequence 4:
Number of A's: 31
Number of C's: 41
Number of G's: 20
Number of T's: 28

>