Question

我的脚本正在运行，但我无法以表格格式打印输出。我有两个文件 - 我读取 file1 序列并查看它们中的任何一个是否与 file2 中的任何序列绑定。

file1.fa

>seq1
ACCGG
>seq2
AATTTC

file2.fa

>seq3
CCGGT
>seq9
GGGGGGCCC

电流输出：

---
file1.seq
>seq1
ACCGG
file2.seq
>seq3
CCGGT

基于文件，只有 seq1 与 seq3 绑定，并且基于我的代码的当前输出是正确的，但随后我必须以显示每个序列 ID 的格式获得我想要的输出- 如果他们交互，则标记为 1，否则标记为 0。

我需要帮助修复 print 部分以得到如下表格格式的输出：

    file2   seq3   seq9
file1 

seq1          1         0

seq2          0         0

我的代码：

from Bio import SeqIO
from Bio.Seq import Seq

records=list(SeqIO.parse("file1.fa","fasta"))

window_size = 5
step_size = 1

target_records=list(SeqIO.parse("file2.fa","fasta"))

for i in records:
    for j in range(0, len(i.seq)-window_size+1):
        for k in (i.seq.reverse_complement()[j: j+5].split()):
            for l in target_records:
                if l.seq.find(k)!=-1:
                    print('---\n{}\n>{}\n{}\n{}\n>{}\n{}'.format ("file1.seq", i.id,i.seq, "file2.seq", l.id,l.seq))

谢谢

Answer 1

当我想获取表格格式时，我想使用pandas：
例如：

from Bio import SeqIO
import pandas as pd


def same_seq(a_record, brecord):
    window_size = 5
    step_size = 1
    for j in range(0, len(a_record.seq) - window_size + 1):
        for k in (a_record.seq.reverse_complement()[j: j + 5].split()):
            return brecord.seq.find(k) != -1


if __name__ == '__main__':
    records = list(SeqIO.parse("file1.fa", "fasta"))
    target_records = list(SeqIO.parse("file2.fa", "fasta"))
    rows_list = []
    for target_record in target_records:
        new_row = {'name': target_record.name}
        for record in records:
            if same_seq(record, target_record):
                new_row[record.name] = 1
            else:
                new_row[record.name] = 0
        rows_list.append(new_row)
    df = pd.DataFrame(rows_list)
    df = df.set_index(["name"])
    print(df)

      seq1  seq2
name            
seq3     1     0
seq9     0     0

如果你想改变seq的比较：
你应该使用 seq[j: j+5].reverse_complement() 但 seq.reverse_complement()[j: j+5] 不知道我理解对不对
file1.fa

>seq1
ACCGG
>seq2
AATTTC
>seqtest1
NNNACCGTGCNN

file1.fa

>seq1
ACCGG
>seq2
AATTTC
>seqtest1
NNNACCGTGCNN

file2.fa

>seq3
CCGGT
>seq9
GGGGGGCCC
>seqtest2
NNCACGGTNN

from Bio import SeqIO
import pandas as pd



def same_seq_window(a_record, b_record, window_size):
    for i in range(len(a_record.seq)- window_size + 1):
        a_seq = a_record.seq[i: i + window_size]
        if b_record.seq.find(a_seq.reverse_complement()) != -1:
            return True
    return False

def same_seq(a_record, b_record):
    window_sizes = range(5, len(a_record.seq)+1)
    for window_size in window_sizes:
        if same_seq_window(a_record, b_record, window_size):
            return True
    return False

if __name__ == '__main__':
    records = list(SeqIO.parse("file1.fa", "fasta"))
    target_records = list(SeqIO.parse("file2.fa", "fasta"))
    rows_list = []
    for target_record in target_records:
        new_row = {'name': target_record.name}
        for record in records:
            if same_seq(record, target_record):
                new_row[record.name] = 1
            else:
                new_row[record.name] = 0
        rows_list.append(new_row)
    df = pd.DataFrame(rows_list)
    df = df.set_index(["name"])
    print(df)

          seq1  seq2  seqtest1
name                          
seq3         1     0         0
seq9         0     0         0
seqtest2     0     0         1

如何使用python以表格格式打印循环输出

1 个答案: