Question

我是python的新手，我有一个python脚本来运行特定文件（input1.txt）并生成一个输出（output1.fasta），但我想为多个文件运行此脚本，例如：input2.txt，input3.txt ...并生成相应的输出：output2.fasta，output3.fasta

vec[vec==''] <- names(vec)[vec=='']

我尝试添加glob函数，但我不知道如何处理输出文件名。

from Bio import SeqIO

fasta_file = "sequences.txt" 
wanted_file = "input1.txt" 
result_file = "output1.fasta" 

wanted = set()
with open(wanted_file) as f:
    for line in f:
        line = line.strip()
        if line != "":
            wanted.add(line)
fasta_sequences = SeqIO.parse(open(fasta_file),'fasta')
with open(result_file, "w") as f:
    for seq in fasta_sequences:
        if seq.id in wanted:
            SeqIO.write([seq], f, "fasta")

错误消息为：NameError：未定义名称“result_file”

Answer 1

您的glob正在拉动您的＆＃34;序列＆＃34;文件以及输入，因为*.txt包含sequences.txt文件。如果＆＃34; fasta＆＃34;文件总是一样的，你只想迭代输入文件，然后你需要

for filename in glob.glob('input*.txt'):

此外，要遍历整个过程，也许您希望将其放在方法中。如果始终创建输出文件名以对应输入，则可以动态创建。

from Bio import SeqIO

def create_fasta_outputs(fasta_file, wanted_file):
    result_file = wanted_file.replace("input","output").replace(".txt",".fasta")

    wanted = set()
    with open(wanted_file) as f:
        for line in f:
            line = line.strip()
            if line != "":
                wanted.add(line)
    fasta_sequences = SeqIO.parse(open(fasta_file),'fasta')
    with open(result_file, "w") as f:
        for seq in fasta_sequences:
            if seq.id in wanted:
                SeqIO.write([seq], f, "fasta")

fasta_file = "sequences.txt"
for wanted_file in glob.glob('input*.txt'):
    create_fasta_outputs(fasta_file, wanted_file)

修改python脚本以运行多个输入文件

1 个答案: