修改python腳本運行多個輸入文件

Question

我是非常新的python，我有一個python腳本運行特定文件（input1.txt）並生成一個輸出（output1.fasta），但我想運行該腳本的多個文件，例如：input2.txt，input3.txt ......併產生相應的輸出：output2.fasta，output3.fasta

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") 

我嘗試添加水珠功能，但我不知道如何處理輸出文件名。

from Bio import SeqIO 
import glob 

fasta_file = "sequences.txt" 

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

    wanted = set() 
    with open(filename) 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") 

的錯誤信息是：NameError：名字「result_file」沒有定義

Answer 1

你glob正在拉你的「序列」的文件，以及輸入，因爲*.txt包括sequences.txt文件。如果「FASTA」文件始終是相同的，你只需要遍歷輸入文件，那麼你需要

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

此外，通過你的整個過程進行迭代，也許你希望把它放在一個方法中。如果輸出文件名總是被創建爲對應於輸入，那麼您可以動態創建它。

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)