python多輸入和多輸出

Question

我已經在python中編寫了一個腳本，它在單個文件上工作。我無法找到答案讓它在多個文件上運行，併爲每個文件單獨提供輸出。

out = open('/home/directory/a.out','w') 
infile = open('/home/directory/a.sam','r') 

for line in infile: 
    if not line.startswith('@'): 
     samlist = line.strip().split() 
     if 'I' or 'D' in samlist[5]: 
      match = re.findall(r'(\d+)I', samlist[5]) # remember to chang I and D here aswell 
      intlist = [int(x) for x in match] 
##   if len(intlist) < 10: 
      for indel in intlist: 
       if indel >= 10: 
##     print indel 
      ###intlist contains lengths of insertions in for each read 
      #print intlist 
        read_aln_start = int(samlist[3]) 
        indel_positions = [] 
        for num1, i_or_d, num2, m in re.findall('(\d+)([ID])(\d+)?([A-Za-z])?', samlist[5]): 
         if num1: 
          read_aln_start += int(num1) 
         if num2: 
          read_aln_start += int(num2) 
         indel_positions.append(read_aln_start) 
       #print indel_positions 
        out.write(str(read_aln_start)+'\t'+str(i_or_d) + '\t'+str(samlist[2])+ '\t' + str(indel) +'\n') 
out.close() 

我想我的腳本拍攝多個文件與像a.sam，b.sam，c.sam名和每個文件給我的輸出：aout.sam，bout.sam，cout.sam

請問您可以通過我的解決方案或提示。

問候， Irek

Answer 1

我建議包裝該腳本的功能，採用def關鍵字，並通過輸入和輸出文件名作爲參數傳遞給該功能。

def do_stuff_with_files(infile, outfile): 
    out = open(infile,'w') 
    infile = open(outfile,'r') 
    # the rest of your script 

現在你可以調用這個函數來輸入和輸出文件名的任意組合。

do_stuff_with_files('/home/directory/a.sam', '/home/directory/a.out') 

如果你想在某個目錄做到這一點的所有文件，使用glob庫。要生成輸出文件名，只需將最後三個字符（「sam」）替換爲「out」。

import glob 
indir, outdir = '/home/directory/', '/home/directory/out/' 
files = glob.glob1(indir, '*.sam') 
infiles = [indir + f    for f in files] 
outfiles = [outdir + f[:-3] + "out" for f in files] 
for infile, outfile in zip(infiles, outfiles): 
    do_stuff_with_files(infile, outfile) 

Answer 2

遍歷文件名。

input_filenames = ['a.sam', 'b.sam', 'c.sam'] 
output_filenames = ['aout.sam', 'bout.sam', 'cout.sam'] 
for infn, outfn in zip(input_filenames, output_filenames): 
    out = open('/home/directory/{}'.format(outfn), 'w') 
    infile = open('/home/directory/{}'.format(infn), 'r') 
    ... 

UPDATE

下面的代碼生成給出input_filenames output_filenames。

import os 

def get_output_filename(fn): 
    filename, ext = os.path.splitext(fn) 
    return filename + 'out' + ext 

input_filenames = ['a.sam', 'b.sam', 'c.sam'] # or glob.glob('*.sam') 
output_filenames = map(get_output_filename, input_filenames) 

Answer 3

以下腳本允許使用輸入和輸出文件。它將使用「.sam」擴展名遍歷給定目錄中的所有文件，對它們執行指定的操作，並將結果輸出到單獨的文件。

Import os 
# Define the directory containing the files you are working with 
path = '/home/directory' 
# Get all the files in that directory with the desired 
# extension (in this case ".sam") 
files = [f for f in os.listdir(path) if f.endswith('.sam')] 
# Loop over the files with that extension 
for file in files: 
    # Open the input file 
    with open(path + '/' + file, 'r') as infile: 
     # Open the output file 
     with open(path + '/' + file.split('.')[0] + 'out.' + 
           file.split('.')[1], 'a') as outfile: 
      # Loop over the lines in the input file 
      for line in infile: 
       # If a line in the input file can be characterized in a 
       # certain way, write a different line to the output file. 
       # Otherwise write the original line (from the input file) 
       # to the output file 
       if line.startswith('Something'): 
        outfile.write('A different kind of something') 
       else: 
        outfile.write(line) 
    # Note the absence of either a infile.close() or an outfile.close() 
    # statement. The with-statement handles that for you